wave optics case study questions

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Case Study on Wave Optics Class 12 Physics PDF

Better preparation of Case Study on Wave Optics Class 12 Physics can help students score good marks in the CBSE Class 12 Board examination. Additionally, it helps build confidence and enables students to deepen their knowledge of Wave Optics. Because case-based questions are equally important for learning and board exam preparation, our team has prepared Case Study on Wave Optics Class 12 Physics in a PDF file for free distribution among students.

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Class 12th Physics - Wave Optics Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 12th Physics Subject - Wave Optics, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Wave optics case study questions with answer key.

12th Standard CBSE

Final Semester - June 2015

(ii) If the width w of one of the slits is increased to 2w, the become the amplitude due to slit

(iii) In YDSE, let A and B be two slits. Films of thicknesses t A and t B and refractive indices m A and m B are placed in front of A and B, respectively. If \(\mu_{\mathrm{A}} t_{A}=\mu_{B} t_{B}\) then the central maxima will

(iv) In Young's double slit experiment, a third slit is made in between the double slits. Then

(v) In Young's double slit experiment, if one of the slits is covered with a microscope cover slip, then

(ii) In Young's double - slit experiment if yellow light is replaced by blue light, the interference fringes become

(iii) In Young's double slit experiment, if the separation between the slits is halved and the distance between the slits and the screen is doubled, then the fringe width compared to the unchanged one will be

(iv) When the complete Young's double slit experiment is immersed in water, the fringes

(v) In a two slit experiment with white light, a white fringe is observed on a screen kept behind the slits. When the screen is moved away by 0.05 m, this white fringe

(ii) In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2 m away from the lens. If wavelength of light used is 5000 \(\lambda\) then the distance between the first minimum on either side the central maximum is

(iii) Light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum)

(iv) A diffraction pattern is obtained by using a beam of red light. What will happen, if the red light is replaced by the blue light?

(v) To observe diffraction, the size of the obstacle

(ii) Find the radius of the n th bright fringe.

(iii) If d = 0.5 mm, \(\lambda\) = 5000 \(\dot A \) and D = 100 em, find the value of n for the closest second bright fringe

(iv) The coherence of two light sources means that the light waves emitted have

(v) The phenomenon of interference is shown by

(ii) If the intensity ratio of two coherent sources used in Young's double slit experiment is 49 : 1, then the ratio between the maximum and minimum intensities in the interference pattern is

(iii) The maximum intensity produced at D is given by

(iv) In a Young's double slit experiment, the intensity at a point where the path difference is \(\lambda\) /6 ( \(\lambda\) - wavelength of the light) is I. If I 0 denotes the maximum intensity, then I/I 0 is equal to

(v) Two identical light waves, propagating in the same direction, have a phase difference d. After they superpose the intensity of the resulting wave will be proportional to

(ii) Find the ratio of the maximum to minimum intensity observed on \(S_{c} \text { if } z=\frac{\lambda D}{d}\)

(iv) In the case of light waves from two coherent sources S 1 and S 2 . there will be constructive interference at an arbitrary point P, if the path difference S 1 P - S 2 P is

(v) Two monochromatic light waves of amplitudes 3A and 2A interfering at a point have a phase difference of 60°. The intensity at that point will be proportional to

(ii) The phases of the light wave at c, d, e and f are \(\phi_{c}, \phi_{d}, \phi_{e}\) and \( \phi_{f}\) respectively. It is given that \(\phi_{c} \neq \phi_{f}\)

(iii) Wavefront is the locus of all points, where the particles of the medium vibrate with the same

(iv) A point source that emits waves uniformly in all directions, produces wavefronts that are

(v) What are the types of wavefronts ?

(ii) According to Huygens Principle, the surface of constant phase is

(iii) As the beam enters the medium, it will

(iv) Two plane wavefronts of ligbt, one incident on a thin convex lens and another on the refracting face of a thin prism. After refraction at them, the emerging wavefronts respectively become

(v) Which of the following phenomena support the wave theory of light? 1. Scattering 2. Interference 3. Diffraction 4. Velocity of light in a denser medium is less than the velocity of light in the rarer medium

(ii) In diffraction from a single slit the angular width of the central maxima does not depends on

(iii) For a diffraction from a single slit, the intensity of the central point is

(iv) .Resolving power of telescope increases when

(v) In a single diffraction pattern observed on a screen placed at D metre distance from the slit of width d metre, the ratio of the width of the ce,ntral maxima to the width of other secondary maxima is

(ii) Which of the following does not show interference?

