review

in young's double slit experiment two coherent sources

Wave Optics

Young’s double slit experiment, learning objectives.

By the end of this section, you will be able to:

  • Explain the phenomena of interference.
  • Define constructive interference for a double slit and destructive interference for a double slit.

Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see Figure 1).

A beam of light strikes a wall through which a pair of vertical slits is cut. On the other side of the wall, another wall shows a pattern of equally spaced vertical lines of light that are of the same height as the slit.

Figure 1. Young’s double slit experiment. Here pure-wavelength light sent through a pair of vertical slits is diffracted into a pattern on the screen of numerous vertical lines spread out horizontally. Without diffraction and interference, the light would simply make two lines on the screen.

Why do we not ordinarily observe wave behavior for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source (the Sun) through a single slit to make the light somewhat coherent. By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships. Why did Young then pass the light through a double slit? The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single λ ) light to clarify the effect. Figure 2 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

Figure a shows three sine waves with the same wavelength arranged one above the other. The peaks and troughs of each wave are aligned with those of the other waves. The top two waves are labeled wave one and wave two and the bottom wave is labeled resultant. The amplitude of waves one and two are labeled x and the amplitude of the resultant wave is labeled two x. Figure b shows a similar situation, except that the peaks of wave two now align with the troughs of wave one. The resultant wave is now a straight horizontal line on the x axis; that is, the line y equals zero.

Figure 2. The amplitudes of waves add. (a) Pure constructive interference is obtained when identical waves are in phase. (b) Pure destructive interference occurs when identical waves are exactly out of phase, or shifted by half a wavelength.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 3a. Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 3b. Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

The figure contains three parts. The first part is a drawing that shows parallel wavefronts approaching a wall from the left. Crests are shown as continuous lines, and troughs are shown as dotted lines. Two light rays pass through small slits in the wall and emerge in a fan-like pattern from two slits. These lines fan out to the right until they hit the right-hand wall. The points where these fan lines hit the right-hand wall are alternately labeled min and max. The min points correspond to lines that connect the overlapping crests and troughs, and the max points correspond to the lines that connect the overlapping crests. The second drawing is a view from above of a pool of water with semicircular wavefronts emanating from two points on the left side of the pool that are arranged one above the other. These semicircular waves overlap with each other and form a pattern much like the pattern formed by the arcs in the first image. The third drawing shows a vertical dotted line, with some dots appearing brighter than other dots. The brightness pattern is symmetric about the midpoint of this line. The dots near the midpoint are the brightest. As you move from the midpoint up, or down, the dots become progressively dimmer until there seems to be a dot missing. If you progress still farther from the midpoint, the dots appear again and get brighter, but are much less bright than the central dots. If you progress still farther from the midpoint, the dots get dimmer again and then disappear again, which is where the dotted line stops.

Figure 3. Double slits produce two coherent sources of waves that interfere. (a) Light spreads out (diffracts) from each slit, because the slits are narrow. These waves overlap and interfere constructively (bright lines) and destructively (dark regions). We can only see this if the light falls onto a screen and is scattered into our eyes. (b) Double slit interference pattern for water waves are nearly identical to that for light. Wave action is greatest in regions of constructive interference and least in regions of destructive interference. (c) When light that has passed through double slits falls on a screen, we see a pattern such as this. (credit: PASCO)

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in Figure 4. Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in Figure 4a. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively as shown in Figure 4b. More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [(1/2) λ , (3/2) λ , (5/2) λ , etc.], then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths ( λ , 2 λ , 3 λ , etc.), then constructive interference occurs.

Both parts of the figure show a schematic of a double slit experiment. Two waves, each of which is emitted from a different slit, propagate from the slits to the screen. In the first schematic, when the waves meet on the screen, one of the waves is at a maximum whereas the other is at a minimum. This schematic is labeled dark (destructive interference). In the second schematic, when the waves meet on the screen, both waves are at a minimum.. This schematic is labeled bright (constructive interference).

Figure 4. Waves follow different paths from the slits to a common point on a screen. (a) Destructive interference occurs here, because one path is a half wavelength longer than the other. The waves start in phase but arrive out of phase. (b) Constructive interference occurs here because one path is a whole wavelength longer than the other. The waves start out and arrive in phase.

Take-Home Experiment: Using Fingers as Slits

Look at a light, such as a street lamp or incandescent bulb, through the narrow gap between two fingers held close together. What type of pattern do you see? How does it change when you allow the fingers to move a little farther apart? Is it more distinct for a monochromatic source, such as the yellow light from a sodium vapor lamp, than for an incandescent bulb?

The figure is a schematic of a double slit experiment, with the scale of the slits enlarged to show the detail. The two slits are on the left, and the screen is on the right. The slits are represented by a thick vertical line with two gaps cut through it a distance d apart. Two rays, one from each slit, angle up and to the right at an angle theta above the horizontal. At the screen, these rays are shown to converge at a common point. The ray from the upper slit is labeled l sub one, and the ray from the lower slit is labeled l sub two. At the slits, a right triangle is drawn, with the thick line between the slits forming the hypotenuse. The hypotenuse is labeled d, which is the distance between the slits. A short piece of the ray from the lower slit is labeled delta l and forms the short side of the right triangle. The long side of the right triangle is formed by a line segment that goes downward and to the right from the upper slit to the lower ray. This line segment is perpendicular to the lower ray, and the angle it makes with the hypotenuse is labeled theta. Beneath this triangle is the formula delta l equals d sine theta.

Figure 5. The paths from each slit to a common point on the screen differ by an amount dsinθ, assuming the distance to the screen is much greater than the distance between slits (not to scale here).

Figure 5 shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle θ between the path and a line from the slits to the screen (see the figure) is nearly the same for each path. The difference between the paths is shown in the figure; simple trigonometry shows it to be d sin  θ , where d is the distance between the slits. To obtain constructive interference for a double slit , the path length difference must be an integral multiple of the wavelength, or  d sin  θ = mλ, for   m = 0, 1, −1, 2, −2, . . . (constructive).

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

[latex]d\sin\theta=\left(m+\frac{1}{2}\right)\lambda\text{, for }m=0,1,-1,2,-2,\dots\text{ (destructive)}\\[/latex],

where λ is the wavelength of the light, d is the distance between slits, and θ is the angle from the original direction of the beam as discussed above. We call m the order of the interference. For example, m  = 4 is fourth-order interference.

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 6. The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation  d sin  θ = mλ, for   m = 0, 1, −1, 2, −2, . . . .

For fixed λ and m , the smaller d is, the larger θ must be, since [latex]\sin\theta=\frac{m\lambda}{d}\\[/latex]. This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance d apart) is small. Small d gives large θ , hence a large effect.

The figure consists of two parts arranged side-by-side. The diagram on the left side shows a double slit arrangement along with a graph of the resultant intensity pattern on a distant screen. The graph is oriented vertically, so that the intensity peaks grow out and to the left from the screen. The maximum intensity peak is at the center of the screen, and some less intense peaks appear on both sides of the center. These peaks become progressively dimmer upon moving away from the center, and are symmetric with respect to the central peak. The distance from the central maximum to the first dimmer peak is labeled y sub one, and the distance from the central maximum to the second dimmer peak is labeled y sub two. The illustration on the right side shows thick bright horizontal bars on a dark background. Each horizontal bar is aligned with one of the intensity peaks from the first figure.

Figure 6. The interference pattern for a double slit has an intensity that falls off with angle. The photograph shows multiple bright and dark lines, or fringes, formed by light passing through a double slit.

Example 1. Finding a Wavelength from an Interference Pattern

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of 10.95º relative to the incident beam. What is the wavelength of the light?

The third bright line is due to third-order constructive interference, which means that m  = 3. We are given d = 0.0100 mm and θ  = 10.95º. The wavelength can thus be found using the equation  d sin  θ = mλ  for constructive interference.

The equation is d sin  θ = mλ . Solving for the wavelength λ gives [latex]\lambda=\frac{d\sin\theta}{m}\\[/latex].

Substituting known values yields

[latex]\begin{array}{lll}\lambda&=&\frac{\left(0.0100\text{ nm}\right)\left(\sin10.95^{\circ}\right)}{3}\\\text{ }&=&6.33\times10^{-4}\text{ nm}=633\text{ nm}\end{array}\\[/latex]

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with λ , so that spectra (measurements of intensity versus wavelength) can be obtained.

Example 2. Calculating Highest Order Possible

Interference patterns do not have an infinite number of lines, since there is a limit to how big m can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation  d sin  θ = mλ ( for   m = 0, 1, −1, 2, −2, . . . ) describes constructive interference. For fixed values of d and λ , the larger m is, the larger sin  θ  is. However, the maximum value that sin  θ  can have is 1, for an angle of 90º. (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which m corresponds to this maximum diffraction angle.

Solving the equation  d sin  θ = mλ  for m gives [latex]\lambda=\frac{d\sin\theta}{m}\\[/latex].

Taking sin  θ = 1 and substituting the values of d and λ from the preceding example gives

[latex]\displaystyle{m}=\frac{\left(0.0100\text{ mm}\right)\left(1\right)}{633\text{ nm}}\approx15.8\\[/latex]

Therefore, the largest integer m can be is 15, or  m  = 15.

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.

Section Summary

  • Young’s double slit experiment gave definitive proof of the wave character of light.
  • An interference pattern is obtained by the superposition of light from two slits.
  • There is constructive interference when  d sin  θ = mλ ( for   m = 0, 1, −1, 2, −2, . . . ), where d  is the distance between the slits, θ is the angle relative to the incident direction, and m  is the order of the interference.
  • There is destructive interference when  d sin  θ = mλ ( for   m = 0, 1, −1, 2, −2, . . . ).

Conceptual Questions

  • Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.
  • Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.
  • Is it possible to create a situation in which there is only destructive interference? Explain.
  • Figure 7 shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

The figure shows a photo of a horizontal line of equally spaced red dots of light on a black background. The central dot is the brightest and the dots on either side of center are dimmer. The dot intensity decreases to almost zero after moving six dots to the left or right of center. If you continue to move away from the center, the dot brightness increases slightly, although it does not reach the brightness of the central dot. After moving another six dots, or twelve dots in all, to the left or right of center, there is another nearly invisible dot. If you move even farther from the center, the dot intensity again increases, but it does not reach the level of the previous local maximum. At eighteen dots from the center, there is another nearly invisible dot.

Figure 7. This double slit interference pattern also shows signs of single slit interference. (credit: PASCO)

Problems & Exercises

  • At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?
  • Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.
  • What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of 30.0º?
  • Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of 45.0º.
  • Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00 μm.
  • What is the wavelength of light falling on double slits separated by 2.00 μm if the third-order maximum is at an angle of 60.0º?
  • At what angle is the fourth-order maximum for the situation in Question 1?
  • What is the highest-order maximum for 400-nm light falling on double slits separated by 25.0 μm?
  • Find the largest wavelength of light falling on double slits separated by 1.20 μm for which there is a first-order maximum. Is this in the visible part of the spectrum?
  • What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?
  • (a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?
  • (a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of 10.0º, at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?

The figure shows a schematic of a double slit experiment. A double slit is at the left and a screen is at the right. The slits are separated by a distance d. From the midpoint between the slits, a horizontal line labeled x extends to the screen. From the same point, a line angled upward at an angle theta above the horizontal also extends to the screen. The distance between where the horizontal line hits the screen and where the angled line hits the screen is marked y, and the distance between adjacent fringes is given by delta y, which equals x times lambda over d.

Figure 8. The distance between adjacent fringes is [latex]\Delta{y}=\frac{x\lambda}{d}\\[/latex], assuming the slit separation d  is large compared with λ .

  • Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in Figure 8.
  • Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see Figure 8).

coherent:  waves are in phase or have a definite phase relationship

constructive interference for a double slit:  the path length difference must be an integral multiple of the wavelength

destructive interference for a double slit:  the path length difference must be a half-integral multiple of the wavelength

incoherent:  waves have random phase relationships

order:  the integer m  used in the equations for constructive and destructive interference for a double slit

Selected Solutions to Problems & Exercises

3. 1.22 × 10 −6 m

9. 1200 nm (not visible)

11. (a) 760 nm; (b) 1520 nm

13. For small angles sin θ − tan θ  ≈ θ  (in radians).

For two adjacent fringes we have,  d  sin θ m = mλ  and  d sin θ m + 1 = ( m  + 1) λ

Subtracting these equations gives

[latex]\begin{array}{}d\left(\sin{\theta }_{\text{m}+1}-\sin{\theta }_{\text{m}}\right)=\left[\left(m+1\right)-m\right]\lambda \\ d\left({\theta }_{\text{m}+1}-{\theta }_{\text{m}}\right)=\lambda \\ \text{tan}{\theta }_{\text{m}}=\frac{{y}_{\text{m}}}{x}\approx {\theta }_{\text{m}}\Rightarrow d\left(\frac{{y}_{\text{m}+1}}{x}-\frac{{y}_{\text{m}}}{x}\right)=\lambda \\ d\frac{\Delta y}{x}=\lambda \Rightarrow \Delta y=\frac{\mathrm{x\lambda }}{d}\end{array}\\[/latex]

  • College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License

Logo for British Columbia/Yukon Open Authoring Platform

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

Chapter 27 Wave Optics

215 27.3 Young’s Double Slit Experiment

  • Explain the phenomena of interference.
  • Define constructive interference for a double slit and destructive interference for a double slit.

Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see Figure 1 ).

A beam of light strikes a wall through which a pair of vertical slits is cut. On the other side of the wall, another wall shows a pattern of equally spaced vertical lines of light that are of the same height as the slit.

Why do we not ordinarily observe wave behavior for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source (the Sun) through a single slit to make the light somewhat coherent. By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships. Why did Young then pass the light through a double slit? The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single [latex]\boldsymbol{\lambda}[/latex]) light to clarify the effect. Figure 2 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

Figure a shows three sine waves with the same wavelength arranged one above the other. The peaks and troughs of each wave are aligned with those of the other waves. The top two waves are labeled wave one and wave two and the bottom wave is labeled resultant. The amplitude of waves one and two are labeled x and the amplitude of the resultant wave is labeled two x. Figure b shows a similar situation, except that the peaks of wave two now align with the troughs of wave one. The resultant wave is now a straight horizontal line on the x axis; that is, the line y equals zero.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 3 (a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 3 (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

The figure contains three parts. The first part is a drawing that shows parallel wavefronts approaching a wall from the left. Crests are shown as continuous lines, and troughs are shown as dotted lines. Two light rays pass through small slits in the wall and emerge in a fan-like pattern from two slits. These lines fan out to the right until they hit the right-hand wall. The points where these fan lines hit the right-hand wall are alternately labeled min and max. The min points correspond to lines that connect the overlapping crests and troughs, and the max points correspond to the lines that connect the overlapping crests. The second drawing is a view from above of a pool of water with semicircular wavefronts emanating from two points on the left side of the pool that are arranged one above the other. These semicircular waves overlap with each other and form a pattern much like the pattern formed by the arcs in the first image. The third drawing shows a vertical dotted line, with some dots appearing brighter than other dots. The brightness pattern is symmetric about the midpoint of this line. The dots near the midpoint are the brightest. As you move from the midpoint up, or down, the dots become progressively dimmer until there seems to be a dot missing. If you progress still farther from the midpoint, the dots appear again and get brighter, but are much less bright than the central dots. If you progress still farther from the midpoint, the dots get dimmer again and then disappear again, which is where the dotted line stops.

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in Figure 4 . Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in Figure 4 (a). If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively as shown in Figure 4 (b). More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [[latex]\boldsymbol{(1/2) \;\lambda}[/latex], [latex]\boldsymbol{(3/2) \;\lambda}[/latex], [latex]\boldsymbol{(5/2) \;\lambda}[/latex], etc.], then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths ([latex]\boldsymbol{\lambda}[/latex], [latex]\boldsymbol{2 \lambda}[/latex], [latex]\boldsymbol{3 \lambda}[/latex], etc.), then constructive interference occurs.

Take-Home Experiment: Using Fingers as Slits

Look at a light, such as a street lamp or incandescent bulb, through the narrow gap between two fingers held close together. What type of pattern do you see? How does it change when you allow the fingers to move a little farther apart? Is it more distinct for a monochromatic source, such as the yellow light from a sodium vapor lamp, than for an incandescent bulb?

Both parts of the figure show a schematic of a double slit experiment. Two waves, each of which is emitted from a different slit, propagate from the slits to the screen. In the first schematic, when the waves meet on the screen, one of the waves is at a maximum whereas the other is at a minimum. This schematic is labeled dark (destructive interference). In the second schematic, when the waves meet on the screen, both waves are at a minimum.. This schematic is labeled bright (constructive interference).

Figure 5 shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle [latex]\boldsymbol{\theta}[/latex] between the path and a line from the slits to the screen (see the figure) is nearly the same for each path. The difference between the paths is shown in the figure; simple trigonometry shows it to be [latex]\boldsymbol{d \;\textbf{sin} \;\theta}[/latex], where [latex]\boldsymbol{d}[/latex] is the distance between the slits. To obtain constructive interference for a double slit , the path length difference must be an integral multiple of the wavelength, or

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

where [latex]\boldsymbol{\lambda}[/latex] is the wavelength of the light, [latex]\boldsymbol{d}[/latex] is the distance between slits, and [latex]\boldsymbol{\theta}[/latex] is the angle from the original direction of the beam as discussed above. We call [latex]\boldsymbol{m}[/latex] the order of the interference. For example, [latex]\boldsymbol{m = 4}[/latex] is fourth-order interference.

The figure is a schematic of a double slit experiment, with the scale of the slits enlarged to show the detail. The two slits are on the left, and the screen is on the right. The slits are represented by a thick vertical line with two gaps cut through it a distance d apart. Two rays, one from each slit, angle up and to the right at an angle theta above the horizontal. At the screen, these rays are shown to converge at a common point. The ray from the upper slit is labeled l sub one, and the ray from the lower slit is labeled l sub two. At the slits, a right triangle is drawn, with the thick line between the slits forming the hypotenuse. The hypotenuse is labeled d, which is the distance between the slits. A short piece of the ray from the lower slit is labeled delta l and forms the short side of the right triangle. The long side of the right triangle is formed by a line segment that goes downward and to the right from the upper slit to the lower ray. This line segment is perpendicular to the lower ray, and the angle it makes with the hypotenuse is labeled theta. Beneath this triangle is the formula delta l equals d sine theta.

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 6 . The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation

For fixed [latex]\boldsymbol{\lambda}[/latex] and [latex]\boldsymbol{m}[/latex], the smaller [latex]\boldsymbol{d}[/latex] is, the larger [latex]\boldsymbol{\theta}[/latex] must be, since [latex]\boldsymbol{\textbf{sin} \;\theta = m \lambda / d}[/latex].

This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance [latex]\boldsymbol{d}[/latex] apart) is small. Small [latex]\boldsymbol{d}[/latex] gives large [latex]\boldsymbol{\theta}[/latex], hence a large effect.

The figure consists of two parts arranged side-by-side. The diagram on the left side shows a double slit arrangement along with a graph of the resultant intensity pattern on a distant screen. The graph is oriented vertically, so that the intensity peaks grow out and to the left from the screen. The maximum intensity peak is at the center of the screen, and some less intense peaks appear on both sides of the center. These peaks become progressively dimmer upon moving away from the center, and are symmetric with respect to the central peak. The distance from the central maximum to the first dimmer peak is labeled y sub one, and the distance from the central maximum to the second dimmer peak is labeled y sub two. The illustration on the right side shows thick bright horizontal bars on a dark background. Each horizontal bar is aligned with one of the intensity peaks from the first figure.

Example 1: Finding a Wavelength from an Interference Pattern

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of [latex]\boldsymbol{10.95 ^{\circ}}[/latex] relative to the incident beam. What is the wavelength of the light?

The third bright line is due to third-order constructive interference, which means that [latex]\boldsymbol{m = 3}[/latex]. We are given [latex]\boldsymbol{d = 0.0100 \;\textbf{mm}}[/latex] and [latex]\boldsymbol{\theta = 10.95^{\circ}}[/latex]. The wavelength can thus be found using the equation [latex]\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda}[/latex] for constructive interference.

The equation is [latex]\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda}[/latex]. Solving for the wavelength [latex]\boldsymbol{\lambda}[/latex] gives

Substituting known values yields

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with [latex]\boldsymbol{\lambda}[/latex], so that spectra (measurements of intensity versus wavelength) can be obtained.

Example 2: Calculating Highest Order Possible

Interference patterns do not have an infinite number of lines, since there is a limit to how big [latex]\boldsymbol{m}[/latex] can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation [latex]\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda \; (\textbf{for} \; m = 0, \; 1, \; -1, \; 2, \; -2, \; \dots)}[/latex] describes constructive interference. For fixed values of [latex]\boldsymbol{d}[/latex] and [latex]\boldsymbol{\lambda}[/latex], the larger [latex]\boldsymbol{m}[/latex] is, the larger [latex]\boldsymbol{\textbf{sin} \;\theta}[/latex] is. However, the maximum value that [latex]\boldsymbol{\textbf{sin} \;\theta}[/latex] can have is 1, for an angle of [latex]\boldsymbol{90 ^{\circ}}[/latex]. (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which [latex]\boldsymbol{m}[/latex] corresponds to this maximum diffraction angle.

Solving the equation [latex]\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda}[/latex] for [latex]\boldsymbol{m}[/latex] gives

Taking [latex]\boldsymbol{\textbf{sin} \;\theta = 1}[/latex] and substituting the values of [latex]\boldsymbol{d}[/latex] and [latex]\boldsymbol{\lambda}[/latex] from the preceding example gives

Therefore, the largest integer [latex]\boldsymbol{m}[/latex] can be is 15, or

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.

Section Summary

  • Young’s double slit experiment gave definitive proof of the wave character of light.
  • An interference pattern is obtained by the superposition of light from two slits.
  • There is constructive interference when [latex]\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda \;(\textbf{for} \; m = 0, \; 1, \; -1, \;2, \; -2, \dots)}[/latex], where [latex]\boldsymbol{d}[/latex] is the distance between the slits, [latex]\boldsymbol{\theta}[/latex] is the angle relative to the incident direction, and [latex]\boldsymbol{m}[/latex] is the order of the interference.
  • There is destructive interference when [latex]\boldsymbol{d \;\textbf{sin} \;\theta = (m+ \frac{1}{2}) \lambda}[/latex] (for [latex]\boldsymbol{m = 0, \; 1, \; -1, \; 2, \; -2, \; \dots}[/latex]).

Conceptual Questions

1: Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.

2: Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.

3: Is it possible to create a situation in which there is only destructive interference? Explain.

4:   Figure 7 shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

The figure shows a photo of a horizontal line of equally spaced red dots of light on a black background. The central dot is the brightest and the dots on either side of center are dimmer. The dot intensity decreases to almost zero after moving six dots to the left or right of center. If you continue to move away from the center, the dot brightness increases slightly, although it does not reach the brightness of the central dot. After moving another six dots, or twelve dots in all, to the left or right of center, there is another nearly invisible dot. If you move even farther from the center, the dot intensity again increases, but it does not reach the level of the previous local maximum. At eighteen dots from the center, there is another nearly invisible dot.

Problems & Exercises

2: Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

3: What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of [latex]\boldsymbol{30.0 ^{\circ}}[/latex]?

4: Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of [latex]\boldsymbol{45.0 ^{\circ}}[/latex].

5: Calculate the wavelength of light that has its third minimum at an angle of [latex]\boldsymbol{30.0 ^{\circ}}[/latex] when falling on double slits separated by [latex]\boldsymbol{3.00 \;\mu \textbf{m}}[/latex]. Explicitly, show how you follow the steps in Chapter 27.7 Problem-Solving Strategies for Wave Optics .

6: What is the wavelength of light falling on double slits separated by [latex]\boldsymbol{2.00 \;\mu \textbf{m}}[/latex] if the third-order maximum is at an angle of [latex]\boldsymbol{60.0 ^{\circ}}[/latex]?

7: At what angle is the fourth-order maximum for the situation in Problems & Exercises 1 ?

8: What is the highest-order maximum for 400-nm light falling on double slits separated by [latex]\boldsymbol{25.0 \;\mu \textbf{m}}[/latex]?

9: Find the largest wavelength of light falling on double slits separated by [latex]\boldsymbol{1.20 \;\mu \textbf{m}}[/latex] for which there is a first-order maximum. Is this in the visible part of the spectrum?

10: What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?

11: (a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?

12: (a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of [latex]\boldsymbol{10.0^{\circ}}[/latex], at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?

