Question

In Young's double slit experiment if the slits widths are in the ratio $1:9$, the ratio of the intensities 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

$\frac{1}{4}$

Solution:

Step 1: Given information

Slits widths are in the ratio $1:9$.

Step 2: Calculate the ratio of the intensities at minima to that at maxima

The breadth of the slits is exactly proportional to the intensity of the fringe.

Therefore,$\frac{I_2}{I_1}=\frac{W_2}{W_1}=1:9$

or$I_1=9I_2$

The maximum intensity and minimum intensity are $I_{max}={(\sqrt{I_1}+\sqrt{I_2})}^2$

and $I_{min}={(\surd I_1–\surd I_2)}^2$respectively.

So, $I_{max}=16I_2$

and $I_{min}=4I_2$

Therefore, $\frac{I_{min}}{I_{max}}=\frac{1}{4}$.

Hence, the correct option is (C).