(iii) In a Young's double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to

(iv) The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double-slit experiment, is

(v) The resultant amplitude of a vibrating particle by the superposition of the two waves \(y_{1}=a \sin \left[\omega t+\frac{\pi}{3}\right] \text { and } y_{2}=a \sin \omega t \text { is }\)

Huygens Wave Theory of Light 1. According to wave theory, light from a source is propagated in the form of longitudinal waves with uniform velocity in a homogeneous medium. 2. To explain the propagation of waves through vacuum, Huygens assumed existance of a hypothetical medium called luminiferous ether. According to Huygens, ether particles are present and possess properties such as inertia, zero density and perfect transparency. 3. On the basis of Huygens wave theory, various colours of light are due to different wavelengths of the light of the waves. (i) Write two merits and two demerits of Huygens wave theory of light. (ii) Write Huygen's postulates to explain wave theory of light. (iii) What are primary source and secondary source of light considered in wave theory?

While comparing the interference pattern with that seen for a coherently illuminated single slit (usually called single slit diffraction pattern), the interference pattern has a number of equally spaced bright and dark bands. The diffraction pattern has a central bright maximum which is twice as wide as the other maxima. The intensity falls as we go to successive maxima away from the centre, on either side. Now answer the following questions: (i) Why interference fringes are equispaced whereas that of diffraction are not? (ii) Why intensity falls as we go to successive maxima away from central bright maximum? (iii) For a single slit of width a, the first null point of the interference pattern occurs at an angle of \(\lambda\) /a. At the same angle of \(\lambda\) /a, we get a maximum (not a null) for two narrow slits separated by a distance a. Explain why?

"According to Richard Feynman, no one has ever been able to distinguish between interference and diffraction satisfactorily. Roughly speaking if the number of sources are few, say two interfering sources then the result is usually interference, but if there is a large number of them, it seems that the word diffraction is more often used". Now explain the following: (i) "In the double-slit experiment, the pattern on the screen is due to superposition of single slit diffraction from each slit." Justify this statement. (il) In a double slit experiment, the wavelength \(\lambda\) of the light source is 400 nm, the slit separation is 20 \(\mu\) m, and the slit width a is 4 \(\mu\) m. Consider the interference of the light from the two slits and also diffraction of the light through each slit (a) Determine how many bright fringes are within the central peak of the diffraction envelope? (b) How many bright fringes are within either of the first side peak of the diffraction envelope?

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Wave optics case study questions with answer key answer keys.

(i) (c) : In Young's double slit experiment, the fringe width is \(\beta=\frac{D \lambda}{d}\) where Dis the distance of the slits from the screen, d is the separation of the slits and \(\lambda\) , the wavelength. Therefore the fringe width \(\beta\) can be changed either by changing the separation between the sources or the distance of the screen from the sources. (ii) (c) : As the width of one of the slits is increased to 2w, the amplitude due to slit become 2a. (iii) (d) : \(\Delta x=\left(\mu_{A}-1\right) t_{A}-\left(\mu_{B}-1\right) t_{B}\) \(=\mu_{A} t_{A}-\mu_{B} t_{B}-t_{A}+t_{B}=t_{B}-t_{A}\) If \(\Delta x\) > 0, then fringe pattern will shift upward. If \(\Delta x\) < 0, then fringe pattern will shift downwards. (iv) (b) : Contrast between the bright and dark fringes will be reduced. (v) (a) : Since, one of the slit is covered, interference will not occur and fringe pattern will disappear.

(i) (b) : The condition for possible interference maxima on the screen is, dsin \(\theta\) = nA where d is slit separation and Ais the wavelength. As d = 2 \(\lambda\) (given) \(\therefore\) 2 \(\lambda\) sin \(\theta\) = n \(\lambda\) or 2sin \(\theta\) = n For number of interference maxima to be maximum, sin \(\theta\) = 1 \(\therefore\) n = 2 The intprference maxima will be forgied when n = 0, ± 1, ± 2 Hence the maximum number of possible maxima is 5. (ii) (c) : Fringe width, \(\beta=\frac{\lambda D}{d}\) \(\therefore\) If we replace yellow light with blue light, i.e., longer wavelength with shorter one, therefore the fringe width decreases. (iii) (d) : \(d^{\prime}=\frac{d}{2} \text { and } D^{\prime}=2 D\) Fringe width, \(\beta=\frac{\lambda D}{d}\) New fringe width \(\beta^{\prime}=\lambda\left(\frac{2 D}{d / 2}\right)=4 \beta\) (iv) (c) : When Young's double slit experiment is repeated in water, instead of air \(\lambda^{\prime}=\frac{\lambda}{\mu}\) i.e., wavelength decreases. \(\beta=\frac{\lambda^{\prime} D}{d}\) i.e..,fringe width decreases. \(\therefore\) The fringe become narrower. (v) (a) : Using white light, we get white fringe at the centre i.e., white fringe is the central maximum. When the screen is moved, its position is not changed.

