The ratio of intensities at amxima and minima is `25 : 16`. What will be ratio of the widths of two slits in YDSE ?

Explanation

Given,
`(I_("max"))/(I_("min")) = 25/16`.
Let r be the amplitude ratio of interfering waves. Then,
`((r + 1)/(r - 1))^(2) = 25/16`
`implies (r + 1)/(r - 1) = 5/4`
`implies 4r + 4 = 5r - 5`
`implies r = 9`
As `r = (a_1)/(a_2) = 9/1 " "` (where `a_1 and a_2` are amplitude of waves)
Width ratio of slits is
`(omega_1)/(omega_2) = (a_1^2)/(a_2^2) = (9^2)/(1^2) = 81 : 1`.