Question

In Young’s double slit experiment, one of the slits is wider than the other, so that the amplitude of the light from one slit is double that from the other slit. If $I_m$ is the maximum intensity, then the resultant intensity when they interfere at a phase difference of $\varphi$ is

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

Answer 1

The correct option is C $\frac{I_m}{9}(1+8co{s}^2\frac{\varphi }{2})$

Given
$A_2=2A_1$
As intensity is proportional to square of amplitude
Ratio of intensity of slits is given by
$\therefore \frac{I_2}{I_1}={\left(\frac{A_2}{A_1}\right)}^2={\left(\frac{2A_1}{A_1}\right)}^2=4\Rightarrow I_2=4I_1$

Maximum intensity,

$I_m=(\sqrt{I_1}+\sqrt{I_2}{)}^2={(\sqrt{I_1}+\sqrt{4I_2})}^2={\left(3\sqrt{I_1}\right)}^2=9I_1\Rightarrow I_1=\frac{I_m}{9}.....(i)$

Resultant intensity due to interference at a phase difference of $\varphi$ is

$I=I_1+I_2+2\sqrt{I_1{I}_2}cos\varphi =I_1+4I_1+2\sqrt{I_1\left(4I_1\right)}cos\varphi =5I_1+4I_{1}cos\varphi =I_1+4I_1+4I_{1}cos\varphi =I_1+8I_{1}co{s}^2\frac{\varphi }{2}(\because 1+cos\theta =2co{s}^2\frac{\varphi }{2})$
$=I_1(1+8co{s}^2\frac{\varphi }{2})$

Substituting the value of $I_1$ from eq $\left(i\right)$,
we get

$I=\frac{I_m}{9}(1+8co{s}^2\frac{\varphi }{2})$