In Young's double slit experiment, if the slit widths are in the ratio `1:9`, then the ratio of the intensity at minima to that at maxima will be

Options

A. `1`
B. `(1)/(9)`
C. `(1)/(4)`
D. `(1)/(3)`

Answer

C

Explanation

Amplitude of the superimposing waves are
`(a_(1))/(a_(2))=((1)/(9))^((1)/(2))=(1)/(3) rArr (I_("minima"))/(I_("maxima"))=((a_(1)-a_(2))^(2))/((a_(1)+a_(2))^(2))=(1)/(4)`