double slit experiment symbol

27.3 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 27.10 ).

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 ) light to clarify the effect. Figure 27.11 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

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.

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). 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 27.13 (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 , ( 3 / 2 ) λ ( 3 / 2 ) λ , ( 5 / 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 λ , 2 λ 2 λ , 3 λ 3 λ , 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?

Figure 27.14 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 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 , where d 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 is the wavelength of the light, d d is the distance between slits, and θ θ is the angle from the original direction of the beam as discussed above. We call m m the order of the interference. For example, m = 4 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 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

For fixed λ λ and m m , the smaller d d is, the larger θ θ must be, since sin θ = mλ / d 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 apart) is small. Small d d gives large θ θ , hence a large effect.

Example 27.1

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 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 . We are given d = 0 . 0100 mm d = 0 . 0100 mm and θ = 10 . 95º θ = 10 . 95º . The wavelength can thus be found using the equation d sin θ = mλ d sin θ = mλ for constructive interference.

The equation is d sin θ = mλ d sin θ = mλ . Solving for the wavelength λ 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 , so that spectra (measurements of intensity versus wavelength) can be obtained.

Example 27.2

Calculating highest order possible.

Interference patterns do not have an infinite number of lines, since there is a limit to how big m m 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 and λ λ , the larger m m is, the larger sin θ sin θ is. However, the maximum value that sin θ sin θ can have is 1, for an angle of 90º 90º . (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which m m corresponds to this maximum diffraction angle.

Solving the equation d sin θ = mλ d sin θ = mλ  for  m m gives

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

Therefore, the largest integer m 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.

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Double-slit Experiment

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Plane wave representing a particle passing through two slits, resulting in an interference pattern on a screen some distance away from the slits. [1] .

The double-slit experiment is an experiment in quantum mechanics and optics demonstrating the wave-particle duality of electrons , photons , and other fundamental objects in physics. When streams of particles such as electrons or photons pass through two narrow adjacent slits to hit a detector screen on the other side, they don't form clusters based on whether they passed through one slit or the other. Instead, they interfere: simultaneously passing through both slits, and producing a pattern of interference bands on the screen. This phenomenon occurs even if the particles are fired one at a time, showing that the particles demonstrate some wave behavior by interfering with themselves as if they were a wave passing through both slits.

Niels Bohr proposed the idea of wave-particle duality to explain the results of the double-slit experiment. The idea is that all fundamental particles behave in some ways like waves and in other ways like particles, depending on what properties are being observed. These insights led to the development of quantum mechanics and quantum field theory , the current basis behind the Standard Model of particle physics , which is our most accurate understanding of how particles work.

The original double-slit experiment was performed using light/photons around the turn of the nineteenth century by Thomas Young, so the original experiment is often called Young's double-slit experiment. The idea of using particles other than photons in the experiment did not come until after the ideas of de Broglie and the advent of quantum mechanics, when it was proposed that fundamental particles might also behave as waves with characteristic wavelengths depending on their momenta. The single-electron version of the experiment was in fact not performed until 1974. A more recent version of the experiment successfully demonstrating wave-particle duality used buckminsterfullerene or buckyballs , the \(C_{60}\) allotrope of carbon.

Waves vs. Particles

Double-slit experiment with electrons, modeling the double-slit experiment.

To understand why the double-slit experiment is important, it is useful to understand the strong distinctions between wave and particles that make wave-particle duality so intriguing.

Waves describe oscillating values of a physical quantity that obey the wave equation . They are usually described by sums of sine and cosine functions, since any periodic (oscillating) function may be decomposed into a Fourier series . When two waves pass through each other, the resulting wave is the sum of the two original waves. This is called a superposition since the waves are placed ("-position") on top of each other ("super-"). Superposition is one of the most fundamental principles of quantum mechanics. A general quantum system need not be in one state or another but can reside in a superposition of two where there is some probability of measuring the quantum wavefunction in one state or another.

