Optics Question 472

Question: In Young’s double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If $ I _{m} $ be the maximum intensity, the resultant intensity $ I $ when they interfere at phase difference $ \phi $ is given by

Options:

A) $ \frac{I _{m}}{9}(4+5cos\phi ) $

B) $ \frac{I _{m}}{3}( 1+2cos^{2}\frac{\phi }{2} ) $

C) $ \frac{I _{m}}{5}( 1+4cos^{2}\frac{\phi }{2} ) $

D) $ \frac{I _{m}}{9}( 1+8cos^{2}\frac{\phi }{2} ) $

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Answer:

Correct Answer: D

Solution:

[d] It is given, $ A _{2}=2A _{1} $

We know, intensity $ \propto {{( Amplitude )}^{2}} $

Hence $ \frac{I _{2}}{I _{1}}={{( \frac{A _{2}}{A _{1}} )}^{2}}={{( \frac{2A _{1}}{A _{1}} )}^{2}}=4 $

$ \Rightarrow I _{2}=4I _{1} $

Maximum intensity, $ I _{m}={{( \sqrt{I _{1}}+\sqrt{I _{2}} )}^{2}} $

$ ={{( \sqrt{I _{1}}+\sqrt{4I _{1}} )}^{2}}={{( 3\sqrt{I _{1}} )}^{2}}=9I _{1} $

Hence $ I _{1}=\frac{I _{m}}{9} $

Resultant intensity, $ I=I _{1}+I _{2}+2\sqrt{I _{1}I _{2}}\cos \phi $

$ =I _{1}+4I _{1}+2\sqrt{I _{1}(4I _{1})}\cos \phi $

$ =5I _{1}+4I _{1}\cos \phi =I _{1}+4I _{1}+4I _{1}\cos \phi $

$ =I _{1}+4I _{1}(1+cos\phi ) $

$ =I _{1}+8I _{1}{{\cos }^{2}}\phi $

$ ( \therefore 1+\cos \phi =2{{\cos }^{2}}\frac{\phi }{2} ) $

$ I=\frac{I _{m}}{9}( 1+8cso^{2}\frac{\phi }{2} ) $ Putting the value of $ I _{1} $ from eqn. (i), we get $ I=\frac{I _{m}}{9}( 1+8{{\cos }^{2}}\frac{\phi }{2} ) $



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