Question

In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit double of that from other slit. If $I_m$ be the maximum intensity, the resultant intensity I. When they interfere at phase difference $\varphi$ is given by

• A. $\frac{I_m}{9}(4+5\cos \varphi )$
• B. $\frac{I_m}{3}(1+2{\cos }^2\frac{\varphi }{2})$
• C. $\frac{I_m}{5}(1+4{\cos }^2\frac{\varphi }{2})$
• D. $\frac{I_m}{9}(1+8{\cos }^2\frac{\varphi }{2})$

Answer 1

The correct option is D $\frac{I_m}{9}(1+8{\cos }^2\frac{\varphi }{2})$

$I=I_0+4I_0+2\sqrt{I_0\times 4I_0}cos\varphi$
$I=I_0+4I_0+4I_{0}cos\varphi$
$I_m$ will be at $cos\varphi =1$,
So, $I_m=9I_0$
$\frac{I}{I_m}=\frac{5I_0+4I_{0}cos\varphi }{9}$
$\frac{I}{I_m}=\frac{1+8co{s}^2\frac{\varphi }{2}}{9}$