Left: example of superposed waves constructively interfering. Right: superposed waves destructively interfering. [2]

If one wave is \(A(x) = \sin (2x)\) and the other is \(B(x) = \sin (2x)\), then they add together to make \(A + B = 2 \sin (2x)\). The addition of two waves to form a wave of larger amplitude is in general known as constructive interference since the interference results in a larger wave.

If one wave is \(A(x) = \sin (2x)\) and the other is \(B(x) = \sin (2x + \pi)\), then they add together to make \(A + B = 0\) \(\big(\)since \(\sin (2x + \pi) = - \sin (2x)\big).\) This is known as destructive interference in general, when adding two waves results in a wave of smaller amplitude. See the figure above for examples of both constructive and destructive interference.

Two speakers are generating sounds with the same phase, amplitude, and wavelength. The two sound waves can make constructive interference, as above left. Or they can make destructive interference, as above right. If we want to find out the exact position where the two sounds make destructive interference, which of the following do we need to know?

a) the wavelength of the sound waves b) the distances from the two speakers c) the speed of sound generated by the two speakers

This wave behavior is quite unlike the behavior of particles. Classically, particles are objects with a single definite position and a single definite momentum. Particles do not make interference patterns with other particles in detectors whether or not they pass through slits. They only interact by colliding elastically , i.e., via electromagnetic forces at short distances. Before the discovery of quantum mechanics, it was assumed that waves and particles were two distinct models for objects, and that any real physical thing could only be described as a particle or as a wave, but not both.

In the more modern version of the double slit experiment using electrons, electrons with the same momentum are shot from an "electron gun" like the ones inside CRT televisions towards a screen with two slits in it. After each electron goes through one of the slits, it is observed hitting a single point on a detecting screen at an apparently random location. As more and more electrons pass through, one at a time, they form an overall pattern of light and dark interference bands. If each electron was truly just a point particle, then there would only be two clusters of observations: one for the electrons passing through the left slit, and one for the right. However, if electrons are made of waves, they interfere with themselves and pass through both slits simultaneously. Indeed, this is what is observed when the double-slit experiment is performed using electrons. It must therefore be true that the electron is interfering with itself since each electron was only sent through one at a time—there were no other electrons to interfere with it!

When the double-slit experiment is performed using electrons instead of photons, the relevant wavelength is the de Broglie wavelength \(\lambda:\)

\[\lambda = \frac{h}{p},\]

where \(h\) is Planck's constant and \(p\) is the electron's momentum.

Calculate the de Broglie wavelength of an electron moving with velocity \(1.0 \times 10^{7} \text{ m/s}.\)

Usain Bolt, the world champion sprinter, hit a top speed of 27.79 miles per hour at the Olympics. If he has a mass of 94 kg, what was his de Broglie wavelength?

Express your answer as an order of magnitude in units of the Bohr radius \(r_{B} = 5.29 \times 10^{-11} \text{m}\). For instance, if your answer was \(4 \times 10^{-5} r_{B}\), your should give \(-5.\)

Image Credit: Flickr drcliffordchoi.

While the de Broglie relation was postulated for massive matter, the equation applies equally well to light. Given light of a certain wavelength, the momentum and energy of that light can be found using de Broglie's formula. This generalizes the naive formula \(p = m v\), which can't be applied to light since light has no mass and always moves at a constant velocity of \(c\) regardless of wavelength.

The below is reproduced from the Amplitude, Frequency, Wave Number, Phase Shift wiki.

In Young's double-slit experiment, photons corresponding to light of wavelength \(\lambda\) are fired at a barrier with two thin slits separated by a distance \(d,\) as shown in the diagram below. After passing through the slits, they hit a screen at a distance of \(D\) away with \(D \gg d,\) and the point of impact is measured. Remarkably, both the experiment and theory of quantum mechanics predict that the number of photons measured at each point along the screen follows a complicated series of peaks and troughs called an interference pattern as below. The photons must exhibit the wave behavior of a relative phase shift somehow to be responsible for this phenomenon. Below, the condition for which maxima of the interference pattern occur on the screen is derived.

