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(a) Can the interference pattern be produced by two independent monochromatic sources of light? Explain.

(b) The intensity at the central maximum (O) in a Young’s double slit experimental set-up shown in the figure is $I_{O}$ . If the distance OP equals one-third of the fringe width of the pattern, show that the intensity at point P, would equal $\frac{I_{O}}{4}$ .

c) In Young’s double-slit experiment, the slits are separated by 0·5 mm and the screen is placed 1·0 m away from the slit. It is found that the 5th bright fringe is at a distance of 4·13 mm from the 2nd 'dark fringe. Find the wavelength of light used.

Answers (1)

(a) No, the sustained interference pattern cant be obtained.

Since light waves emitted from a source undergoes abrupt phase changes in times of the order of $10^{-10}s$ .

So, light from two independent sources will not have a fixed phase relationship and it will be incoherent

(b) given

$x=\frac{\beta }{3}$ and

$\Delta x=\frac{xd}{D}$

the phase difference

$=\frac{2\pi}{\lambda}\Delta x=\frac{2\pi}{\lambda}\times \frac{xd}{D}\\\\=\frac{2\pi}{\lambda}\times \frac{\beta d}{3D}=\frac{2\pi}{\lambda}\times \frac{\lambda D d}{3Dd}= \frac{2\pi}{3}$

$=\frac{2\pi }{3}$

Now, the intensity at the point P will be

$I=I_{0}\cos ^{2}\frac{\phi }{2}$

$I=I_{0}\cos ^{2}\left ( \frac{2\pi }{3\times 2} \right )\; \; \; \; \left ( \therefore \phi =\frac{2\pi }{3} \right )$

$I=I_{0}\cos ^{2}\left ( \frac{\pi }{ 3} \right )$

$I=I_{0}\left ( \frac{1}{4} \right )=\frac{I_{0}}{4}$

Hence, the intensity at point P is equal to $\frac{I_{0}}{4}$

(c) the distance of 5th bright fringe from 2nd dark fringe is given :

So, calculation the wavelength $(\lambda )$ of light as ;

$x=\frac{5\lambda D}{d}-\frac{3\lambda D}{2d}=\frac{7}{2}\frac{\lambda D}{d}$

$\lambda =\frac{2xd}{7D}=\frac{2\times 4.13\times 10^{-3}\times 10^{-5}\times 10^{-3}}{7\times 1}$

$\lambda =0.59\times 10^{-6}nm$ or $\lambda =5900A^{\circ}$ is the wavelength of light.

Posted by

Safeer PP

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