Question

In Young's double slit experiment if the slit widths are in the ratio $1:9$. The ratio of the intensity at minima to that at maxima will be :

• A. $1$
• B. $\frac{1}{9}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{4}$

Intensity of a fringe is directly proportional to width of the slits i.e $I\propto W$

$\therefore$ $\frac{I_2}{I_1}=\frac{W_2}{W_1}$

$\frac{I_2}{I_1}=\frac{1}{9}$ $⟹I_1=9I_2$

Maximum intensity $I_{max}=(\sqrt{I_1}+\sqrt{I_2}{)}^2$

$\therefore$ $I_{max}=(\sqrt{9I_2}+\sqrt{I_2}{)}^2=(3\sqrt{I_2}+\sqrt{I_2}{)}^2=(4\sqrt{I_2}{)}^2$

$⟹$ $I_{max}=16I_2$

Minimum intensity $I_{min}=(\sqrt{I_1}-\sqrt{I_2}{)}^2$

$\therefore$ $I_{min}=(\sqrt{9I_2}-\sqrt{I_2}{)}^2=(3\sqrt{I_2}-\sqrt{I_2}{)}^2=(2\sqrt{I_2}{)}^2$

$⟹$ $I_{min}=4I_2$

$\therefore$ $\frac{I_{min}}{I_{max}}=\frac{4I_2}{16I_2}=\frac{1}{4}$