Question

In a 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

• A. $\frac{I_{m}}{9}\left(4 + 5 \cos\phi\right)$
• B. $\frac{I_{m}}{3}\left(1+2 \cos^2\, \frac{\phi}{2}\right)$
• C. $\frac{I_{m}}{5}\left(1+4 \cos^2\, \frac{\phi}{2}\right)$
• D. $\frac{I_{m}}{9}\left(1+8 \cos^2\, \frac{\phi}{2}\right)$

Answer 1

Answer: (D)

Given, $a_2 = 2a_1 \Rightarrow I_1 = 4 I_2 = 4I_0$
$ \therefore \:\:\: I_m = ( \sqrt{I_1} + \sqrt{I_2})^2 = (3 \sqrt{I_2})^2 = 9 I_2 = 9I_0$
= $I_0 = \frac{I_m}{9}$
Now, Resultant intensity
$I = I_1 + I_2 +2 \sqrt{I_1 I_2} \cos \phi$
$ = 4I_0 + I_0 + 2 \sqrt{4 I_0 I_0} \, \cos \phi$
$ = 5I_0 + 4I_0 \cos \phi = \frac{I_m}{9} (5 + 4 \cos \phi)$
$ \frac{I_m}{9} [ 1 + 4 (1 + \cos \, \phi)]$
$ = \frac{I_{m}}{9 } \left( 1+ \frac{8 \cos^{2} \phi}{2}\right) \left[\left(1 +\cos \phi\right)=2 \cos^{2} \frac{\phi}{2}\right]$