Left: actual experimental two-slit interference pattern of photons, exhibiting many small peaks and troughs. Right: schematic diagram of the experiment as described above. [3]

Since \(D \gg d\), the angle from each of the slits is approximately the same and equal to \(\theta\). If \(y\) is the vertical displacement to an interference peak from the midpoint between the slits, it is therefore true that

\[D\tan \theta \approx D\sin \theta \approx D\theta = y.\]

Furthermore, there is a path difference \(\Delta L\) between the two slits and the interference peak. Light from the lower slit must travel \(\Delta L\) further to reach any particular spot on the screen, as in the diagram below:

Light from the lower slit must travel further to reach the screen at any given point above the midpoint, causing the interference pattern.

The condition for constructive interference is that the path difference \(\Delta L\) is exactly equal to an integer number of wavelengths. The phase shift of light traveling over an integer \(n\) number of wavelengths is exactly \(2\pi n\), which is the same as no phase shift and therefore constructive interference. From the above diagram and basic trigonometry, one can write

\[\Delta L = d\sin \theta \approx d\theta = n\lambda.\]

The first equality is always true; the second is the condition for constructive interference.

Now using \(\theta = \frac{y}{D}\), one can see that the condition for maxima of the interference pattern, corresponding to constructive interference, is

\[n\lambda = \frac{dy}{D},\]

i.e. the maxima occur at the vertical displacements of

\[y = \frac{n\lambda D}{d}.\]

The analogous experimental setup and mathematical modeling using electrons instead of photons is identical except that the de Broglie wavelength of the electrons \(\lambda = \frac{h}{p}\) is used instead of the literal wavelength of light.

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Young's Double-Slit Experiment ( AQA A Level Physics )

Revision note.

Double Slit Interference

• The interference of two coherent wave sources 
• A single wave source passing through a double slit
• The laser light source is placed behind the single slit
• So the light is diffracted, producing two light sources at slits A  and B
• The light from the double slits is then diffracted, producing a diffraction pattern made up of bright and dark fringes on a screen

The typical arrangement of Young's double-slit experiment

Diffraction Pattern

• Constructive interference between light rays forms bright strips, also called fringes , interference fringes or maxima , on the screen
• Destructive interference forms dark strips, also called dark fringes or minima , on the screen

Young's double slit experiment and the resulting diffraction pattern

• Each bright fringe is identical and has the same width and intensity
• The fringes are all separated by dark narrow bands of destructive interference

The constructive and destructive interference of laser light through a double slit creates bright and dark strips called fringes on a screen placed far away

Interference Pattern

• The Young's double slit interference pattern shows the regions of constructive and destructive interference:
• Each bright fringe is a peak of equal maximum intensity
• Each dark fringe is a a trough or minimum of zero intensity
• The maxima are formed by the constructive interference of light
• The minima are formed by the destructive interference of light

The interference pattern of Young's double-slit diffraction of light

• When two waves interfere, the resultant wave depends on the path difference between the two waves
• This extra distance is the path difference

The path difference between two waves is determined by the number of wavelengths that cover their difference in length

• For constructive interference (or maxima), the difference in wavelengths will be an integer number of whole wavelengths
• For destructive interference (or minima) it will be an integer number of whole wavelengths plus a half wavelength
• There is usually more than one produced
• n is the order of the maxima or minima; which represents the position of the maxima away from the central maximum
• n  = 0 is the central maximum
• n  = 1 represents the first maximum on either side of the central,  n  = 2 the next one along....

Worked example

Determine which orders of maxima are detected at M as the wavelength is increased from 3.5 cm to 12.5 cm.

Examiner Tip

The path difference is more specifically how much longer, or shorter, one path is than the other. In other words, the difference in the distances. Make sure not to confuse this with the distance between the two paths.

Fringe Spacing Equation

• The spacing between the bright or dark fringes in the diffraction pattern formed on the screen can be calculated using the double slit equation:

Double slit interference equation with w, s and D represented on a diagram

• D  is much bigger than any other dimension, normally several metres long
• s is the separation between the two slits and is often the smallest dimension, normally in mm
• w  is the distance between the fringes on the screen, often in cm. This can be obtained by measuring the distance between the centre of each consecutive bright spot.
• The wavelength ,  λ of the incident light increases
• The distance , s  between the screen and the slits increases
• The separation , w between the slits decreases

Calculate the separation of the two slits.

