Wave Optics

1. 1Refraction and Reflection of Plane Waves using Huygens Principle16 board questions
2. 2Coherent and Incoherent Addition of Waves1 board question
3. 3Interference of Light Waves and Young's Experiment23 board questions
4. 4Diffraction24 board questions

These are being added chapter by chapter. They are solved numericals from the prescribed book and the reference guides — not previous board questions, which live under Practice — so that you can work through the method before meeting it in a real paper.

1. 1

   Derivation 1

   In interference of light, write the expression for the intensity of the resultant wave if $I_0$ is the intensity of the light wave from each slit. Hence obtain an expression for the intensity of the resultant wave if the two sources are (i) incoherent, and (ii) coherent.

   Asked in: CBSE Class 12 Physics — SA · 2021C · Section B, 2 marks

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   Resultant intensity when two waves of intensities $I_1,I_2$ (from each slit) superpose with a phase difference $\phi$: $\;I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\phi$ (i) For incoherent sources, the phase difference varies randomly and rapidly, so the interference (cosine) term averages to zero over time: $\;I_R=I_1+I_2$. With $I_1=I_2=I_0$: $\;I_R=2I_0$ (ii) For coherent sources, $\phi$ is constant, so the cosine term persists: $\;I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\phi=2I_0+2I_0\cos\phi=2I_0(1+\cos\phi)$ (with $I_1=I_2=I_0$)

2. 2

   Derivation 2

   Briefly explain how bright and dark fringes are formed on the screen in Young's double slit experiment. Hence derive the expression for the fringe width.

   Asked in: CBSE Class 12 Physics — SA · Term II 2021-22 · Section C, 3 marks

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   A narrow slit $S$, illuminated by a monochromatic source, sends out a cylindrical wavefront that falls symmetrically on two fine, closely-spaced slits $S_1$ and $S_2$ (separation $d$). These act as coherent secondary sources, sending out overlapping spherical wavefronts. At any point $P$ on a screen at distance $D$ from the slits, the path difference between the waves from $S_1$ and $S_2$ is $p=S_2P-S_1P$. Using the right triangles formed with the screen (and $D\gg d$, so $S_1P\approx S_2P\approx D$): $\;p\approx\dfrac{xd}D$, where $x$ is the distance of $P$ from the centre $O$. Where the path difference is an integral multiple of $\lambda$ (i.e. $p=n\lambda$), the waves arrive in phase and reinforce — producing a BRIGHT fringe. Where $p=(n+\tfrac12)\lambda$, they arrive out of phase and cancel — producing a DARK fringe. Position of the $n$th bright fringe: $\;x_n=\dfrac{n\lambda D}d$; position of the $n$th dark fringe: $\;x_n'=\dfrac{(2n-1)\lambda D}{2d}$ Fringe width (separation between two consecutive bright, or two consecutive dark, fringes): $\;\beta=x_{n+1}-x_n=\dfrac{(n+1)\lambda D}d-\dfrac{n\lambda D}d=\dfrac{\lambda D}d$

3. 3

   Derivation 3

   (a) Two monochromatic waves emanating from two coherent sources have the displacements represented by $y_1=a\cos\omega t$ and $y_2=a\cos(\omega t+\phi)$, where $\phi$ is the phase difference between the two displacements. Show that the resultant intensity at a point due to their superposition is given by $I=4I_0\cos^2(\phi/2)$, where $I_0=a^2$. (b) Hence obtain the conditions for constructive and destructive interference.

