Question

In a Young's double slit experiment, the width of one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

• A. $4:1$
• B. $2:1$
• C. $3:1$
• D. $1:4$

Answer 1

The correct option is A $4:1$

Given:
Slit width, $d_2=3d_1$
Also, $A\propto d$

$\therefore \frac{A_1}{A_2}=\frac{1}{3}$

Assuming, $A_1$$=$ $x$ , $A_2$ $=$ $3x$

We know that,

$I_{\text{max}}={(A_1+A_2)}^2=16x^2$

$I_{\text{min}}={(A_1-A_2)}^2=4x^2$

Now,
$\frac{I_{\text{max}}}{I_{\text{min}}}=\frac{16x^2}{4x^2}$

$\Rightarrow \frac{I_{\text{max}}}{I_{\text{min}}}=\frac{4}{1}=4:1$