13:   Figure 8 shows a double slit located a distance [latex]\boldsymbol{x}[/latex] from a screen, with the distance from the center of the screen given by [latex]\boldsymbol{y}[/latex]. When the distance [latex]\boldsymbol{d}[/latex] between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, for small angles (where [latex]\boldsymbol{\textbf{sin} \;\theta \approx \theta}[/latex], with [latex]\boldsymbol{\theta}[/latex] in radians), the distance between fringes is given by [latex]\boldsymbol{\Delta y = x \lambda /d}[/latex].

The figure shows a schematic of a double slit experiment. A double slit is at the left and a screen is at the right. The slits are separated by a distance d. From the midpoint between the slits, a horizontal line labeled x extends to the screen. From the same point, a line angled upward at an angle theta above the horizontal also extends to the screen. The distance between where the horizontal line hits the screen and where the angled line hits the screen is marked y, and the distance between adjacent fringes is given by delta y, which equals x times lambda over d.

14: Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in Figure 8 .

15: Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see Figure 8 ).

1:  [latex]\boldsymbol{0.516 ^{\circ}}[/latex]

3:  [latex]\boldsymbol{1.22 \times 10^{-6} \;\textbf{m}}[/latex]

7: [latex]\boldsymbol{2.06 ^{\circ}}[/latex]

9: 1200 nm (not visible)

11: (a) 760 nm

(b) 1520 nm

13: For small angles [latex]\boldsymbol{\textbf{sin} \;\theta - \;\textbf{tan} \;\theta \approx \theta}[/latex] (in radians).

For two adjacent fringes we have,

Subtracting these equations gives

College Physics: OpenStax Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

Share This Book

Logo for Open Textbook Publishing

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

154 Young’s Double Slit Experiment

[latexpage]

Learning Objectives

  • Explain the phenomena of interference.
  • Define constructive interference for a double slit and destructive interference for a double slit.

Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see (Figure) ).

A beam of light strikes a wall through which a pair of vertical slits is cut. On the other side of the wall, another wall shows a pattern of equally spaced vertical lines of light that are of the same height as the slit.

Why do we not ordinarily observe wave behavior for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source (the Sun) through a single slit to make the light somewhat coherent. By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships. Why did Young then pass the light through a double slit? The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single \(\lambda \)) light to clarify the effect. (Figure) shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

Figure a shows three sine waves with the same wavelength arranged one above the other. The peaks and troughs of each wave are aligned with those of the other waves. The top two waves are labeled wave one and wave two and the bottom wave is labeled resultant. The amplitude of waves one and two are labeled x and the amplitude of the resultant wave is labeled two x. Figure b shows a similar situation, except that the peaks of wave two now align with the troughs of wave one. The resultant wave is now a straight horizontal line on the x axis; that is, the line y equals zero.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in (Figure) (a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in (Figure) (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

The figure contains three parts. The first part is a drawing that shows parallel wavefronts approaching a wall from the left. Crests are shown as continuous lines, and troughs are shown as dotted lines. Two light rays pass through small slits in the wall and emerge in a fan-like pattern from two slits. These lines fan out to the right until they hit the right-hand wall. The points where these fan lines hit the right-hand wall are alternately labeled min and max. The min points correspond to lines that connect the overlapping crests and troughs, and the max points correspond to the lines that connect the overlapping crests. The second drawing is a view from above of a pool of water with semicircular wavefronts emanating from two points on the left side of the pool that are arranged one above the other. These semicircular waves overlap with each other and form a pattern much like the pattern formed by the arcs in the first image. The third drawing shows a vertical dotted line, with some dots appearing brighter than other dots. The brightness pattern is symmetric about the midpoint of this line. The dots near the midpoint are the brightest. As you move from the midpoint up, or down, the dots become progressively dimmer until there seems to be a dot missing. If you progress still farther from the midpoint, the dots appear again and get brighter, but are much less bright than the central dots. If you progress still farther from the midpoint, the dots get dimmer again and then disappear again, which is where the dotted line stops.

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in (Figure) . Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in (Figure) (a). If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively as shown in (Figure) (b). More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [\(\left(1/2\right)\lambda \), \(\left(3/2\right)\lambda \), \(\left(5/2\right)\lambda \), etc.], then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths (\(\lambda \), \(2\lambda \), \(3\lambda \), etc.), then constructive interference occurs.

Look at a light, such as a street lamp or incandescent bulb, through the narrow gap between two fingers held close together. What type of pattern do you see? How does it change when you allow the fingers to move a little farther apart? Is it more distinct for a monochromatic source, such as the yellow light from a sodium vapor lamp, than for an incandescent bulb?

Both parts of the figure show a schematic of a double slit experiment. Two waves, each of which is emitted from a different slit, propagate from the slits to the screen. In the first schematic, when the waves meet on the screen, one of the waves is at a maximum whereas the other is at a minimum. This schematic is labeled dark (destructive interference). In the second schematic, when the waves meet on the screen, both waves are at a minimum.. This schematic is labeled bright (constructive interference).

(Figure) shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle \(\theta \) between the path and a line from the slits to the screen (see the figure) is nearly the same for each path. The difference between the paths is shown in the figure; simple trigonometry shows it to be \(d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \), where \(d\) is the distance between the slits. To obtain constructive interference for a double slit , the path length difference must be an integral multiple of the wavelength, or

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

where \(\lambda \) is the wavelength of the light, \(d\) is the distance between slits, and \(\theta \) is the angle from the original direction of the beam as discussed above. We call \(m\) the order of the interference. For example, \(m=4\) is fourth-order interference.

The figure is a schematic of a double slit experiment, with the scale of the slits enlarged to show the detail. The two slits are on the left, and the screen is on the right. The slits are represented by a thick vertical line with two gaps cut through it a distance d apart. Two rays, one from each slit, angle up and to the right at an angle theta above the horizontal. At the screen, these rays are shown to converge at a common point. The ray from the upper slit is labeled l sub one, and the ray from the lower slit is labeled l sub two. At the slits, a right triangle is drawn, with the thick line between the slits forming the hypotenuse. The hypotenuse is labeled d, which is the distance between the slits. A short piece of the ray from the lower slit is labeled delta l and forms the short side of the right triangle. The long side of the right triangle is formed by a line segment that goes downward and to the right from the upper slit to the lower ray. This line segment is perpendicular to the lower ray, and the angle it makes with the hypotenuse is labeled theta. Beneath this triangle is the formula delta l equals d sine theta.

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in (Figure) . The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation

For fixed \(\lambda \) and \(m\), the smaller \(d\) is, the larger \(\theta \) must be, since \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }/d\). This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance \(d\) apart) is small. Small \(d\) gives large \(\theta \), hence a large effect.

The figure consists of two parts arranged side-by-side. The diagram on the left side shows a double slit arrangement along with a graph of the resultant intensity pattern on a distant screen. The graph is oriented vertically, so that the intensity peaks grow out and to the left from the screen. The maximum intensity peak is at the center of the screen, and some less intense peaks appear on both sides of the center. These peaks become progressively dimmer upon moving away from the center, and are symmetric with respect to the central peak. The distance from the central maximum to the first dimmer peak is labeled y sub one, and the distance from the central maximum to the second dimmer peak is labeled y sub two. The illustration on the right side shows thick bright horizontal bars on a dark background. Each horizontal bar is aligned with one of the intensity peaks from the first figure.

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of \(\text{10}\text{.}\text{95º}\) relative to the incident beam. What is the wavelength of the light?

The third bright line is due to third-order constructive interference, which means that \(m=3\). We are given \(d=0\text{.}\text{0100}\phantom{\rule{0.25em}{0ex}}\text{mm}\) and \(\theta =\text{10}\text{.}\text{95º}\). The wavelength can thus be found using the equation \(d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\) for constructive interference.

The equation is \(d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\). Solving for the wavelength \(\lambda \) gives

Substituting known values yields

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with \(\lambda \), so that spectra (measurements of intensity versus wavelength) can be obtained.

Interference patterns do not have an infinite number of lines, since there is a limit to how big \(m\) can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation \(d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\phantom{\rule{0.25em}{0ex}}\text{(for}\phantom{\rule{0.25em}{0ex}}m=0,\phantom{\rule{0.25em}{0ex}}1,\phantom{\rule{0.25em}{0ex}}-1,\phantom{\rule{0.25em}{0ex}}2,\phantom{\rule{0.25em}{0ex}}-2,\phantom{\rule{0.25em}{0ex}}\dots \right)\) describes constructive interference. For fixed values of \(d\) and \(\lambda \), the larger \(m\) is, the larger \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \) is. However, the maximum value that \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \) can have is 1, for an angle of \(\text{90º}\). (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which \(m\) corresponds to this maximum diffraction angle.

Solving the equation \(d\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\) for \(m\) gives

Taking \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1\) and substituting the values of \(d\) and \(\lambda \) from the preceding example gives

Therefore, the largest integer \(m\) can be is 15, or

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.

Section Summary

  • Young’s double slit experiment gave definitive proof of the wave character of light.
  • An interference pattern is obtained by the superposition of light from two slits.
  • There is constructive interference when \(d\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\phantom{\rule{0.25em}{0ex}}\text{(for}\phantom{\rule{0.25em}{0ex}}m=0,\phantom{\rule{0.25em}{0ex}}1,\phantom{\rule{0.25em}{0ex}}-1,\phantom{\rule{0.25em}{0ex}}2,\phantom{\rule{0.25em}{0ex}}-2,\phantom{\rule{0.25em}{0ex}}\dots \right)\), where \(d\) is the distance between the slits, \(\theta \) is the angle relative to the incident direction, and \(m\) is the order of the interference.
  • There is destructive interference when \(d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\left(m+\frac{1}{2}\right)\lambda \phantom{\rule{0.25em}{0ex}}\text{(for}\phantom{\rule{0.5em}{0ex}}m=0,\phantom{\rule{0.25em}{0ex}}1,\phantom{\rule{0.25em}{0ex}}-1,\phantom{\rule{0.25em}{0ex}}2,\phantom{\rule{0.25em}{0ex}}-2,\phantom{\rule{0.25em}{0ex}}\dots \right)\).

Conceptual Questions

Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.

Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.

Is it possible to create a situation in which there is only destructive interference? Explain.

(Figure) shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

The figure shows a photo of a horizontal line of equally spaced red dots of light on a black background. The central dot is the brightest and the dots on either side of center are dimmer. The dot intensity decreases to almost zero after moving six dots to the left or right of center. If you continue to move away from the center, the dot brightness increases slightly, although it does not reach the brightness of the central dot. After moving another six dots, or twelve dots in all, to the left or right of center, there is another nearly invisible dot. If you move even farther from the center, the dot intensity again increases, but it does not reach the level of the previous local maximum. At eighteen dots from the center, there is another nearly invisible dot.

Problems & Exercises

At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?

\(0\text{.}\text{516º}\)

Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of \(\text{30}\text{.}0º\)?

\(1\text{.}\text{22}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{m}\)

Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of \(\text{45}\text{.}0º\).

Calculate the wavelength of light that has its third minimum at an angle of \(\text{30}\text{.}0º\) when falling on double slits separated by \(3\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μm}\). Explicitly, show how you follow the steps in Problem-Solving Strategies for Wave Optics .

What is the wavelength of light falling on double slits separated by \(2\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μm}\) if the third-order maximum is at an angle of \(\text{60}\text{.}0º\)?

At what angle is the fourth-order maximum for the situation in (Figure) ?

\(2\text{.}\text{06º}\)

What is the highest-order maximum for 400-nm light falling on double slits separated by \(\text{25}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{μm}\)?

Find the largest wavelength of light falling on double slits separated by \(1\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\text{μm}\) for which there is a first-order maximum. Is this in the visible part of the spectrum?

1200 nm (not visible)

What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?

(a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?

(b) 1520 nm

(a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of \(\text{10}\text{.}0º\), at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?

(Figure) shows a double slit located a distance \(x\) from a screen, with the distance from the center of the screen given by \(y\). When the distance \(d\) between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, for small angles (where \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \approx \theta \), with \(\theta \) in radians), the distance between fringes is given by \(\text{Δ}y=\mathrm{x\lambda }/d\).