   Asked in: CBSE Class 12 Physics — SA · AI 2014C, An · Section E, 5 marks

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   (a) Resultant displacement: $\;y=y_1+y_2=a\cos\omega t+a\cos(\omega t+\phi)=a[\cos\omega t+\cos(\omega t+\phi)]$ Using the sum-to-product identity: $\;y=2a\cos\left(\dfrac\phi2\right)\cos\left(\omega t+\dfrac\phi2\right)$ Let $A=2a\cos(\phi/2)$ be the amplitude of the resultant wave, so $y=A\cos(\omega t+\phi/2)$. $\;A^2=4a^2\cos^2(\phi/2)$; since intensity $\propto$ (amplitude)$^2$, and with $I_0=a^2$ (intensity due to each source alone): $\;I=4I_0\cos^2\left(\dfrac\phi2\right)$ (b) Constructive interference: $I$ is maximum when $\cos^2(\phi/2)=1$, i.e. $\phi=0,2\pi,4\pi,\dots=2n\pi$, equivalently the path difference $\Delta x=n\lambda$ ($n=0,1,2,\dots$). Destructive interference: $I=0$ when $\cos^2(\phi/2)=0$, i.e. $\phi=\pi,3\pi,5\pi,\dots=(2n-1)\pi$, equivalently the path difference $\Delta x=(2n-1)\dfrac\lambda2$ ($n=1,2,3,\dots$).

4. 4

   Derivation 4

   In Young's double slit experiment, deduce the conditions for (a) constructive, and (b) destructive interference at a point on the screen. Draw a graph showing the variation of intensity in the interference pattern against position '$x$' on the screen.

   Asked in: CBSE Class 12 Physics — Long Answer · 4/5, Delhi 2016, Ev · Section E, 5 marks

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   Same derivation as Q37 above: with $y_1=a\cos\omega t$, $y_2=a\cos(\omega t+\phi)$ from the two coherent slits, resultant intensity $\;I=4I_0\cos^2(\phi/2)$, giving: (a) Constructive interference: $\;\cos(\Delta\phi)=+1\;\Rightarrow\;\Delta x=n\lambda$ ($n=0,1,2,\dots$), giving maximum intensity $I_{max}=4I_0$. (b) Destructive interference: $\;\cos(\Delta\phi)=-1\;\Rightarrow\;\Delta x=(2n-1)\dfrac\lambda2$ ($n=1,2,3,\dots$), giving zero intensity. Graph: plotting intensity $I$ against position $x$ on the screen gives a series of equally-spaced peaks of height $4I_0$ (bright fringes) separated by zero-intensity troughs (dark fringes), with the pattern symmetric about the central bright fringe at $x=0$.

5. 5

   Derivation 5

   (a) Consider two coherent sources $S_1$ and $S_2$ producing monochromatic waves to produce an interference pattern. Let the displacement of the wave produced by $S_1$ be given by $y_1=a\cos\omega t$ and the displacement by $S_2$ be $y_2=a\cos(\omega t+\phi)$. Find out the expression for the amplitude of the resultant displacement at a point, and show that the intensity at that point will be $I=4a^2\cos^2(\phi/2)$. Hence establish the conditions for constructive and destructive interference. (b) What is the effect on the interference fringes in Young's double slit experiment when (i) the width of the source slit is increased; (ii) the monochromatic source is replaced by a source of white light?

   Asked in: CBSE Class 12 Physics — Long Answer · Section E, 5 marks

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   (a) Same derivation as Q37 above: resultant amplitude $A=2a\cos(\phi/2)$, giving $I=4a^2\cos^2(\phi/2)$; constructive interference at $\Delta x=n\lambda$, destructive interference at $\Delta x=(2n-1)\lambda/2$. (b)(i) Increasing the width of the source slit reduces the spatial coherence of the light reaching $S_1,S_2$ — beyond a certain width, the fringes start to get blurred and the contrast between bright and dark fringes decreases; eventually the fringe pattern disappears if the slit is made too wide. (ii) With white light, the central fringe remains white, while a few coloured fringes (violet nearest the centre, red farthest) appear on either side before the pattern washes out into general illumination — the same effect as described in Q33(b).

6. 6

   Derivation 6

   (a) (i) Two independent monochromatic sources of light cannot produce a sustained interference pattern. Give reason. (ii) Light waves, each of amplitude '$a$' and frequency '$\omega$', emanating from two coherent light sources, superpose at a point. If the displacements due to these waves is given by $y_1=a\cos\omega t$ and $y_2=a\cos(\omega t+\phi)$, where $\phi$ is the phase difference between the two, obtain the expression for the resultant intensity at the point. (b) In Young's double slit experiment, using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where the path difference is $\lambda$ is $K$ units. Find out the intensity of light at a point where the path difference is $\lambda/3$.

