# (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.

(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.

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