(i) (c) : Here \(d=0.1 \mathrm{~mm}, \lambda=6000 \dot A, D=0.5 \mathrm{~m}\) For third dark band \(d \sin \theta=3 \lambda ; \sin \theta=\frac{3 \lambda}{d}=\frac{y}{D}\) \(y=\frac{3 D \lambda}{d}=\frac{3 \times 0.5 \times 6 \times 10^{-7}}{0.1 \times 10^{-3}}=9 \times 10^{-3} \mathrm{~m}=9 \mathrm{mn}\) (ii) (b) : Given \(d=0.2 \mathrm{~mm}=0.2 \times 10^{-3} \mathrm{~m}, D=2 \mathrm{~m}\) \(\lambda=5000 \dot A=5 \times 10^{-7} \mathrm{~m}\) The distance between the first minimum on other side of the central maximum \(x=\frac{2 \lambda D}{d}=\frac{2 \times 5 \times 10^{-7} \times 2}{0.2 \times 10^{-3}} \Rightarrow x=10^{-2} \mathrm{~m}\) (iii) (a) : Here \(\lambda=600 \mathrm{nm}=6 \times 10^{-7} \mathrm{~m}\) \(a=0.2 \mathrm{~mm}=2 \times 10^{-4} \mathrm{~m}, \theta=?\) Angular width of central maxima \(\theta=\frac{2 \lambda}{a}=\frac{2 \times 6 \times 10^{-7}}{2 \times 10^{-4}}=6 \times 10^{-3} \mathrm{rad}\) (iv) (d) : When red light is replaced by blue light ( \(\lambda\) B < \(\lambda\) R) the diffraction pattern bands becomes narrow and crowded together. (v) (b) : To observe diffraction, the size of the obstacle should be of the order of wavelength.

(i) (c) :The optical path difference at P is \(\Delta x=S_{1} P-S_{2} P=d \cos \theta\) \(\because \quad \cos \theta=1-\frac{\theta^{2}}{2} \text { for small } \theta\) \(\therefore \quad \Delta x=d\left(1-\frac{\theta^{2}}{2}\right)=d\left[1-\frac{y^{2}}{2 D^{2}}\right], \text { where } D+d=D\) (ii) (b) : \(\text { For } n^{\text {th }} \text { maxima, }\) \(\Rightarrow \Delta x=n \lambda\) \(d\left[1-\frac{y^{2}}{2 D^{2}}\right]=n \lambda\) Y = radius of the n th bright ring \(=D \sqrt{2\left(1-\frac{n \lambda}{d}\right)}\) (iii) (d) : At the central maxima, \(\theta\) = 0. \(\Delta x=d=n \lambda \) \(\Rightarrow n=\frac{d}{\lambda}=\frac{0.5}{0.5 \times 10^{-3}}=1000\) Hence, for the closet second bright fringe, n = 998. (iv) (c) : Light waves from two coherent sources must have a constant phase difference. (v) (d) : Interference is shown by transverse as well as mechanical waves.

(i) (d) : Path difference produced is \(\Delta x=\frac{3}{2} \pi R-\frac{\pi}{2} R=\pi R\) For maxima \(\Delta x=n \lambda\) \(\begin{array}{l} \therefore \quad n \lambda=\pi R \\ \Rightarrow \lambda=\frac{\pi R}{n}, n=1,2,3, \ldots . \end{array}\) Thus, the possible values of \(\lambda \text { are } \pi R, \frac{\pi R}{2}, \frac{\pi R}{3}, \ldots\) (ii) (d) (iii) (b) : Maximum intensity \(I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}\) \(\text { Here, } I_{1}=I_{2}=\frac{I_{0}}{2} \text { (given) }\) \(\therefore \quad I_{\max }=\left(\sqrt{\frac{I_{0}}{2}}+\sqrt{\frac{I_{0}}{2}}\right)^{2}=2 I_{0}\) (iv) (d) : Phase difference \(\phi=\frac{2 \pi}{\lambda}\) x path diffrerence \(\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{\pi}{3}=60^{\circ} \text { As } I=I_{\max } \cos ^{2} \frac{\phi}{2}\) \(\therefore \quad I=I_{0} \cos ^{2} \frac{60^{\circ}}{2}=I_{0} \times\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{3}{4} I_{0} \Rightarrow \frac{I}{I_{0}}=\frac{3}{4}\) (v) (c) : HereA 2 = \(a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \delta \because a_{1}=a_{2}=a\) \(\therefore A^{2}=2 a^{2}(1+\cos \delta)=2 a^{2}\left(1+2 \cos ^{2} \frac{\delta}{2}-1\right)\) \(\text { or } A^{2} \propto \cos ^{2} \frac{\delta}{2}\) \(\text { Now } I \propto A^{2} \quad \therefore I \propto A^{2} \propto \cos ^{2} \frac{\delta}{2} \Rightarrow I \propto \cos ^{2} \frac{\delta}{2}\)