The figure shows a schematic of a double slit experiment. A double slit is at the left and a screen is at the right. The slits are separated by a distance d. From the midpoint between the slits, a horizontal line labeled x extends to the screen. From the same point, a line angled upward at an angle theta above the horizontal also extends to the screen. The distance between where the horizontal line hits the screen and where the angled line hits the screen is marked y, and the distance between adjacent fringes is given by delta y, which equals x times lambda over d.

For small angles \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta -\text{tan}\phantom{\rule{0.25em}{0ex}}\theta \approx \theta \phantom{\rule{0.25em}{0ex}}\left(\text{in radians}\right)\).

For two adjacent fringes we have,

Subtracting these equations gives

Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in (Figure) .

Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see (Figure) ).

ISP209: The Mystery of the Physical World Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

Share This Book

Logo for University of Central Florida Pressbooks

Chapter 27 Wave Optics

27.3 Young’s Double Slit Experiment

  • Explain the phenomena of interference.
  • Define constructive interference for a double slit and destructive interference for a double slit.

Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see Figure 1 ).

A beam of light strikes a wall through which a pair of vertical slits is cut. On the other side of the wall, another wall shows a pattern of equally spaced vertical lines of light that are of the same height as the slit.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 3 (a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 3 (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

The figure contains three parts. The first part is a drawing that shows parallel wavefronts approaching a wall from the left. Crests are shown as continuous lines, and troughs are shown as dotted lines. Two light rays pass through small slits in the wall and emerge in a fan-like pattern from two slits. These lines fan out to the right until they hit the right-hand wall. The points where these fan lines hit the right-hand wall are alternately labeled min and max. The min points correspond to lines that connect the overlapping crests and troughs, and the max points correspond to the lines that connect the overlapping crests. The second drawing is a view from above of a pool of water with semicircular wavefronts emanating from two points on the left side of the pool that are arranged one above the other. These semicircular waves overlap with each other and form a pattern much like the pattern formed by the arcs in the first image. The third drawing shows a vertical dotted line, with some dots appearing brighter than other dots. The brightness pattern is symmetric about the midpoint of this line. The dots near the midpoint are the brightest. As you move from the midpoint up, or down, the dots become progressively dimmer until there seems to be a dot missing. If you progress still farther from the midpoint, the dots appear again and get brighter, but are much less bright than the central dots. If you progress still farther from the midpoint, the dots get dimmer again and then disappear again, which is where the dotted line stops.

Take-Home Experiment: Using Fingers as Slits

Look at a light, such as a street lamp or incandescent bulb, through the narrow gap between two fingers held close together. What type of pattern do you see? How does it change when you allow the fingers to move a little farther apart? Is it more distinct for a monochromatic source, such as the yellow light from a sodium vapor lamp, than for an incandescent bulb?

Both parts of the figure show a schematic of a double slit experiment. Two waves, each of which is emitted from a different slit, propagate from the slits to the screen. In the first schematic, when the waves meet on the screen, one of the waves is at a maximum whereas the other is at a minimum. This schematic is labeled dark (destructive interference). In the second schematic, when the waves meet on the screen, both waves are at a minimum.. This schematic is labeled bright (constructive interference).

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

\boldsymbol{d \;\textbf{sin} \;\theta = m +}

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 6 . The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation

\boldsymbol{d \;\textbf{sin} \;\theta = m \theta , \;\textbf{for} \; m = 0, \;1, \; -1, \; 2, \; -2, \; \dots}.

Example 1: Finding a Wavelength from an Interference Pattern

\boldsymbol{10.95 ^{\circ}}

Substituting known values yields

$\begin{array}{r @{{}={}}l} \boldsymbol{\lambda} & \boldsymbol{\frac{(0.0100 \;\textbf{mm})(\textbf{sin} 10.95^{\circ})}{3}} \\[1em] & \boldsymbol{6.33 \times 10^{-4} \;\textbf{mm} = 633 \;\textbf{nm}}. \end{array}

Example 2: Calculating Highest Order Possible

Strategy and Concept

\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda \; (\textbf{for} \; m = 0, \; 1, \; -1, \; 2, \; -2, \; \dots)}

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.

Section Summary

  • Young’s double slit experiment gave definitive proof of the wave character of light.
  • An interference pattern is obtained by the superposition of light from two slits.

\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda \;(\textbf{for} \; m = 0, \; 1, \; -1, \;2, \; -2, \dots)}

Conceptual Questions

1: Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.

2: Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.

3: Is it possible to create a situation in which there is only destructive interference? Explain.

4:   Figure 7 shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

The figure shows a photo of a horizontal line of equally spaced red dots of light on a black background. The central dot is the brightest and the dots on either side of center are dimmer. The dot intensity decreases to almost zero after moving six dots to the left or right of center. If you continue to move away from the center, the dot brightness increases slightly, although it does not reach the brightness of the central dot. After moving another six dots, or twelve dots in all, to the left or right of center, there is another nearly invisible dot. If you move even farther from the center, the dot intensity again increases, but it does not reach the level of the previous local maximum. At eighteen dots from the center, there is another nearly invisible dot.

Problems & Exercises

2: Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

\boldsymbol{30.0 ^{\circ}}

7: At what angle is the fourth-order maximum for the situation in Problems & Exercises 1 ?

\boldsymbol{25.0 \;\mu \textbf{m}}

10: What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?

11: (a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?

\boldsymbol{10.0^{\circ}}

14: Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in Figure 8 .

15: Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see Figure 8 ).

\boldsymbol{0.516 ^{\circ}}

9: 1200 nm (not visible)

11: (a) 760 nm

(b) 1520 nm

\boldsymbol{\textbf{sin} \;\theta - \;\textbf{tan} \;\theta \approx \theta}

For two adjacent fringes we have,

\boldsymbol{d \;\textbf{sin} \;\theta _{\textbf{m}} = m \lambda}

Subtracting these equations gives

$\begin{array}{r @{{}={}}l} \boldsymbol{d (\textbf{sin} \; \theta _{\textbf{m} + 1} - \textbf{sin} \; \theta _{\textbf{m}})} & \boldsymbol{[(m + 1) - m] \lambda} \\[1em] \boldsymbol{d(\theta _{{\textbf{m}} + 1} - \theta _{\textbf{m}})} & \boldsymbol{\lambda} \end{array}$

College Physics Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

Share This Book

Encyclopedia Britannica

  • History & Society
  • Science & Tech
  • Biographies
  • Animals & Nature
  • Geography & Travel
  • Arts & Culture
  • Games & Quizzes
  • On This Day
  • One Good Fact
  • New Articles
  • Lifestyles & Social Issues
  • Philosophy & Religion
  • Politics, Law & Government
  • World History
  • Health & Medicine
  • Browse Biographies
  • Birds, Reptiles & Other Vertebrates
  • Bugs, Mollusks & Other Invertebrates
  • Environment
  • Fossils & Geologic Time
  • Entertainment & Pop Culture
  • Sports & Recreation
  • Visual Arts
  • Demystified
  • Image Galleries
  • Infographics
  • Top Questions
  • Britannica Kids
  • Saving Earth
  • Space Next 50
  • Student Center
  • Introduction & Top Questions
  • Ray theories in the ancient world
  • Early particle and wave theories
  • Reflection and refraction
  • Total internal reflection
  • Characteristics of waves

Young’s double-slit experiment

  • Thin-film interference
  • Diffraction
  • Poisson’s spot
  • Circular apertures and image resolution
  • Atmospheric diffraction effects
  • Doppler effect
  • Electric and magnetic fields
  • Maxwell’s equations
  • Electromagnetic waves
  • The electromagnetic spectrum
  • Sources of electromagnetic waves
  • Early measurements
  • The Michelson-Morley experiment
  • Fundamental constant of nature
  • Transverse waves
  • Unpolarized light
  • Sources of polarized light
  • Radiation pressure
  • Interactions of light with matter
  • Nonlinear interactions
  • Blackbody radiation
  • Wave-particle duality
  • Quantum optics
  • Spontaneous emission
  • Stimulated emission
  • Quantum electrodynamics

visible spectrum of light

  • Why is light important for life on Earth?

Light refraction image with a cube. Prism

Our editors will review what you’ve submitted and determine whether to revise the article.

  • Berkleley Lab Environment, Health & Safety - Light and Infrared Radiation
  • Chemistry LibreTexts - The Nature of Light
  • NeoK12 - Educational Videos and Games for School Kids - Light and Optics
  • University of Colorado Boulder - Department of Physics - Light and Optics
  • Khan Academy - Introduction to light
  • light - Children's Encyclopedia (Ages 8-11)
  • light - Student Encyclopedia (Ages 11 and up)
  • Table Of Contents

The discovery of light's wave-particle duality

The observation of interference effects definitively indicates the presence of overlapping waves. Thomas Young postulated that light is a wave and is subject to the superposition principle; his great experimental achievement was to demonstrate the constructive and destructive interference of light (c. 1801). In a modern version of Young’s experiment, differing in its essentials only in the source of light, a laser equally illuminates two parallel slits in an otherwise opaque surface. The light passing through the two slits is observed on a distant screen. When the widths of the slits are significantly greater than the wavelength of the light, the rules of geometrical optics hold—the light casts two shadows, and there are two illuminated regions on the screen. However, as the slits are narrowed in width, the light diffracts into the geometrical shadow, and the light waves overlap on the screen. (Diffraction is itself caused by the wave nature of light, being another example of an interference effect—it is discussed in more detail below.)

in young's double slit experiment two coherent sources

The superposition principle determines the resulting intensity pattern on the illuminated screen. Constructive interference occurs whenever the difference in paths from the two slits to a point on the screen equals an integral number of wavelengths (0, λ, 2λ,…). This path difference guarantees that crests from the two waves arrive simultaneously. Destructive interference arises from path differences that equal a half-integral number of wavelengths (λ/2, 3λ/2,…). Young used geometrical arguments to show that the superposition of the two waves results in a series of equally spaced bands, or fringes, of high intensity, corresponding to regions of constructive interference, separated by dark regions of complete destructive interference.

An important parameter in the double-slit geometry is the ratio of the wavelength of the light λ to the spacing of the slits d . If λ/ d is much smaller than 1, the spacing between consecutive interference fringes will be small, and the interference effects may not be observable. Using narrowly separated slits, Young was able to separate the interference fringes. In this way he determined the wavelengths of the colours of visible light. The very short wavelengths of visible light explain why interference effects are observed only in special circumstances—the spacing between the sources of the interfering light waves must be very small to separate regions of constructive and destructive interference.

Observing interference effects is challenging because of two other difficulties. Most light sources emit a continuous range of wavelengths, which result in many overlapping interference patterns, each with a different fringe spacing. The multiple interference patterns wash out the most pronounced interference effects, such as the regions of complete darkness. Second, for an interference pattern to be observable over any extended period of time, the two sources of light must be coherent with respect to each other. This means that the light sources must maintain a constant phase relationship. For example, two harmonic waves of the same frequency always have a fixed phase relationship at every point in space, being either in phase, out of phase, or in some intermediate relationship. However, most light sources do not emit true harmonic waves; instead, they emit waves that undergo random phase changes millions of times per second. Such light is called incoherent . Interference still occurs when light waves from two incoherent sources overlap in space, but the interference pattern fluctuates randomly as the phases of the waves shift randomly. Detectors of light, including the eye, cannot register the quickly shifting interference patterns, and only a time-averaged intensity is observed. Laser light is approximately monochromatic (consisting of a single wavelength) and is highly coherent; it is thus an ideal source for revealing interference effects.

After 1802, Young’s measurements of the wavelengths of visible light could be combined with the relatively crude determinations of the speed of light available at the time in order to calculate the approximate frequencies of light. For example, the frequency of green light is about 6 × 10 14 Hz ( hertz , or cycles per second). This frequency is many orders of magnitude larger than the frequencies of common mechanical waves. For comparison, humans can hear sound waves with frequencies up to about 2 × 10 4 Hz. Exactly what was oscillating at such a high rate remained a mystery for another 60 years.

3.1 Young's Double-Slit Interference

Learning objectives.

By the end of this section, you will be able to:

  • Explain the phenomenon of interference
  • Define constructive and destructive interference for a double slit

The Dutch physicist Christiaan Huygens (1629–1695) thought that light was a wave, but Isaac Newton did not. Newton thought that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous reputation, his view generally prevailed; the fact that Huygens’s principle worked was not considered direct evidence proving that light is a wave. The acceptance of the wave character of light came many years later in 1801, when the English physicist and physician Thomas Young (1773–1829) demonstrated optical interference with his now-classic double-slit experiment.

