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(a) Explain how a diffraction pattern is formed due to interference of secondary wavelets of light waves from a slit.

(b) Sodium light consists of two wavelengths, $5900\dot{A}$ and $5960\dot{A}$ . If a slit width $2\times 10^{-4}m$ is liumated by sodium light, find the separation between the first secondary maxima of the diffraction pattern of the two wavelength on a screen placed 1.5m away.

Answers (1)

A plane wavefront is incident on a narrow slit AB of width 'd'. The secondary wavelet from A and B reach the point 'O'. There is no path difference between them. So that will interfere constructively and that point 'O' is bright. The secondary wavelet reach the point 'P' also. The path difference is BN.

from figure,

$\sin \theta = \frac{BN}{AB}$

$BN = AB\sin \theta$

$BN =a \sin \theta\approx a\theta$

At points P on the screen for which path difference $n\lambda(n=1,2,.....)$ wavelets will cancel each other in pairs and produce minima.
At points P on the screen for which path difference $(2n+1)\lambda/2 (n=1,2,....)$ wavelets produce a maxima

(b) Given

The wavelength of first sodium light, $\lambda _{1}= 5900\AA =5900\times 10^{-10}m$

The wavelength of second sodium light, $\lambda _{2}= 5960\AA =5960\times 10^{-10}m$

Width of slit, $a = 20\times 10^{-9}m$

Distance of screen D = 1.3

We know that , the condition for n^th maxima is,

$a \sin \theta = \left ( 2n+1 \right )\frac{\lambda }{2}$

Here n=1

Therefore, $\sin \theta = \frac{3\lambda _{1}}{2a}=\frac{x_{1}}{D}$

Similarly $= \frac{3\lambda _{2}}{2a}=\frac{x_{2}}{D}$

$x_{1}=\frac{3\lambda _{1}D}{2a};x_{2}=\frac{3\lambda _{2}D}{2a}$

Separation between two cases= x₁ - x₂

$= \frac{3D}{2a}(\lambda _{1}-\lambda _{2})$

$= \frac{3\times 1.5}{2\times 2\times 10^{-4}}(590-596)\times 10^{-9}$

$=$ $6.75\times 10^{-5}m$

The separation between the front secondary maxima is $6.75\times 10^{-5}m$ .

Posted by

Safeer PP

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