Since w , s and D are all distances, it's easy to mix up which they refer to. Labelling the double-slit diagram with each of these quantities can help ensure you don't use the wrong variable for a quantity.

Interference Patterns

• It is different to that produced by a single slit or a diffraction grating

The interference pattern produced when white light is diffracted through a double slit

• Each maximum is of roughly equal width
• There are two dark narrow destructive interference fringes on either side
• All other maxima are composed of a   spectrum
• The shortest wavelength (violet / blue) would appear   nearest   to the central maximum because it is diffracted the least
• The longest wavelength (red) would appear   furthest   from the central maximum because it is diffracted the most
• As the maxima move   further away   from the central maximum, the wavelengths of   blue   observed   decrease   and the wavelengths of   red   observed   increase

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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 ).

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.

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?

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

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

Example 1: Finding a Wavelength from an Interference Pattern

Substituting known values yields

Example 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.

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.

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.

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.

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

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?

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 ).

9: 1200 nm (not visible)

11: (a) 760 nm

(b) 1520 nm

For two adjacent fringes we have,

Subtracting these equations gives

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Chapter 12 – Wave Optics

12.3 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 colour, 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 ).

Why do we not ordinarily observe wave behaviour 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 colour, single wavelength,   λ ) light to clarify the effect. Figure 2 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

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.

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 ( 1/2λ, 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 (1 λ, 2λ, 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 vapour lamp, than for an incandescent bulb?

Figure 5 below 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 dsinθ, 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  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 centre. 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   and m , the smaller  d is, the larger  θ must be, since dsinθ= mλ.

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.

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 10.95 o 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 = 10.95 o . The wavelength can thus be found using the equation d sinθ = m λ for constructive interference.

The equation is  d sinθ = m λ.  Solving for the wavelength  λ  gives

Substituting known values yields

λ = (d sinθ) / m  = (0.0100 x 10 -3 m) (sin 10.95 o )  /  3 = 6.33 x 10 -7 m  or 

λ  =  633 nm

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red colour 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 sinθ = mλ  for  m = 0,  1, -1, 2, -2, 3, -3, 4, -4, …  describes constructive interference. For fixed values of  d and λ , the larger  m is, the larger  sinθ is. However, the maximum value that sinθ can have is 1, for an angle of 90 o . (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 sinθ = mλ   for  m gives

Taking sinθ =sin90 o = 1 and substituting the values of  d and  λ, and remembering the values for the metric prefixes, 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 centre. 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 sinθ = mλ  for  m = 0, 1, -1, 2, -2, 3, -3, 4, -4, ….(constructive) , 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 sinθ =  ( m + 1 / 2 )λ   for  m = 0, 1, -1, 2, -2, 3, -3, 4, -4, ….(constructive)

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 colour 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 centre. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

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 30.0  degrees?
• Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of 45.0  degrees.
•  Calculate the wavelength of light that has its third minimum at an angle of 30.0 o when falling on double slits separated by 3.00 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 2.00 μm if the third-order maximum is at an angle of 60.0 o ?
• At what angle is the fourth-order maximum for the situation in Problems & Exercises 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?
• Figure 8 shows a double slit located a distance x from a screen, with the distance from the centre 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 sinθ≈tanθ, with θ in radians), the distance between fringes is given by Δy = x λ /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:  0.516 o

3:  1.22 x10 -6 m

4: 0.290 μm

9: 1200 nm (not visible)

10 :  1.44 μm

11: (a) 760 nm (b) 1520 nm

13: For small angles sinθ ≈ tanθ≈θ when the angle is in radians

For two adjacent fringes we have d sinθ m = m λ

and d sinθ m +1 =  (m+1)  λ

Subtracting these equations gives

14: 2.37 cm

Douglas College Physics 1207 Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.