If there were not one but two sources of waves, the waves could be made to interfere, as in the case of waves on water ( Figure 3.2 ). If light is an electromagnetic wave, it must therefore exhibit interference effects under appropriate circumstances. In Young’s experiment, sunlight was passed through a pinhole on a board. The emerging beam fell on two pinholes on a second board. The light emanating from the two pinholes then fell on a screen where a pattern of bright and dark spots was observed. This pattern, called fringes, can only be explained through interference, a wave phenomenon.

We can analyze double-slit interference with the help of Figure 3.3 , which depicts an apparatus analogous to Young’s. Light from a monochromatic source falls on a slit S 0 S 0 . The light emanating from S 0 S 0 is incident on two other slits S 1 S 1 and S 2 S 2 that are equidistant from S 0 S 0 . A pattern of interference fringes on the screen is then produced by the light emanating from S 1 S 1 and S 2 S 2 . All slits are assumed to be so narrow that they can be considered secondary point sources for Huygens’ wavelets ( The Nature of Light ). Slits S 1 S 1 and S 2 S 2 are a distance d apart ( d ≤ 1 mm d ≤ 1 mm ), and the distance between the screen and the slits is D ( ≈ 1 m ) D ( ≈ 1 m ) , which is much greater than d.

Since S 0 S 0 is assumed to be a point source of monochromatic light, the secondary Huygens wavelets leaving S 1 S 1 and S 2 S 2 always maintain a constant phase difference (zero in this case because S 1 S 1 and S 2 S 2 are equidistant from S 0 S 0 ) and have the same frequency. The sources S 1 S 1 and S 2 S 2 are then said to be coherent. By coherent waves , we mean the waves are in phase or have a definite phase relationship. The term incoherent means the waves have random phase relationships, which would be the case if S 1 S 1 and S 2 S 2 were illuminated by two independent light sources, rather than a single source S 0 S 0 . Two independent light sources (which may be two separate areas within the same lamp or the Sun) would generally not emit their light in unison, that is, not coherently. Also, because S 1 S 1 and S 2 S 2 are the same distance from S 0 S 0 , the amplitudes of the two Huygens wavelets are equal.

Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. In the following discussion, we illustrate the double-slit experiment with monochromatic light (single λ λ ) to clarify the effect. Figure 3.4 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

When light passes through narrow slits, the slits act as sources of coherent waves and light spreads out as semicircular waves, as shown in Figure 3.5 (a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 3.2 . Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

To understand the double-slit interference pattern, consider how two waves travel from the slits to the screen ( Figure 3.6 ). Each slit is a different distance from a given point on the screen. Thus, different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively. More generally, if the path length difference Δ l Δ l between the two waves is any half-integral number of wavelengths [(1 / 2) λ λ , (3 / 2) λ λ , (5 / 2) λ λ , etc.], then destructive interference occurs. Similarly, if the path length difference is any integral number of wavelengths ( λ λ , 2 λ λ , 3 λ λ , etc.), then constructive interference occurs. These conditions can be expressed as equations:

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/university-physics-volume-3/pages/1-introduction
  • Authors: Samuel J. Ling, Jeff Sanny, William Moebs
  • Publisher/website: OpenStax
  • Book title: University Physics Volume 3
  • Publication date: Sep 29, 2016
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/university-physics-volume-3/pages/1-introduction
  • Section URL: https://openstax.org/books/university-physics-volume-3/pages/3-1-youngs-double-slit-interference

© Jul 23, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

PHYS102: Introduction to Electromagnetism

Young's double slit experiment.

As the reading illustrates, Huygens' principle is not just a philosophical interpretation – it is also a computational tool . In particular, the idea of circular (or spherical ) elementary waves makes it relatively easy to explain how a wave can bend around corners and spread out after passing through a constriction. This is called diffraction because it allows wave energy to go around corners in directions that the rays of geometric optics (or the trajectories of classical particles) would not be permitted to go.

Read about the proof that light is a wave in this experiment Thomas Young gave using diffraction by a pair of closely spaced slits.

Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see Figure 27.10 ).

A beam of light strikes a wall through which a pair of vertical slits is cut. On the other side of the wall, another wall shows a pattern of equally spaced vertical lines of light that are of the same height as the slit.

Figure 27.10 Young’s double slit experiment. Here pure-wavelength light sent through a pair of vertical slits is diffracted into a pattern on the screen of numerous vertical lines spread out horizontally. Without diffraction and interference, the light would simply make two lines on the screen.

Why do we not ordinarily observe wave behavior for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source (the Sun) through a single slit to make the light somewhat coherent.

By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships.

Figure a shows three sine waves with the same wavelength arranged one above the other. The peaks and troughs of each wave are aligned with those of the other waves. The top two waves are labeled wave one and wave two and the bottom wave is labeled resultant. The amplitude of waves one and two are labeled x and the amplitude of the resultant wave is labeled two x. Figure b shows a similar situation, except that the peaks of wave two now align with the troughs of wave one. The resultant wave is now a straight horizontal line on the x axis; that is, the line y equals zero.

Figure 27.11 The amplitudes of waves add. (a) Pure constructive interference is obtained when identical waves are in phase. (b) Pure destructive interference occurs when identical waves are exactly out of phase, or shifted by half a wavelength.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 27.12 (a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 27.12 (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

The figure contains three parts. The first part is a drawing that shows parallel wavefronts approaching a wall from the left. Crests are shown as continuous lines, and troughs are shown as dotted lines. Two light rays pass through small slits in the wall and emerge in a fan-like pattern from two slits. These lines fan out to the right until they hit the right-hand wall. The points where these fan lines hit the right-hand wall are alternately labeled min and max. The min points correspond to lines that connect the overlapping crests and troughs, and the max points correspond to the lines that connect the overlapping crests. The second drawing is a view from above of a pool of water with semicircular wavefronts emanating from two points on the left side of the pool that are arranged one above the other. These semicircular waves overlap with each other and form a pattern much like the pattern formed by the arcs in the first image. The third drawing shows a vertical dotted line, with some dots appearing brighter than other dots. The brightness pattern is symmetric about the midpoint of this line. The dots near the midpoint are the brightest. As you move from the midpoint up, or down, the dots become progressively dimmer until there seems to be a dot missing. If you progress still farther from the midpoint, the dots appear again and get brighter, but are much less bright than the central dots. If you progress still farther from the midpoint, the dots get dimmer again and then disappear again, which is where the dotted line stops.

Figure 27.12 Double slits produce two coherent sources of waves that interfere. (a) Light spreads out (diffracts) from each slit, because the slits are narrow. These waves overlap and interfere constructively (bright lines) and destructively (dark regions). We can only see this if the light falls onto a screen and is scattered into our eyes. (b) Double slit interference pattern for water waves are nearly identical to that for light. Wave action is greatest in regions of constructive interference and least in regions of destructive interference. (c) When light that has passed through double slits falls on a screen, we see a pattern such as this. (credit: PASCO)

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in Figure 27.13 . Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in Figure 27.13 (a).

Take-Home Experiment: Using Fingers as Slits

Look at a light, such as a street lamp or incandescent bulb, through the narrow gap between two fingers held close together. What type of pattern do you see? How does it change when you allow the fingers to move a little farther apart? Is it more distinct for a monochromatic source, such as the yellow light from a sodium vapor lamp, than for an incandescent bulb?

Both parts of the figure show a schematic of a double slit experiment. Two waves, each of which is emitted from a different slit, propagate from the slits to the screen. In the first schematic, when the waves meet on the screen, one of the waves is at a maximum whereas the other is at a minimum. This schematic is labeled dark (destructive interference). In the second schematic, when the waves meet on the screen, both waves are at a minimum.. This schematic is labeled bright (constructive interference).

Figure 27.13 Waves follow different paths from the slits to a common point on a screen. (a) Destructive interference occurs here, because one path is a half wavelength longer than the other. The waves start in phase but arrive out of phase. (b) Constructive interference occurs here because one path is a whole wavelength longer than the other. The waves start out and arrive in phase.

Similarly, to obtain destructive interference for a double slit, the path length difference must be a half-integral multiple of the wavelength, or

The figure is a schematic of a double slit experiment, with the scale of the slits enlarged to show the detail. The two slits are on the left, and the screen is on the right. The slits are represented by a thick vertical line with two gaps cut through it a distance d apart. Two rays, one from each slit, angle up and to the right at an angle theta above the horizontal. At the screen, these rays are shown to converge at a common point. The ray from the upper slit is labeled l sub one, and the ray from the lower slit is labeled l sub two. At the slits, a right triangle is drawn, with the thick line between the slits forming the hypotenuse. The hypotenuse is labeled d, which is the distance between the slits. A short piece of the ray from the lower slit is labeled delta l and forms the short side of the right triangle. The long side of the right triangle is formed by a line segment that goes downward and to the right from the upper slit to the lower ray. This line segment is perpendicular to the lower ray, and the angle it makes with the hypotenuse is labeled theta. Beneath this triangle is the formula delta l equals d sine theta.

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 27.15 . The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation

The figure consists of two parts arranged side-by-side. The diagram on the left side shows a double slit arrangement along with a graph of the resultant intensity pattern on a distant screen. The graph is oriented vertically, so that the intensity peaks grow out and to the left from the screen. The maximum intensity peak is at the center of the screen, and some less intense peaks appear on both sides of the center. These peaks become progressively dimmer upon moving away from the center, and are symmetric with respect to the central peak. The distance from the central maximum to the first dimmer peak is labeled y sub one, and the distance from the central maximum to the second dimmer peak is labeled y sub two. The illustration on the right side shows thick bright horizontal bars on a dark background. Each horizontal bar is aligned with one of the intensity peaks from the first figure.

Figure 27.15 The interference pattern for a double slit has an intensity that falls off with angle. The photograph shows multiple bright and dark lines, or fringes, formed by light passing through a double slit.

Example 27.1 Finding a Wavelength from an Interference Pattern

Substituting known values yields

Example 27.2 Calculating Highest Order Possible

Strategy and concept.

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.

Creative Commons License

Young's Double Slit Interference

 


Waves can be added together either constructively or destructively. The result of adding two waves of the same frequency depends on the value of the phase of the wave at the point in which the waves are added.

Electromagnetic waves are subject to interference. For two sources of electromagnetic waves to interfere:

In the double slit experiment, a single source is split in two, to generate two coherent sources. When the light from the two sources is projected on a screen, an interference pattern is observed.

To explain the origin of the interference pattern, consider the distance traveled from the two sources. At the center of the screen the waves from the two sources are in phase. As we move away from the center, the path traveled by the light from one source is larger than that traveled by the light from the other source. When the difference in path is equal to half a wavelength, destructive interference occurs. Instead, when the difference in path length is equal to a wavelength, constructive interference occurs.

This is a classic example of interference effects in light waves. Two light rays pass through two slits, separated by a distance and strike a screen a distance, , from the slits. You can change this parameters (drag scrollbars to do it) and you see the result of interference on the screen.


Wave Optics
Copyright © 1996, 1997 Sergey G. Vtorov. All rights reserved.

 




TXLS logo

10.3 Young’s Double Slit Experiment

Learning objectives.

By the end of this section, you will be able to do the following:

  • Explain the phenomena of interference
  • Define constructive interference for a double slit and destructive interference for a double slit

The information presented in this section supports the following AP® learning objectives and science practices:

  • 6.C.3.1 The student is able to qualitatively apply the wave model to quantities that describe the generation of interference patterns to make predictions about interference patterns that form when waves pass through a set of openings whose spacing and widths are small, but larger than the wavelength. (S.P. 1.4, 6.4)
  • 6.D.1.1 The student is able to use representations of individual pulses and construct representations to model the interaction of two wave pulses to analyze the superposition of two pulses. (S.P. 1.1, 1.4)
  • 6.D.1.3 The student is able to design a plan for collecting data to quantify the amplitude variations when two or more traveling waves or wave pulses interact in a given medium. (S.P. 4.2)
  • 6.D.2.1 The student is able to analyze data or observations or evaluate evidence of the interaction of two or more traveling waves in one or two dimensions (i.e., circular wave fronts) to evaluate the variations in resultant amplitudes. (S.P. 5.1)

Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see Figure 10.10 ).

A beam of light strikes a wall through which a pair of vertical slits is cut. On the other side of the wall, another wall shows a pattern of equally spaced vertical lines of light that are of the same height as the slit.

