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In Young's double slit experiment, one of the slit is wider than other, so that the 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 :
Option: 1
$\frac{I_{m}}{9}\left ( 4+5\cos \phi \right )$

Option: 2
$\frac{I_{m}}{3}\left ( 1+2\cos^{2}\frac{\phi }{2} \right )$

Option: 3
$\frac{I_{m}}{5}\left ( 1+4\cos^{2}\frac{\phi }{2} \right )$

Option: 4
$\frac{I_{m}}{9}\left ( 1+8\cos^{2}\frac{\phi }{2} \right )$

Answers (1)

best_answer

Here , $A_{2}= 2A_{1}$

$\because \: \: Intensity\alpha \left ( Amplitude \right ) ^{2}$

$\therefore \: \: \: \frac{I_{2}}{I_{1}}= \left ( \frac{A_{2}}{A_{1}} \right )^{2}= \left ( \frac{2A_{1}}{A_{1}} \right )^{2}= 4$

$I_{2}= 4I_{1}$

Maximum intensity, $I_{m}= \left ( \sqrt{I_{1}} +\sqrt{I_{2}}\right )^{2}$

$= \left ( \sqrt{I_{1}} +\sqrt{4I_{2}} \right )^{2}= \left ( 3\sqrt{I_{1}} \right )^{2}= 9I_{1}$

$or\: \: I_{1}= \frac{I_{m}}{9}\cdots \cdots \cdots (i)$

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

$I_{1}+4I_{1}+2\sqrt{I_{1}\left ( 4I_{1} \right )}\cos \phi$

$5I_{1}+4I_{1}\cos \phi = I_{1}+4I_{1}+4I_{1}\cos \phi$

$I_{1}+4I_{1}\left ( 1+\cos \phi \right )$

$I_{1}+8I_{1}\cos ^{2}\frac{\phi }{2}\: \: \: \: \: \left ( \because 1+\cos \phi = 2\cos ^{2}\frac{\phi }{2} \right )$

$= I_{1}\left ( 1+8\cos ^{2} \frac{\phi }{2}\right )$

Putting the value of $I_{1}$ from eqn. (i), we get

$I=\frac{I_{m}}{9}\left ( 1+8\cos ^{2} \frac{\phi }{2}\right )$

Posted by

Divya Prakash Singh

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