Interference of Light Waves and Young's Experiment

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   Fringe width $\;\beta=\dfrac{\lambda D}d\;\Rightarrow\;\beta\propto\lambda$. Among red, green and blue, red light has the longest wavelength. Hence the fringe width will be maximum for RED light.

2. 2.↑ back to the question

   For a dark fringe, path difference $=(n+\tfrac12)\lambda$ (with $n=0,1,2,\dots$ giving the 1st, 2nd, 3rd$\dots$ dark fringes respectively). $\left(n+\dfrac12\right)\lambda=\dfrac{5\lambda}2\;\Rightarrow\;n+\dfrac12=\dfrac52\;\Rightarrow\;n=2$ Since $n=0$ gives the 1st dark fringe, $n=2$ corresponds to the THIRD dark fringe.

3. 3.↑ back to the question

   With one slit closed, the setup becomes a single-slit diffraction experiment instead of a double-slit interference experiment. The central maximum of a single-slit diffraction pattern is much wider (angularly) than the central fringe of a double-slit interference pattern. So the new pattern has a LARGER (wider) central maximum in angular size.

4. 4.↑ back to the question

   (i) Constructive interference (bright fringe): path difference $\;\Delta x=n\lambda$, where $n=0,\pm1,\pm2,\dots$ (ii) Destructive interference (dark fringe): path difference $\;\Delta x=\left(n+\dfrac12\right)\lambda$, where $n=0,\pm1,\pm2,\dots$

5. 5.↑ back to the question

   (a) $\;\dfrac{I_1}{I_2}=\dfrac91\;\Rightarrow\;\dfrac{a_1}{a_2}=\dfrac{\sqrt{I_1}}{\sqrt{I_2}}=\sqrt{\dfrac91}=\dfrac31$ (since $I\propto a^2$) $\;\dfrac{I_{max}}{I_{min}}=\dfrac{(a_1+a_2)^2}{(a_1-a_2)^2}=\dfrac{(3+1)^2}{(3-1)^2}=\dfrac{16}4=4$ Ratio of intensity of bright fringe to dark fringe $=4:1$ (b) $\;n_1\lambda_1=n_2\lambda_2$ (frequency unchanged, so $n\lambda=$ speed and using refractive-index relation): with $n_1=1$, $\lambda_1=600$ nm, $n_2=\mu_w=\dfrac43$: $\;\lambda_2=\dfrac{n_1\lambda_1}{n_2}=\dfrac{1\times600}{4/3}=\dfrac{600\times3}4=450\text{ nm}$

6. 6.↑ back to the question

   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 $I_1=I_2=I_0$)

7. 7.↑ back to the question

   Path difference between the two waves from $S_1$ and $S_2$ reaching a point $P$ on the screen (at distance $y$ from $O$) is: $\;\Delta x=(SS_2+S_2P)-(SS_1+S_1P)=(SS_2-SS_1)+(S_2P-S_1P)$ $\;\Delta x=\dfrac\lambda4+\dfrac{yd}D$, where $d$ is the slit separation and $D$ the screen distance (using the standard YDSE path-difference result $S_2P-S_1P\approx yd/D$). For constructive interference at $P$: $\;\Delta x=n\lambda\;\Rightarrow\;\dfrac\lambda4+\dfrac{yd}D=n\lambda\;\Rightarrow\;\dfrac{yd}D=\left(n-\dfrac14\right)\lambda$, where $n=0,\pm1,\pm2,\dots$

8. 8.↑ back to the question

   For the single slit of width $a$, at angle $\theta$ such that $a\sin\theta=\lambda$, the slit can be divided into two halves such that light from corresponding points in the two halves has a path difference of $\lambda/2$ — every ray from the upper half is cancelled by a corresponding ray from the lower half. This destructive interference of the wavelets ACROSS the width of a SINGLE slit produces the first minimum. For the two narrow slits (each effectively a point source) separated by distance $a$, at the SAME angle $\theta=\lambda/a$, the path difference between the two slits' waves is simply $a\sin\theta=\lambda$ — a full wavelength — which is exactly the condition for CONSTRUCTIVE interference (a bright fringe), since here there is only one path difference to consider (between the two point sources), not a continuous range of path differences across a finite slit width. So the same geometric angle gives a minimum for single-slit diffraction (from the extended width of one slit) but a maximum for double-slit interference (from two narrow, point-like slits) — because the two situations involve fundamentally different physical origins of path difference.