Why do we not ordinarily observe wave behavior for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source—the sun—through a single slit to make the light somewhat coherent. By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships. Why did Young then pass the light through a double slit? The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single λ ) λ ) size 12{λ} {} light to clarify the effect. Figure 10.11 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

Figure a shows three sine waves with the same wavelength arranged one above the other. The peaks and troughs of each wave are aligned with those of the other waves. The top two waves are labeled wave one and wave two and the bottom wave is labeled resultant. The amplitude of waves one and two are labeled x and the amplitude of the resultant wave is labeled two x. Figure b shows a similar situation, except that the peaks of wave two now align with the troughs of wave one. The resultant wave is now a straig

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 10.12 (a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 10.12 (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

The figure contains three parts. The first part is a drawing that shows parallel wavefronts approaching a wall from the left. Crests are shown as continuous lines, and troughs are shown as dotted lines. Two light rays pass through small slits in the wall and emerge in a fan-like pattern from two slits. These lines fan out to the right until they hit the right-hand wall. The points where these fan lines hit the right-hand wall are alternately labeled min and max. The min points correspond to lines that con

Making Connections: Interference

In addition to light waves, the phenomenon of interference also occurs in other waves, including water and sound waves. You will observe patterns of constructive and destructive interference if you throw two stones in a lake simultaneously. The crests and troughs of the two waves interfere constructively whereas the crest of a wave interferes destructively with the trough of the other wave. Similarly, sound waves traveling in the same medium interfere with each other. Their amplitudes add if they interfere constructively or subtract if there is destructive interference.

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in Figure 10.13 . Each slit is a different distance from a given point on the screen. Thus, different numbers of wavelengths fit into each path. Waves start out from the slits in phase—crest to crest—but they may end up out of phase—crest to trough—at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in Figure 10.13 (a). If the paths differ by a whole wavelength, then the waves arrive in phase—crest to crest—at the screen, interfering constructively as shown in Figure 10.13 (b). More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [ ( 1 / 2 ) λ , [ ( 1 / 2 ) λ , size 12{ \( 1/2 \) λ} {} ( 3 / 2 ) λ , ( 3 / 2 ) λ , size 12{ \( 3/2 \) λ} {} ( 5 / 2 ) λ , ( 5 / 2 ) λ , size 12{ \( 5/2 \) λ} {} etc.], then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths ( λ , λ , size 12{λ} {} 2 λ , 2 λ , size 12{2λ} {} 3 λ , 3 λ , size 12{3λ} {} etc.), then constructive interference occurs.

Take-Home Experiment: Using Fingers as Slits

Look at a light, such as a street lamp or incandescent bulb, through the narrow gap between two fingers held close together. What type of pattern do you see? How does it change when you allow the fingers to move a little farther apart? Is it more distinct for a monochromatic source, such as the yellow light from a sodium vapor lamp, than for an incandescent bulb?

Both parts of the figure show a schematic of a double slit experiment. Two waves, each of which is emitted from a different slit, propagate from the slits to the screen. In the first schematic, when the waves meet on the screen, one of the waves is at a maximum whereas the other is at a minimum. This schematic is labeled dark (destructive interference). In the second schematic, when the waves meet on the screen, both waves are at a minimum.. This schematic is labeled bright (constructive interference).

Figure 10.14 shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle θ θ size 12{θ} {} between the path and a line from the slits to the screen (see the figure) is nearly the same for each path. The difference between the paths is shown in the figure; simple trigonometry shows it to be d sin θ , d sin θ , size 12{d`"sin"θ} {} where d d size 12{d} {} is the distance between the slits. To obtain constructive interference for a double slit , the path length difference must be an integral multiple of the wavelength, or

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

where λ λ size 12{λ} {} is the wavelength of the light, d d size 12{d} {} is the distance between slits, and θ θ size 12{θ} {} is the angle from the original direction of the beam as discussed above. We call m m size 12{m} {} the order of the interference. For example, m = 4 m = 4 size 12{m=4} {} is fourth-order interference.

The figure is a schematic of a double slit experiment, with the scale of the slits enlarged to show the detail. The two slits are on the left, and the screen is on the right. The slits are represented by a thick vertical line with two gaps cut through it a distance d apart. Two rays, one from each slit, angle up and to the right at an angle theta above the horizontal. At the screen, these rays are shown to converge at a common point. The ray from the upper slit is labeled l sub one, and the ray from the l

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 10.15 . The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation

For fixed λ λ size 12{λ} {} and m , m , size 12{m} {} the smaller d d size 12{d} {} is, the larger θ θ size 12{θ} {} must be, since sin θ = mλ / d . sin θ = mλ / d . size 12{"sin"θ=mλ/d} {} This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance d d size 12{d} {} apart) is small. Small d d size 12{d} {} gives large θ , θ , size 12{θ} {} hence, a large effect.

The figure consists of two parts arranged side-by-side. The diagram on the left side shows a double slit arrangement along with a graph of the resultant intensity pattern on a distant screen. The graph is oriented vertically, so that the intensity peaks grow out and to the left from the screen. The maximum intensity peak is at the center of the screen, and some less intense peaks appear on both sides of the center. These peaks become progressively dimmer upon moving away from the center, and are symmetric

Making Connections: Amplitude of Interference Fringe

The amplitude of the interference fringe at a point depends on the amplitudes of the two coherent waves ( A 1 and A 2 ) arriving at that point and can be found using the relationship

where δ is the phase difference between the arriving waves.

This equation is also applicable for Young's double slit experiment. If the two waves come from the same source or two sources with the same amplitude, then A 1 = A 2 , and the amplitude of the interference fringe can be calculated using

The amplitude will be maximum when cosδ = 1 or δ = 0. This means the central fringe has the maximum amplitude. Also the intensity of a wave is directly proportional to its amplitude (i.e., I ∝ A 2 ) and consequently the central fringe also has the maximum intensity.

Example 10.1 Finding a Wavelength from an Interference Pattern

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of 10 . 95º 10 . 95º size 12{"10" "." "95"°} {} relative to the incident beam. What is the wavelength of the light?

The third bright line is due to third-order constructive interference, which means that m = 3 . m = 3 . size 12{m=3} {} We are given d = 0 . 0100 mm d = 0 . 0100 mm size 12{d=0 "." "0100"`"mm"} {} and θ = 10 . 95º . θ = 10 . 95º . size 12{θ="10" "." "95"°} {} The wavelength can thus be found using the equation d sin θ = mλ d sin θ = mλ size 12{d`"sin"θ=mλ} {} for constructive interference.

The equation is d sin θ = mλ . d sin θ = mλ . size 12{d`"sin"θ=mλ} {} Solving for the wavelength λ λ size 12{λ} {} gives

Substituting known values yields

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with λ , λ , size 12{λ} {} so that spectra—measurements of intensity versus wavelength—can be obtained.

Example 10.2 Calculating Highest Order Possible

Interference patterns do not have an infinite number of lines, since there is a limit to how big m m size 12{m} {} can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation d sin θ = mλ (for m = 0, 1, − 1, 2, − 2, … ) d sin θ = mλ (for m = 0, 1, − 1, 2, − 2, … ) describes constructive interference. For fixed values of d d size 12{d} {} and λ , λ , size 12{λ} {} the larger m m size 12{m} {} is, the larger sin θ sin θ size 12{"sin"`θ} {} is. However, the maximum value that sin θ sin θ size 12{"sin"θ} {} can have is 1, for an angle of 90º . 90º . size 12{"90"°} {} Larger angles imply that light goes backward and does not reach the screen at all. Let us find which m m size 12{m} {} corresponds to this maximum diffraction angle.

Solving the equation d sin θ = mλ d sin θ = mλ size 12{d`"sin"θ=mλ} {}  for  m m size 12{m} {} gives

Taking sin θ = 1 sin θ = 1 size 12{"sin"θ=1} {} and substituting the values of d d size 12{d} {} and λ λ size 12{m} {} from the preceding example gives

Therefore, the largest integer m m size 12{m} {} can be is 15, or

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.

Applying the Science Practices: Double Slit Experiment

Design an Experiment

Design a double slit experiment to find the wavelength of a He-Ne laser light. Your setup may include the He-Ne laser, a glass plate with two slits, paper, measurement apparatus, and a light intensity recorder. Write a step-by-step procedure for the experiment, draw a diagram of the set-up, and describe the steps followed to calculate the wavelength of the laser light.

Analyze Data

A double slit experiment is performed using three lasers. The table below shows the locations of the bright fringes that are recorded in meters on a screen.

Fringe Location for Laser 1 Location for Laser 2 Location for Laser 3
30.3710.3440.395
20.3140.2960.330
10.2570.2480.265
00.2000.2000.200
-10.1430.1520.135
-20.0860.1040.070
-30.0290.0560.005
  • Assuming the screen is 2.00 m away from the slits, find the angles for the first, second, and third bright fringes for each laser.
  • If the distance between the slits is 0.02 mm, calculate the wavelengths of the three lasers used in the experiment.
  • If the amplitudes of the three lasers are in the ratio 1:2:3, find the ratio of intensities of the central bright fringes formed by the three lasers.

Copy and paste the link code above.

Related Items

In Young's experiment two coherent sources are placed 0.90 m m apart and fringe are observed one metre away. If it produces second dark fringes at a distance of 1 m m from central fringe., the wavelength of monochromatic light is used would be 60 × 10 − 4 c m 10 × 10 − 4 c m 10 × 10 − 5 c m 6 × 10 − 5 c m

Formula, x = ( 2 n − 1 ) λ d 2 d ⇒ λ = 2 x d ( 2 n − 1 ) d = 2 × 10 − 3 × 0.9 × 10 − 3 ( 2 × 2 − 1 ) × 1 = 6 × 10 − 5 c m.

Stack Exchange Network

Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Q&A for work

Connect and share knowledge within a single location that is structured and easy to search.

Is coherent light required for interference in Young's double slit experiment?

In this Veritasium video , a home experiment is presented which appears to produce a very good double-slit interference pattern with normal sunlight.

The experiment is an empty cardboard box with a visor and a placeholder for a microscope slide with two slits on one side. This is arranged with the slits and visor facing the Sun, so the interference forms on the bottom of the box.

enter image description here

They claim to observe a good interference pattern from the two slits:

enter image description here

Discussions of interference in optics textbooks often stress that coherent light is needed to produce such patterns, and that sunlight and other thermal sources of light do not have such coherence. How, then, is this possible?

  • visible-light
  • double-slit-experiment
  • interference
  • home-experiment
  • popular-science

benrg's user avatar

10 Answers 10

Yes coherent light is required. The important thing to realize is that coherent light is not something that is magically created by lasers. Sunlight is somewhat coherent and it's easy to make it as coherent as you like.

What do people mean when they say "coherent light"? Well, it can be a few different things, but the relevant criteria in this context are:

  • The light is all travelling more-or-less in the same direction ("spatial coherence" or "collimation")
  • The light is more-or-less the same frequency ("temporal coherence" or "monochromaticity")

(See Footnote.)

I say "more or less" to emphasize the fact that it is never 100% coherent, (even from a laser), and it is never 0% coherent (even from a lightbulb or sunlight)

The way to think about it is, the light travelling towards the double-slits coming from a certain direction (e.g. 10 degrees away from normal incidence) create a really nice sharp double-slit pattern. The light travelling towards the double-slits from a different direction (e.g. 20 degrees away from normal incidence) also creates a really nice sharp double-slit pattern, but shifted !

So if you have light coming from every direction between 10 degrees and 20 degrees, you see a blurry composite of all those different double-slit patterns . It's possible that it will be so blurry that you can't even see that there's any pattern there -- it's just blurred out into a smooth line. But it's also possible that it will be only a little bit blurred out and the pattern is still recognizable.

The reason there's a cardboard box in the youtube video is to ensure that all the light from the sky that makes it to the slit is travelling in more-or-less the same direction. (Do you see how that could be done? Take a cardboard box, poke a small hole in it, and then put a double-slit far from the hole ... all the light at the double-slit is now coming in the same direction, i.e. from the hole.)

Frequency (or wavelength) is basically the same: Different frequencies of light make different interference patters, and we see a blurry composite of all those different patterns at once. If more monochromatic light was used (e.g. red laser light), the pattern would be much less blurry and easier to see, especially far from the center of the pattern. Luckily we have color vision, so we can (to some extent) recognize the composite pattern for what it is -- we see rainbows near the center, not just a blur.

