Question

In Young's double slit experiment , i the widths of the slits are in the ratio 4:9, the ratio of the intensity at 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 A 169:25

Given ratio:

$\begin{array}{c}{I}_1=16I\\ I_2=81I\\ I_{max}={\left(4\sqrt{I}+9\sqrt{I}\right)}^2\\ =169I\\ I_{min}={\left(9\sqrt{I}-4\sqrt{I}\right)}^2\\ =84I\\ \therefore \frac{I_{max}}{I_{min}}=\frac{169}{25}\end{array}$

Then,

Option $A$ is correct answer.