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

`(W_(1))/(W_(2))=(a_(1)^(2))/(a_(2)^(2))=(1)/(9)`
`:. (a_(1))/(a_(2))=(1)/(3) " " :. a_(2)=3a_(1)`
Now, `(I_("max"))/(I_("min"))=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))=((a_(1)+3a_(1))^(2))/((a_(1)-3a_(1))^(2))=(4)/(1)`
Thus, `(I_("min"))/(I_("max"))=(1)/(4)`