Footnote: In comments, people are complaining that the term "coherent light" should refer only to spatial coherence, not temporal coherence. I disagree: The term can refer to either of these, depending on context. For example, in the context of optical coherence tomography, or in the context of "coherence length", or in the context of Michelson interferometers, people routinely use the phrase "coherent light" to mean temporal coherence.

Steve Byrnes's user avatar

  • 21 $\begingroup$ This is not a correct definition of coherent light. Light of a single frequency is monochromatic. Coherence means that the phase is correlated over large distances, or maybe in pop-sci terms that the wave trains are fairly long. $\endgroup$ –  user4552 Commented Sep 8, 2013 at 22:17
  • 7 $\begingroup$ When light is almost-monochromic, it's temporally coherent. When it's traveling more-or-lesss the same direction, it's spatially coherent. Do you agree? As I said, "coherence" means different things in different contexts, and there are certainly times when "coherence" specifically means only "spatial coherence" (as you propose). But I think I'm entitled to refer to temporal coherence as a kind of coherence. $\endgroup$ –  Steve Byrnes Commented Sep 8, 2013 at 22:29
  • 8 $\begingroup$ No, sorry, but this is just wrong. $\endgroup$ –  user4552 Commented Sep 8, 2013 at 22:40
  • 10 $\begingroup$ I'm with @Ben on this. Monochromaticity and coherence are different things. Lasers are monochromatic (except the ultra-short pulsed ones) and coherent. Raw sunlight is neither. The light from a sodium vapor lamp is fairly monochromatic but not coherent. Light from a broadband source that has been passed through two pinholes in series is polychromatic and coherent. $\endgroup$ –  dmckee --- ex-moderator kitten Commented Sep 8, 2013 at 23:12
  • 7 $\begingroup$ I edited to clarify that "coherent" usually means "spatially coherent". (I do NOT believe that it ALWAYS means spatially coherent. For example, read some descriptions of a Michelson interferometer...) dmckee -- I wish you would not describe things as "coherent" or "incoherent", as if coherence were all-or-nothing. That's what caused the questioner to be confused in the first place. For example, sunlight is not totally incoherent, it's pretty close to collimated, at least when there are no clouds. $\endgroup$ –  Steve Byrnes Commented Sep 9, 2013 at 1:29

“Interference is observed only when the light from the slits is coherent” (by the way, coherent light is defined as having all photons in the same phase, not just about the same wavelength and direction, as one answer here seems to suggest.) The statement can be challenged on three grounds:

  • Experiment. The Young's double slit experiment predates the laser. Light from a filament lamp produces a satisfactory interference pattern, provided it is approximately monochromatic, and provided that it is nearly parallel. It is a straightforward matter to replicate the experiment, without the need for a laser.
  • Theory. I quote the great physicist Paul Dirac (The Principles of Quantum Mechanics, Oxford Science Publications, Fourth Edition, p 9) “If the two components are now made to interfere, we should require a photon in one component to be able to interfere with one in the other. Sometimes these two photons would have to annihilate one another and sometimes they would have to produce four photons. This would contradict the conservation of energy. The new theory, which connects the wave function with the probabilities for one photon, gets over the difficulty by making each photon go partly into each of the two components. Each photon then interferes only with itself. Interference between two different photons never occurs.”
  • More experiment. This latter statement has been tested experimentally by performing the double slit experiment with photographic film and light of a very low intensity. The intensity is so small that the photons pass through the apparatus effectively one-at-a-time, with the average interval between two emitted photons much greater than the time required to pass through the apparatus, so that the probability of two photons “meeting” at the slits, although not zero, is very small. The interference pattern which builds up on the film is precisely the same as when high intensity light is used.

It has been argued that light from say a filament lamp is usually passed through a single narrow slit (as well as a colour filter) before arriving at the double slits. Without this “coherer”, the interference pattern is not observed. While experimentally true, the explanation is erroneous. Two incoherent photons arriving at this slit do not suddenly become coherent because they pass though a small hole together.

The whole thing is resolved as follows: A. Light for the double slit experiment must be nearly monochromatic so that the fringe separation is about the same for all photons, otherwise the interference patterns will form an indiscriminate jumble. B. Light for the double slit experiment must be nearly unidirectional (parallel) otherwise the interference patterns formed in all the slightly different directions behind the two slits will form an indiscriminate jumble.

These two conditions can be met by passing light from an incandescent lamp through a colour filter and a small hole, or by using a laser. The fact that laser light is also coherent is quite unimportant.

Bill Dixon's user avatar

  • 2 $\begingroup$ @Numrok and Bill Dixon: this answer is not very different from that of Steve B. But this one is too extreme to be correct. "Light for the double slit experiment must be nearly monochromatic (...), otherwise the interference patterns will form an indiscriminate jumble": the picture in the question shows exactly the opposite, the pattern is absolutely not "an indiscriminate jumble". Polychromatic sources give very nice rainbow patterns: the first two orders of interference do not overlap in visible light. $\endgroup$ –  L. Levrel Commented May 25, 2016 at 12:43
  • 1 $\begingroup$ Secondly, mentioning photons gives an interesting discussion, but it's not very relevant to the question. (Also I fear it may sustain the frequent misconception of photons being another name for classical wavepackets.) Thirdly, "coherent" means "with constant phase difference", not "with zero phase difference". $\endgroup$ –  L. Levrel Commented May 25, 2016 at 12:50
  • $\begingroup$ @Numrok: about coherence, I just meant that coherence (in a weak form of the word maybe) only indicates points have a definite phase relation, but they are not necessarily in phase. Points which are constantly in opposing phase are coherent. $\endgroup$ –  L. Levrel Commented May 25, 2016 at 18:00
  • $\begingroup$ OK, so we agree, then I don't see why @SteveB's answer is "flat out wrong". See also his comment "not describe things as 'coherent' or 'incoherent', as if coherence were all-or-nothing". As regards colors, I understand your radiometric point of view. However, in principle quantum detectors are not required to distinguish colors: in the microwaves or radio waves regimes, a classical LC circuit does the filtering job. (Apologies to everyone for the chatty comments, but it's relevant to the subject, and useful to understanding. Maybe we can remove them when Numrok posts his answer.) $\endgroup$ –  L. Levrel Commented May 25, 2016 at 19:24
  • $\begingroup$ @L.Levrel I finally got around to writing the answer, see below. hope you like it. $\endgroup$ –  Wolpertinger Commented Aug 17, 2016 at 15:38

Here is Young's original experiment with light ( after having studied water waves)

young dbslit

The first screen generates a point source, so as to create a coherent wave . If it is a pin hole the geometry assures that all the photons come from the same original tiny source of light. Nice illustration here , page 5. Coherent means that the phases describing the mathematical form of the wave are not randomized.

In the video above the slits must be narrow enough and the distance between them small enough so that the wavefront arriving at them is similar to a point source wavefront.In any case the interference pattern is sort of blurred due to the many frequencies.

anna v's user avatar

  • 1 $\begingroup$ anna v How the first pinhole makes a coherent wave from a thermic source? $\endgroup$ –  HolgerFiedler Commented Sep 6, 2018 at 7:29
  • $\begingroup$ @HolgerFiedler it is the mathematics of the maxwell equation for a point source. At the photon level one would have to solve the quantized maxwell for a photon passing through a point source. I suspect it is because of the constraint of having a specific x,y,z,t . $\endgroup$ –  anna v Commented Sep 6, 2018 at 7:58
  • $\begingroup$ you had picture from wiki article and the text under picture says "Modern illustration", the video of OP claims to study original Young notes. Where have you got necessity of S1 screen except from that "Modern illustration"? $\endgroup$ –  Martian2020 Commented Nov 9, 2022 at 14:22
  • $\begingroup$ example researchgate.net/figure/… $\endgroup$ –  anna v Commented Nov 9, 2022 at 14:35
  • $\begingroup$ Anna, I'm not arguing this modern setup won't work, I'm arguing you had not proven original one had extra (compared to setup in the video) screen. $\endgroup$ –  Martian2020 Commented Nov 9, 2022 at 15:01

This question is answered with the Van-Cittert Zernike theorem . When the double slit experiment is performed with incoherent light the fringes get blurred. How much the fringes get blurred depend of how big is the light source and how far is the light source from the double slit.

The intensity pattern on the screen by the an incoherent source is given by:

$I ∝ sinc^2( \frac{𝜋 a}{z λ} x ) ( 1 + γ \cos(\frac{2𝜋D}{zλ} x))$

$D$ = distance between the slits

$a$ = slits width

$γ$ is the degree of spatial coherence: $γ = sinc(\frac{2 𝜋 D M}{L λ})$

$M$ = width of the light source

$z$ = distance from the screen to the double slit

$L$ = distance from the light source to the double slit

When the experiment is performed with sunlight this formula gives for the coherence of the sunlight:

$L = 150.17$ million of km (distance from sun to earth)

$M = 1.3927$ million of km (diameter of the sun)

$D = 15 μm$

$λ = 500 nm$ (peak wavelength of the spectrum of the sunlight)

$γ = sin(\frac{2\pi DM}{Lλ}) ≈ 0.4$

So with this slits separation the light coming to the sun is partially coherent .

Plotting the pattern with the formula indicated above for z = 1m and a = 1 mm we get:

Partially coherent light

You can see in the plot that the fringes are still visible.

I made a video explaining and simulating how the double slit experiment with incoherent light works. Hope that it helps!

Rafael's user avatar

  • $\begingroup$ Your video looks great. BUT the original question is probably better worded as " Is only collimated light created in a double slit apparatus? ... the answer being YES. What your video really shows (or should show) are the allowed paths that photons would take, the requirement being consistent with Feynman theory that boils down to photons only travel (highest probability) allowed paths that are n (n=integer) times their wavelength. This is what happens in laser cavities and dichroic filters for example. The DSE places tight requirements on optical paths or "modes" ( like in a fiber optic). $\endgroup$ –  PhysicsDave Commented Sep 25, 2020 at 0:14
  • $\begingroup$ Another concept is that photons find this path before they become an actual photon! (Maybe its virtual to start). Also single photons behave the same way as multiple photons, i.e. they still make the pattern ... the singles have no other photons to "interfere" with. $\endgroup$ –  PhysicsDave Commented Sep 25, 2020 at 0:18
  • $\begingroup$ @PhysicsDave, "they still make the pattern". as I understand we see interference pattern because photos interfere only with themselves, not others. Photos are each in different phases, the sums of wave functions of each two of them have different patterns. P.S. I hope I wrote correctly after several edits. $\endgroup$ –  Martian2020 Commented Nov 9, 2022 at 14:53

If the source is far away, light acquires a certain degree of coherence. Have a look at the Van Cittert–Zernike theorem , as pointed in wikipedia:

[...]the wavefront from an incoherent source will appear mostly coherent at large distances

The resulting fringes are different for different colors, but any color is maximum for straightforward direction. So, you see the bright spot at the center.

Then, the wavelengths our eyes are sensible to are not very different for this experiment. In other words, you may choose a distance between the slits such that wavelength/distance is the approx the same for all the frequencies your eye is sensible to (from red to blue), i.e. you choose a large distance. Then, the all frequencies between red and blue will approx peak at the same position. Blue will peak slightly before than red. From the figure, you indeed see the overlapping fringes given by the highest frequency you can see with your eyes (blue light) and the lowest (red light) soon afterwards.

Wizzerad's user avatar

This answer is in the domain of classical electrodynamics. For a quantum description, look at Bill Dixon’s answer .

Let us first see when an interference pattern occurs. On the screen, the intensity of light is given by the resultant field which is the sum of (to a great approximation) two fields. One from each slit. Assuming equal intensity of light from both the slits, the governing factor for the intensity at the screen is phase difference between the two fields.

To ensure a stable interference pattern all one needs to do is to ensure the phase difference between the fields of the two slits at the screen remain fixed. This is easy to ensure. If the field before passing through the slits is a plane wave, then the phase difference is fixed at all points on the screen.

enter image description here

To get plane waves (approximately) one uses a pinhole sufficiently far before the double slit that acts as a collimator. So finally you’ll see an overlap of $N$ double slit pattens for each of the $N$ different frequencies present.

Superfast Jellyfish's user avatar

The answer is no. Coherent light would be required for different photons to interfere with each other constructively, but as the quantum double slit experiment shows, individual photons effectively interfere with themselves, they do not need to interfere with other photons so coherent light is not required. Just as well since Young did not have coherent light.

