Question

In Young's double slit experiment two coherent sources of intensity ratio of $64:1$ produce interference fringes. Calculate the ratio of maximum and minimum intensities.

Answer 1

Given data :
$I_1:I_2::64:1$ $\frac{I_{max}}{I_{min}}=$?
Solution:
$\frac{I_1}{I_2}=\frac{a_1^2}{a_2^2}=\frac{64}{1}$
$\therefore \frac{a_1}{a_2}=\frac{8}{1}$ ; $a_1=8a_2$
$\frac{I_{max}}{I_{min}}=\frac{{(a_1+a_2)}^2}{{(a_1-a_2)}^2}=\frac{{(8a_2+a_2)}^2}{{(8a_2-a_2)}^2}=\frac{{\left(9a_2\right)}^2}{{\left(7a_2\right)}^2}=\frac{81}{49}$
$I_{max}:I_{min}::81:49$