9. 9.↑ back to the question

   (i) Path difference at $P$: $\;\Delta x=(SS_2-SS_1)+(S_2P-S_1P)=\dfrac\lambda4+\dfrac{yd}D$ (same setup as Q28 above). For constructive interference: $\;\Delta x=n\lambda\;\Rightarrow\;\dfrac{yd}D=\left(n-\dfrac14\right)\lambda,\ n=0,\pm1,\pm2,\dots$ For destructive interference: $\;\Delta x=\left(n+\dfrac12\right)\lambda\;\Rightarrow\;\dfrac\lambda4+\dfrac{yd}D=\left(n+\dfrac12\right)\lambda\;\Rightarrow\;\dfrac{yd}D=\left(n+\dfrac14\right)\lambda,\ n=0,\pm1,\pm2,\dots$ (ii) With white light, the central fringe (zero path difference) remains white (all colours reinforce there), but on either side, the fringes of different colours (having different wavelengths, hence different fringe widths $\beta\propto\lambda$) overlap and produce a few coloured fringes before merging into a general uniform illumination — a clear, sharp interference pattern is not sustained far from the centre.

10. 10.↑ back to the question

    Fringe width for $\lambda=500$ nm: $\;\beta_{500}=\dfrac{\lambda D}d=\dfrac{500\times10^{-9}\times0.6}{1.0\times10^{-3}}=3\times10^{-4}\text{ m}=0.30\text{ mm}$ Fringe width for $\lambda=600$ nm: $\;\beta_{600}=\dfrac{600\times10^{-9}\times0.6}{1.0\times10^{-3}}=3.6\times10^{-4}\text{ m}=0.36\text{ mm}$ (i) Position of the 2nd bright fringe for $\lambda=500$ nm: $\;y_2=2\beta_{500}=2\times0.30=0.60\text{ mm}=6\times10^{-4}\text{ m}$ (ii) Bright fringes of the two wavelengths coincide when $\;m\beta_{600}=n\beta_{500}$ (integers $m,n$): $\;\dfrac mn=\dfrac{\beta_{500}}{\beta_{600}}=\dfrac{0.30}{0.36}=\dfrac56$ The smallest integers satisfying this are $m=5$, $n=6$: distance $=m\beta_{600}=5\times0.36=1.8\text{ mm}$ (checking: $n\beta_{500}=6\times0.30=1.8\text{ mm}$ — consistent). So the least distance from the central maximum where the bright fringes coincide is $1.8$ mm.

11. 11.↑ back to the question

    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): $\;\beta=x_{n+1}-x_n=\dfrac{(n+1)\lambda D}d-\dfrac{n\lambda D}d=\dfrac{\lambda D}d$

12. 12.↑ back to the question

    (a) Since $I\propto a^2$ and $I_2=0.5I_1$: $\;a_2^2=0.5a_1^2\;\Rightarrow\;a_2=\dfrac{a_1}{\sqrt2}$ $\;\dfrac{I_{max}}{I_{min}}=\dfrac{(a_1+a_2)^2}{(a_1-a_2)^2}=\dfrac{\left(1+\dfrac1{\sqrt2}\right)^2}{\left(1-\dfrac1{\sqrt2}\right)^2}=\left(\dfrac{\sqrt2+1}{\sqrt2-1}\right)^2\approx34$ (b) With white light, the central fringe is white (all wavelengths reinforce at zero path difference), and a few coloured fringes appear on either side (with the violet fringe closest to the centre, since it has the smallest wavelength and hence smallest fringe width, and red farthest out); beyond a few fringes, the overlap of different colours' fringes washes out into general illumination.

13. 13.↑ back to the question

    (a) Angular fringe width $\;\theta=\dfrac\lambda d\;\Rightarrow\;d=\dfrac\lambda\theta$ $\lambda=600$ nm $=6\times10^{-7}$ m; $\theta=0.1°=\dfrac{0.1\times\pi}{180}=\dfrac\pi{1800}$ rad $\;d=\dfrac{6\times10^{-7}\times1800}\pi\approx3.44\times10^{-4}\text{ m}$ (b) Frequency of light depends only on its source, not on the medium, so the frequencies of both the reflected and refracted light remain the same as the incident frequency. The reflected light stays in the same medium (air), so its wavelength also remains unchanged, at 500 Å. The refracted light enters water (denser, $\mu_w>1$), so its wavelength DECREASES, to $\lambda_r=\lambda/\mu_w$.

