Question

In Young's double slit experiment, if the slit widths are in the ratio 1 : 9, then the ratio of the intensity at minima to that at maxima will be [MP PET 1987]
A) 1 B) 1/9 C) 1/4 D) 1/3

Answer 1

Option c is your answer of a question

Answer 2

Answer:

(C). The ratio of the minimum and maximum intensity is $\dfrac{1}{4}$.

Explanation:

Given that,

The slit widths are in the ratio 1 : 9,

We know that,

Width is directly proportional to the intensity

$\dfrac{\beta_{1}}{\beta_{2}}=\dfrac{I_{1}}{I_{2}}$

$\dfrac{I_{1}}{I_{2}}=\dfrac{1}{9}$

The intensity is directly proportional to square of the amplitude.

$\dfrac{I_{1}}{I_{2}}=(\dfrac{A_{1}^2}{A_{2}^2})$

$\dfrac{A_{1}}{A_{2}}=\dfrac{1}{3}$

The maximum intensity is

$I_{max}=(A_{1}+A_{2})^2$

$I_{max}=(1+3)^2$

$I_{max}=16$

The minimum intensity is

$I_{min}=(A_{1}-A_{2})^2$

$I_{min}=(1-3)^2$

$I_{min}=4$

The ratio of the minimum and maximum intensity will be

$\dfrac{I_{min}}{I_{max}}=\dfrac{A_{2}}{A_{1}}$

$\dfrac{I_{min}}{I_{max}}=\dfrac{4}{16}$

$\dfrac{I_{min}}{I_{max}}=\dfrac{1}{4}$

Hence, The ratio of the minimum and maximum intensity is $\dfrac{1}{4}$.