In Young's double slit experiment, if the widths of the slit are in the ratio 4:9, ratio of intensity of maxima to intensity of minima will be
Answer & explanation
Correct answer: option 1
$ Intensity \propto \text{ slit width}$
$\frac{I_1}{I_2} = \frac{4}{9} = \frac{a_1^2}{a_2^2}$
$\Rightarrow \frac{a_1}{a_2} = \frac{2}{3}$
$ \frac{I_{max}}{I_{min}} = \frac{(a_1+a_2)^2}{(a_1-a_2)^2} = \frac{5^2}{1^2} = 25:1$