Question

In Young's double slit experiment , if the width of the slits are in the ratio $4:9$ the ratio of the intensity of maxima to the intensity at minima will be

• A. $169:25$
• B. $81:16$
• C. $25:1$
• D. $9:4$

Answer 1

The correct option is C $25:1$

Slit width ratio $=4:9$, hence $I_1:I_2=4:9$

$\therefore \frac{a_1^2}{a_2^2}=\frac{4}{9}\Rightarrow \frac{a_1}{a_2}=\frac{2}{3}$

$\therefore \frac{I_{max}}{I_{min}}=\frac{(a_1+a_2{)}^2}{(a_1-a_2{)}^2}=\frac{25}{1}$