In Young's experiment the ratio of intensity at the maxima and minima in the interference patner is `3:16` . What is the ratio of the widths of the two slits

Options

A. `(4)/(1)`
B. `(2)/(3)`
C. `(5)/(1)`
D. `(1)/(5)`

Answer

C

Explanation

`(I_(max))/(I_(min))=(36)/(16),(W_(1))/(W_(2))=?`
`(I_(max))/(I_(min))=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))`
`(36)/(16)=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))`
`(6)/(4)=(a_(1)+a_(2))/(a_(1)-a_(2))`
`(3+2)/(3-2)=(a_(1)+a_(2)+a_(1)-a_(2))/(a_(1)+a_(2)-a_(1)+a_(2))`
`(5)/(1)=(a_(1))/(a_(2))`
But , `(W_(1))/(W_(2))=(a_(1))/(a_(2))=(5)/(1)`