   Asked in: CBSE Class 12 Physics — Long Answer · Section E, 5 marks

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   (a)(i) Same reasoning as Q35 above: the phase difference between two independent sources keeps changing continuously (randomly), so no sustained interference pattern is possible. (ii) Same derivation as Q37 above: $\;I=4I_0\cos^2(\phi/2)$, where $I_0=a^2$. (b) At path difference $\lambda$: phase difference $\;\phi=\dfrac{2\pi}\lambda\times\lambda=2\pi\;\Rightarrow\;K=4I_0\cos^2(\pi)=4I_0\qquad\text{...(i)}$ At path difference $\lambda/3$: phase difference $\;\phi'=\dfrac{2\pi}\lambda\times\dfrac\lambda3=\dfrac{2\pi}3$ Intensity: $\;I'=4I_0\cos^2\left(\dfrac\pi3\right)=4I_0\times\left(\dfrac12\right)^2=I_0=\dfrac K4\;$ (using (i), $I_0=K/4$)

7. 7

   Derivation 7

   How can you differentiate whether a pattern is produced by a single slit or by double slits? Derive the expression for the angular position of (i) bright, and (ii) dark fringes produced in single slit diffraction.

   Asked in: CBSE Class 12 Physics — Long Answer · Term II 2021-22 · Section E, 5 marks

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   A single-slit diffraction pattern has a tall central maximum (twice the width of the other maxima) with much weaker, rapidly-decreasing secondary maxima on either side, and NON-uniform-width fringes overall; a double-slit interference pattern instead shows a series of EQUALLY bright, equally-spaced fringes — this contrast in appearance distinguishes the two. Consider a plane wavefront incident normally on a slit $AB$ of width $a$. Every point of the unblocked wavefront sends out secondary wavelets (Huygens' principle). Consider the wavelets travelling at angle $\theta$ to the normal, converging (via a lens, or at a large distance) to a point $P$ on the screen. The wavelets from $A$ and $B$ have a path difference $BN=a\sin\theta$ (where $N$ is the foot of the perpendicular from $A$ onto the ray from $B$). Dark fringes (secondary minima): if $BN=\lambda$, the slit can be divided into two halves, each pair of corresponding points having path difference $\lambda/2$, causing complete destructive interference. In general, for the $n$th minimum: $\;a\sin\theta_n=n\lambda\;\Rightarrow\;\sin\theta_n=\dfrac{n\lambda}a$, and for small $\theta_n$: $\;\theta_n=\dfrac{n\lambda}a$ ($n=\pm1,\pm2,\dots$) Bright fringes (secondary maxima): if $BN=(2n+1)\dfrac\lambda2$, the slit can be divided into $(2n+1)$ equal parts; all but one part cancel out in pairs (each pair contributing a path difference of $\lambda/2$), leaving one part's contribution unbalanced, producing a (weak) secondary maximum. In general: $\;a\sin\theta_n=(2n+1)\dfrac\lambda2\;\Rightarrow\;\theta_n=(2n+1)\dfrac\lambda{2a}$ (for small $\theta_n$), $n=\pm1,\pm2,\dots$

8. 8

   Derivation 8

   (a) The intensity pattern on the screen is shown in the given figure: intensity peaks at angular positions $0,\pm\lambda/a,\pm2\lambda/a,\pm3\lambda/a,\pm4\lambda/a$, with a tall central maximum at $\theta=0$ and much smaller secondary maxima at $\theta=\pm\lambda/a,\pm2\lambda/a,\dots$, decreasing in height further from the centre. Show that the width of the central maximum is $\;\dfrac{2D\lambda}a$. (b) What is the effect on the angular width of the central maximum, and its intensity, when the slit width $a$ is increased? Justify.

   Asked in: CBSE Class 12 Physics — Long Answer · Section E, 5 marks

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   (a) The first minima on either side of the central maximum occur at $\;\sin\theta=\pm\dfrac\lambda a$; the angular width of the central maximum (between the two first minima) is thus $\;2\theta_0=\dfrac{2\lambda}a$ (for small angles). The corresponding linear width on a screen at distance $D$ is $\;\beta_0=D\times2\theta_0=\dfrac{2D\lambda}a$ (b)(i) Since $\beta_0\propto1/a$, increasing the slit width $a$ DECREASES the angular (and linear) width of the central maximum. (ii) Since more light energy passes through a wider slit, and this energy is now concentrated into a narrower central band, the intensity of the central maximum INCREASES as the slit width is increased.