14. 14.↑ back to the question

    Two independent monochromatic sources, however identical in wavelength, cannot maintain a fixed phase relationship with each other — the phase difference between the light waves emitted by two independent sources keeps changing randomly and continuously (since each source's atoms emit light independently, in random bursts). As a result, the positions of constructive and destructive interference on the screen shift too rapidly to be observed, and the pattern averages out to uniform illumination rather than a sustained (stable) interference pattern.

15. 15.↑ back to the question

    (a) Since intensity is proportional to slit width: $\;\dfrac{I_1}{I_2}=\dfrac{w_1}{w_2}=\dfrac41\;\Rightarrow\;\dfrac{a_1^2}{a_2^2}=\dfrac41\;\Rightarrow\;\dfrac{a_1}{a_2}=2\;\Rightarrow\;a_1=2a_2$ $\;\dfrac{I_{max}}{I_{min}}=\dfrac{(a_1+a_2)^2}{(a_1-a_2)^2}=\dfrac{(2a_2+a_2)^2}{(2a_2-a_2)^2}=\dfrac{9a_2^2}{a_2^2}=9:1$ (b) No, it does not violate the conservation of energy. In interference, energy is only REDISTRIBUTED — it disappears from the regions of destructive interference (dark fringes) and reappears, in exactly compensating amount, at the regions of constructive interference (bright fringes), so the total (average) energy over the whole pattern remains conserved.

16. 16.↑ back to the question

    (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 $\Delta x=(2n-1)\dfrac\lambda2$ ($n=1,2,3,\dots$).

17. 17.↑ back to the question

    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 $x=0$.

18. 18.↑ back to the question

    (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).

19. 19.↑ back to the question

    (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), $I_0=K/4$)

20. 20.↑ back to the question

    $\;\dfrac{I_{min}}{I_{max}}=\dfrac{(a_1-a_2)^2}{(a_1+a_2)^2}=\dfrac9{25}\;\Rightarrow\;\dfrac{a_1-a_2}{a_1+a_2}=\dfrac35$ $\Rightarrow\;5(a_1-a_2)=3(a_1+a_2)\;\Rightarrow\;5a_1-5a_2=3a_1+3a_2\;\Rightarrow\;2a_1=8a_2\;\Rightarrow\;a_1=4a_2$ Since $I\propto a^2$ and width $\propto I$: $\;\dfrac{w_1}{w_2}=\dfrac{a_1^2}{a_2^2}=\dfrac{(4a_2)^2}{a_2^2}=16$ So the ratio of the widths of the two slits is $16:1$.

21. 21.Answer: (C)↑ back to the question

    Angular separation between fringes $=\dfrac\lambda d$, which depends only on $\lambda$ and the slit separation $d$ — it is independent of the screen distance $D$, so it remains constant when the screen is moved. Fringe width $\;\beta=\dfrac{\lambda D}d$; since $\beta\propto D$, moving the screen further away (increasing $D$) increases the fringe width. Option (c).

22. 22.Answer: (A)↑ back to the question

    Position of the $n$th bright fringe: $\;x_n=\dfrac{n\lambda D}d$. For the same point on the screen to be the 3rd bright fringe of $\lambda=700$ nm and the 5th bright fringe of $\lambda'$: $\;x_3=x_5$ $\Rightarrow\;\dfrac{3\times700\times D}d=\dfrac{5\times\lambda'\times D}d\;\Rightarrow\;\lambda'=\dfrac{3\times700}5=\dfrac{2100}5=420\text{ nm}$ Option (a).

23. 23.Answer: (C)↑ back to the question

    Fringe width in air: $\;\beta=\dfrac{\lambda D}d$. When immersed in a liquid of refractive index $\mu$, the wavelength changes to $\;\lambda'=\dfrac\lambda\mu$, while $d$ and $D$ remain unchanged. New fringe width: $\;\beta'=\dfrac{\lambda' D}d=\dfrac{\lambda D}{\mu d}=\dfrac\beta\mu$ Option (c).