Question

In the young’s double slit experiment, the slit widths are in the ratio of 9:1. The ratio of the intensities at the maxima and minima will be

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

Answer 1

The correct option is D 4 :1

$\frac{I_1}{I_2}=\frac{w_1}{w_2}$ [$\because I\propto w$ (slit width)] = $\frac{9}{1}$
$\frac{I_{max}}{I_{min}}={\left[\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-{\sqrt{I}}_2}\right]}^2={\left[\frac{\sqrt{9}+1}{\sqrt{9}-1}\right]}^2=2^2$
$I_{max}:I_{min}=4:1$