(i) (b) : \(\text { As } z=\frac{\lambda D}{2 d}\) \(\text { At } S_{4}: \frac{\Delta x}{d}=\frac{z}{D}\) \(\Rightarrow \Delta x=\frac{\lambda D}{2 d} \frac{d}{d}=\frac{\lambda}{2}\) (ii) (c) : \(z=\frac{\lambda D}{d}\) \(\Delta x \text { at } S_{4}: \Delta x=\frac{\lambda D}{d} \frac{d}{d}=\lambda\) Hence, maxima at S 4 as well as S 3 ' Resultant intensity at S 4 I = 4I 0 \(\therefore \quad \frac{I_{\max }}{I_{\min }}=\frac{\left[\left(4 I_{0}\right)^{1 / 2}+4\left(4 I_{0}\right)^{1 / 2}\right]^{2}}{\left[\left(4 I_{0}\right)^{1 / 2}-\left(4 I_{0}\right)^{1 / 2}\right]^{2}}=\infty\) (iii) (a) : When the screen is placed perpendicular to the line joining the 'sources, the fringes will be concentric circles. (iv) (b) : Constructive interference occurs when the path difference (S 1 P - S 2 P) is an integral multiple of \(\lambda\) .or S 1 P - S 2 P = n \(\lambda\) , where n = 0,1,2,3, ..... (v) (d) : Here, A 1 = 3A, A 2 = 2A and \(\varphi\) = 60° The resultant amplitude at a point is \(R =\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi} \) \(=\sqrt{(3 A)^{2}+(2 A)^{2}+2 \times 3 A \times 2 A \times \cos 60^{\circ}} \) \(=\sqrt{9 A^{2}+4 A^{2}+6 A^{2}}=A \sqrt{19}\) As, Intensity \(\infty\) (Amplituder) 2 Therefore, intensity at the same point is \(I \propto 19 A^{2}\)

(i) (a) : Since the path difference between two waveform is equal, light traves as parallel beam in each medium. (ii) (c): Since all points on the wavefront are in the same phase, \(\phi_{d}=\phi_{c} \text { and } \phi_{f}=\phi_{e} \) \(\therefore \phi_{d}-\phi_{f}=\phi_{c}-\phi_{e^{-}}\) (iii) (a) : Wavefront is the locus of all points, where the particles of the medium vibrate with the same phase (iv) (a) (v) (d)

(i) (c) : Given \(I_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \text { and } I_{2}=25 \mathrm{~W} / \mathrm{m}^{2}\) \(\frac{I_{1}}{I_{2}}=\frac{a_{1}^{2}}{a_{2}^{2}}=\frac{10}{25} \Rightarrow \frac{a_{1}}{a_{2}}=\frac{3.16}{5} \text { or } a_{1}=\frac{3.16}{5} a_{2}=0.6324 a_{2}\) \(\frac{I_{\max }}{I_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}=\frac{\left[0.6324 a_{2}+a_{2}\right]^{2}}{\left[0.6324 a_{2}-a_{2}\right]^{2}}=19.724\) (ii) (b) : In an excessively thin film, the thickness of the film is negligible. Thus the path difference between the reflected rays becomes \(\lambda\) /2 which produces a minima. (iii) (a) : Since , \(\beta=\frac{\lambda D}{d} \text { for } d=2 d\) \(\beta^{\prime}=\frac{\lambda D^{\prime}}{2 d}=\beta(\text { Gives })\) \(\therefore \quad D_{1}=2 D\) (iv) (b) : The condition for possible interference maxima on the screen is, dsin \(\theta\) = n \(\lambda\) where d is slit separation and Ais the wavelength. \(\text { As } d=2 \lambda \text { (given) } \quad \therefore 2 \lambda \sin \theta=n \lambda \text { or } 2 \sin \theta=n\) For number of interference maxima to be maximum,sin \(\theta\) = 1 :. n = 2 The interference maxima will be formed when n = 0, ± 1, ± 2 Hence the maximum number of possible maxima is 5. (v) (d) : \(y_{1}=a \sin \left(\omega t+\frac{\pi}{3}\right) \text { and } y_{2}=a \sin \omega t\) \(A=\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi}, \text { where } \phi=\frac{\pi}{3}\) \(=\sqrt{a^{2}+a^{2}+2 a a \cos \frac{\pi}{3}}=\sqrt{3} a\)

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