Question

The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9:25. Find the ratio of the widths of the two slits.

Answer 1

If the width of the slits are A₁ and A₂, the amplitude will also be $A_1$ and$A_2$.

The maximum amplitude, A' will happen when constructive interference happens-

$A^{\prime }=(A_1+A_2)$

The maximum amplitude, A will happen when destructive interference happens-

$A=(A_1-A_2)$

We know that intensity is proportional to the square of amplitude.

$\frac{I_1}{I_2}=\frac{A^{\prime 2}}{A^2}=\frac{(A_1+A_2{)}^2}{(A_1-A_2{)}^2}\frac{9}{25}=(\frac{A_1+A_2}{A_1-A_2}{)}^2\sqrt{\frac{9}{25}}=\frac{A_1+A_2}{A_1-A_2}\frac{A_1+A_2}{A_1-A_2}=\frac{3}{5}\frac{A_1}{A_2}=\frac{5+3}{5-3}=\frac{8}{2}{A}_1:A_2=4:1$