9. 9

   Derivation 9

   Describe any two characteristic features which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young's double slit experiment.

   Asked in: CBSE Class 12 Physics — Long Answer · Section E, 5 marks

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   Two differences: (1) in interference, all bright fringes are of the same intensity, whereas in diffraction, successive bright fringes decrease sharply in intensity away from the central maximum; (2) in interference, all bright and dark fringes are of equal width, whereas in diffraction, the central bright fringe is twice as wide as the other (secondary) maxima. Derivation of intensity: let the waves from the two coherent sources be $\;y_1=a\sin\omega t\;$ and $\;y_2=b\sin(\omega t+\phi)$, where $a,b$ are their amplitudes and $\phi$ the constant phase difference. By superposition, resultant displacement $\;y=y_1+y_2=a\sin\omega t+b\sin(\omega t+\phi)=a\sin\omega t+b\sin\omega t\cos\phi+b\cos\omega t\sin\phi$ $\;y=\sin\omega t\,(a+b\cos\phi)+\cos\omega t\,(b\sin\phi)$ Let $\;a+b\cos\phi=R\cos\theta\;$...(i) and $\;b\sin\phi=R\sin\theta$...(ii), so that $\;y=R\sin\omega t\cos\theta+R\cos\omega t\sin\theta=R\sin(\omega t+\theta)$ — a harmonic wave of amplitude $R$. Squaring and adding (i) and (ii): $\;R^2(\cos^2\theta+\sin^2\theta)=(a+b\cos\phi)^2+(b\sin\phi)^2=a^2+b^2\cos^2\phi+2ab\cos\phi+b^2\sin^2\phi$ $\;R^2=a^2+b^2(\cos^2\phi+\sin^2\phi)+2ab\cos\phi=a^2+b^2+2ab\cos\phi\;\Rightarrow\;R=\sqrt{a^2+b^2+2ab\cos\phi}$ Since intensity $\propto$ (amplitude)$^2$: $\;I_1=Ka^2,\;I_2=Kb^2,\;I_R=KR^2=K(a^2+b^2+2ab\cos\phi)$ $\therefore\;I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\phi$

10. 10

    Derivation 10

    (a) Using Huygens' construction of secondary wavelets, explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a monochromatic beam of light is incident normally. (b) Show that the angular width of the first diffraction fringe is half that of the central fringe. (c) Explain why the maxima at $\theta=\left(n+\dfrac12\right)\dfrac\lambda a$ become weaker and weaker with increasing $n$.

    Asked in: CBSE Class 12 Physics — Long Answer · Delhi 2015 · Section E, 5 marks

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    (a) A wavefront striking a barrier with a narrow slit sends secondary wavelets, per Huygens' principle, from every unblocked point of the wavefront. Away from the slit, most of the wavefront is blocked, but the points at (and within) the slit continue emitting forward-moving wavelets that bend ('diffract') around the edges of the slit rather than continuing in a straight-edged beam. The superposition of all these wavelets, arriving at a point on a distant screen with various path differences depending on the diffraction angle $\theta$, builds up the characteristic diffraction pattern: an intense central maximum together with much weaker secondary maxima, separated by dark fringes. (b) The central maximum spans from $\theta=-\lambda/a$ to $\theta=+\lambda/a$, i.e. angular width $=2\lambda/a$. The first secondary (diffraction) fringe lies between the first minimum ($\theta=\lambda/a$) and the second minimum ($\theta=2\lambda/a$), giving it an angular width of $\;\dfrac{2\lambda}a-\dfrac\lambda a=\dfrac\lambda a$. $\therefore\;$ angular width of first diffraction fringe $\left(\dfrac\lambda a\right)=\dfrac12\times$ angular width of central fringe $\left(\dfrac{2\lambda}a\right)$ — i.e. exactly HALF. (c) Same reasoning as in Q49 above: at $\theta=(n+\tfrac12)\lambda/a$, the slit divides into $(2n+1)$ equal parts, of which all but one pair cancel; the uncancelled contributing fraction of the slit shrinks as $n$ increases, so the intensity of successive maxima keeps falling.