JEE PYQ: Wave Optics Question 4
Question 4 - 2021 (24 Feb Shift 1)
In a Young’s double slit experiment, the width of one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
(1) 4 : 1
(2) 2 : 1
(3) 3 : 1
(4) 1 : 4
Show Answer
Answer: (1)
Solution
Given: $\omega_2 = 3\omega_1$
Also, $A \propto \omega$
$\frac{\omega_1}{\omega_2} = \frac{1}{3}$ …(1)
Assume $\omega_1 = x$, $\omega_2 = 3x$
We know that $I_{\max} = (A_1 + A_2)^2$, and $I_{\min} = (A_1 - A_2)^2$
$\frac{A_1}{A_2} = \frac{\omega_1}{\omega_2}$ …(2)
From equation (2) we can say that $A_1 = A$ and $A_2 = 3A$
Now, $\frac{I_{\max}}{I_{\min}} = \frac{(A + 3A)^2}{(A - 3A)^2} = \frac{16A^2}{4A^2} = \frac{4}{1}$
$\Rightarrow \frac{I_{\max}}{I_{\min}} = \frac{4}{1}$
BETA
