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} ) $
