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 $\varphi$, is given by

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

Answer 1

Answer: (D)

Given, $a_2=2a_1\Rightarrow I_1=4I_2=4I_0$
$\therefore I_m=(\sqrt{I_1}+\sqrt{I_2}{)}^2=(3\sqrt{I_2}{)}^2=9I_2=9I_0$
= $I_0=\frac{I_m}{9}$
Now, Resultant intensity
$I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$
$=4I_0+I_0+2\sqrt{4I_0{I}_0}\cos \varphi$
$=5I_0+4I_0\cos \varphi =\frac{I_m}{9}(5+4\cos \varphi )$
$\frac{I_m}{9}[1+4(1+\cos \varphi )]$
$=\frac{I_m}{9}\left(1+\frac{8{\cos }^2\varphi }{2}\right)\left[\left(1+\cos \varphi \right)=2{\cos }^2\frac{\varphi }{2}\right]$