But to get a good pattern of fringes requires monochromatic light, a filter, and all of the photons arriving at the twin slits coming from the same direction; an initial single slit achieves this. Young's experiment had the perfect set up.

Before reading this question I was unaware of the difference between spatial and temporal coherence, but it is not relevant here. Spatial coherence definitely does not just mean 'coming from the same direction'.

John Hobson's user avatar

Newton and Young didn't have coherent light, they worked with white light and saw color fringes. But there is a second condition. The dimension of the light source has to be very small or the source has to be at a big distance (in relation to its dimension). In this cases the photons propagates parallel to each other and do not overlap each other on the observation screen.

HolgerFiedler's user avatar

  • 1 $\begingroup$ Newton and Young did have spatially coherent light ...but not monochromatic light. Pinhole-light is coherent. Passing sunlight through a small pinhole or slit makes it spatially-coherent but not monochromatic. (Slits produce coherent light in one transverse dimension.) Young's experiment can employ pinholes alone, but if slits are employed the pattern becomes several orders brighter. However, because sunlight lacks temporal coherence, only the first few interference nodes are easily seen (but with colored edges,) while the higher order narrow fringes are blurred into uniform white. $\endgroup$ –  wbeaty Commented Sep 6, 2018 at 6:54
  • $\begingroup$ @wbeaty Thanks for specifying. Could you explain, what makes the light from a thermic source coherent during the passage of the pinhole? $\endgroup$ –  HolgerFiedler Commented Sep 6, 2018 at 7:26
  • 2 $\begingroup$ @HolgerFieldler light of perfect spatial-coherence is defined as ideal sphere-wave EM, and with sphere-waves, phase between widely separate points on the sphere is constant (transverse coherence length is long.) If we pass broadband EM through a "small" pinhole (diameter << 1/4 wavelength,) then only spherical EM waves will exit the far side. (One source must be coherent with itself!) Yet "white" light is limited to a 2:1 frequency spread. If using spatially-coherent sunlight, the observed 2-slit node pattern will have a 2:1 color-blur caused by this partial temporal incoherence. $\endgroup$ –  wbeaty Commented Sep 6, 2018 at 7:41
  • $\begingroup$ @HolgerFieldler also: temporal coherence or "monochromaticity" is measured along the axis of propagation, while spatial coherence (often called "Coherence") is measured at 90deg: across the wavefront. The pinhole only creates spatial coherence; changing a source with multiple uncorrelated points to a single point. Thermic light through a pinhole has fixed phase only when measured transverse to propagation axis. Search: "spatial filter," a lens and pinhole. See electron6.phys.utk.edu/optics421/modules/m5/Coherence.htm , en.wikipedia.org/wiki/Van_Cittert%E2%80%93Zernike_theorem $\endgroup$ –  wbeaty Commented Sep 6, 2018 at 8:06

This is possible because the pattern you see is made of coherent light. The source of the coherent light is the hole. Each color hits the screen where it is supposed to. If you only had one color you would notice the repeating pattern and it would be easier to realize that it was coherent. If you look close you will see where the colors repeat themselves over and over and over.

Bill Alsept's user avatar

  • $\begingroup$ "The source of the coherent light is the hole.". Do you mean the slits? Cause in the video holes are slits plus one for viewing with an eye. $\endgroup$ –  Martian2020 Commented Nov 9, 2022 at 14:33

The following equations predict the phase difference between to light sources, when the light reaches a screen. These equations assume coherent light. In other words, the photons having the same phase when the photons leave the slit.

Constructive interference: d sin θ = mλ

Destructive interference: d sin θ = (m+1/2) λ For m = 0, 1, -1, 2, -2, … and d = distance between slits.

Compliance with these equations (a clean interference pattern) can indicate a high coherent light to random light ratio.

Eileen Gimper's user avatar

Your Answer

Sign up or log in, post as a guest.

Required, but never shown

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy .

Not the answer you're looking for? Browse other questions tagged visible-light double-slit-experiment interference home-experiment popular-science or ask your own question .

  • Featured on Meta
  • Join Stack Overflow’s CEO and me for the first Stack IRL Community Event in...
  • User activation: Learnings and opportunities

Hot Network Questions

  • Alice splits the bill not too generously with Bob
  • (How) is it possible to let portable communication devices detonate via software?
  • Could a Gamma Ray Burst knock a Space Mirror out of orbit?
  • Count squares in my pi approximation
  • Smallest prime q such that concatenation (p+q)"q is a prime
  • Can turbo trainers be easily damaged when instaling a cassette?
  • Rules on Feats When Creating Custom Backgrounds with Player's Handbook (2024)
  • LTspice Pulse Help
  • Was the total glaciation of the world, a.k.a. snowball earth, due to Bok space clouds?
  • Fantasy novel where a brother and sister go to a fantasy and travel with a one-armed warrior
  • How to translate the letter Q to Japanese?
  • Can I use a Forward Transformer in a Flyback Converter Circuit?
  • Apple IIgs to VGA adapter
  • What is an apologetic to confront Schellenberg's non-resistant divine hiddenness argument?
  • Class and macro with same name from different libraries
  • Determining Entropy in PHP
  • What can I do to limit damage to a ceiling below bathroom after faucet leak?
  • My math professor is Chinese. Is it okay for me to speak Chinese to her in office hours?
  • A word like "science/scientific" that can be used for ALL academic fields?
  • Counting the number of meetings
  • Why does fdisk create a 512B partition when I enter +256K?
  • how does the US justice system combat rights violations that happen when bad practices are given a new name to avoid old rulings?
  • Raster to polygon in QGIS - How to Poligonize using a limit of vertices
  • Can the signing of a bill by an acting governor be overturned by the actual governor?

in young's double slit experiment two coherent sources

IMAGES

  1. Double-Slit Experiment: Explanation, Diagram, and Equation

    in young's double slit experiment two coherent sources

  2. Young's Double Slit Experiment

    in young's double slit experiment two coherent sources

  3. Young’s Double Slits Experiment Derivation

    in young's double slit experiment two coherent sources

  4. Young's Double Slit Experiment

    in young's double slit experiment two coherent sources

  5. 27.3 Young’s Double Slit Experiment

    in young's double slit experiment two coherent sources

  6. Young's Double Slit Experiment (9.2.4)

    in young's double slit experiment two coherent sources

VIDEO

  1. 6 Youngs Double Slit Experiment & Coherent Source 1

  2. Chapter 9, 9.4 Young's double slit experiment, First year physics

  3. In Young's double slit experiment, two slits \(S_1\) and \(S_2\) are ' \(d\) ' distance apart an

  4. 38.Chapter

  5. Young's Double Slit Experiment derivation| Wave Optics class 12| boards|JEE|maxima and minima points

  6. Conceptual Questions on Wave optics

COMMENTS

  1. 27.3 Young's Double Slit Experiment

    Figure 27.10 Young's double slit experiment. Here pure-wavelength light sent through a pair of vertical slits is diffracted into a pattern on the screen of numerous vertical lines spread out horizontally. ... The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively ...

  2. Young's Double-Slit Experiment

    Revision Notes. BiologyFirst Exams 2025HL. Topic Questions. Revision Notes. Chemistry. ChemistryLast Exams 2024SL. Topic Questions. Revision Notes. Revision notes on 3.3.3 Young's Double-Slit Experiment for the AQA A Level Physics syllabus, written by the Physics experts at Save My Exams.

  3. Young's Double Slit Experiment

    The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single λ) light to clarify the effect ...

  4. Young's Double Slit Experiment

    The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773-1829) did his now-classic double slit experiment (see (Figure)). Young's double slit experiment. Here pure-wavelength light sent through a pair of vertical slits is diffracted into a pattern on the screen ...

  5. 27.3 Young's Double Slit Experiment

    The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773-1829) did his now-classic double slit experiment (see Figure 1). Figure 1. Young's double slit experiment. Here pure-wavelength light sent through a pair of vertical slits is diffracted into a pattern on ...

  6. 27.3 Young's Double Slit Experiment

    Section Summary. Young's double slit experiment gave definitive proof of the wave character of light. An interference pattern is obtained by the superposition of light from two slits. There is constructive interference when d sin θ = mλ (for m = 0, 1, −1, 2, −2,…) d sin θ = m λ (for m = 0, 1, − 1, 2, − 2, …), where d d is the ...

  7. 154 Young's Double Slit Experiment

    The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single \(\lambda \)) light to clarify the ...

  8. 27.3 Young's Double Slit Experiment

    Double slits produce two coherent sources of waves that interfere. (a) Light spreads out (diffracts) from each slit, because the slits are narrow. These waves overlap and interfere constructively (bright lines) and destructively (dark regions). ... Young's double slit experiment breaks a single light beam into two sources. Would the same ...

  9. Light

    Young's double-slit experiment When monochromatic light passing through two narrow slits illuminates a distant screen, a characteristic pattern of bright and dark fringes is observed. This interference pattern is caused by the superposition of overlapping light waves originating from the two slits. ... the two sources of light must be coherent ...

  10. 3.1 Young's Double-Slit Interference

    The acceptance of the wave character of light came many years later in 1801, when the English physicist and physician Thomas Young (1773-1829) demonstrated optical interference with his now-classic double-slit experiment. If there were not one but two sources of waves, the waves could be made to interfere, as in the case of waves on water ...

  11. Double-slit experiment

    In the basic version of this experiment, a coherent light source, such as a laser beam, illuminates a plate pierced by two parallel slits, and the light passing through the slits is observed on a screen behind the plate. [8] [9] The wave nature of light causes the light waves passing through the two slits to interfere, producing bright and dark bands on the screen - a result that would not ...

  12. Young's Double Slit Experiment

    Young's double slit experiment uses two coherent sources of light placed at a small distance apart. Usually, only a few orders of magnitude greater than the wavelength of light are used. Young's double slit experiment helped in understanding the wave theory of light, which is explained with the help of a diagram. As shown, a screen or ...

  13. PHYS102: Young's Double Slit Experiment

    The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773-1829) did his now-classic double slit experiment (see Figure 27.10). Figure 27.10 Young's double slit experiment. Here pure-wavelength light sent through a pair of vertical slits is diffracted into a ...

  14. PDF Chapter 14 Interference and Diffraction

    14.2 Young's Double-Slit Experiment In 1801 Thomas Young carried out an experiment in which the wave nature of light was demonstrated. The schematic diagram of the double-slit experiment is shown in Figure ... Iinc =I1 +I2 (14.3.3) For coherent sources, the cross term is non-zero. In fact, for constructive interference, E1=E GG =41 2, and

  15. Young's Double Slit Interference

    For two sources of electromagnetic waves to interfere: the sources must have the same frequency and polarization, the sources must be coherent, and; the superposition principle must apply. In the double slit experiment, a single source is split in two, to generate two coherent sources.

  16. 10.3 Young's Double Slit Experiment

    Figure 10.12 Double slits produce two coherent sources of waves that interfere. (a) Light spreads out—diffracts—from each slit, because the slits are narrow. ... This equation is also applicable for Young's double slit experiment. If the two waves come from the same source or two sources with the same amplitude, ...

  17. In Young's experiment two coherent sources are placed

    In Young's double slit experiment, the two slits 0.15 mm apart are illuminated by monochromatic light of wavelength 450 nm. The screen is 1.0 m away from the slits. (a) Find the distance of the second. (I) bright fringe, (ii) dark fringe from the central maximum, (b) How will the fringe pattern change if the screen is moved away from the slits?

  18. Interference of light

    Young's Double Slit Experiment. The great scientist Young performed an experiment to prove the wave nature of light by explaining the phenomenon of interference of light. In Young's double slit experiment, two coherent sources were generated using diffracted light from a single slit. Note that the waves must have a constant phase difference ...

  19. Is coherent light required for interference in Young's double slit

    The Young's double slit experiment predates the laser. Light from a filament lamp produces a satisfactory interference pattern, provided it is approximately monochromatic, and provided that it is nearly parallel. It is a straightforward matter to replicate the experiment, without the need for a laser. ... The source of the coherent light is the ...

  20. In young's double slit experiment, the two slits act as coherent

    In Young's double slit experiment, the two slits act as coherent sources of light waves of equal amplitude A and of wavelength λ. In another experiment with the same setup, the two slits are source of light waves of equal amplitude A and wavelength λ but are incoherent. The ratio of the intensity of light at the mid-point of the screen in ...