In Young's double slits experiment, if the ratio of the width of the two slits is 25 : 1, what is the ratio of intensity at the maxima and minima in the interference pattern?

9 : 4

The correct answer is Option (4) → 9 : 4

Amplitude, $A∝\sqrt{Width\,of\,slits}$

$⇒A_1∝\sqrt{5},A_2∝\sqrt{1}$

$⇒A_1:A_2=5:1$

Now,

$A_{max}=A_1+A_2=5+1=6$

$A_{min}=A_1-A_2=5-1=4$

and,

$\frac{I_{max}}{I_{min}}∝\frac{{A_{max}}^2}{{A_{min}}^2}$

[$I_{max}$ = Intensity of Maxima]

[$I_{min}$ = Intensity of Minima]

$\frac{I_{max}}{I_{min}}∝\frac{(6)^2}{(4)^2}=\frac{36}{16}$

$⇒\frac{I_{max}}{I_{min}}=\frac{9}{4}$