In Young's double slit experiment, the widths of two slits are n the ratio `4 : 1`. The ratio of maximum and minimum intensity in the interference pattern will be :

Answer

`9 : 1`

Explanation

Here, `(w_(1))/(w_(2)) = (4)/(1) = (a^(2))/(b^(2)) :. (a)/(b) = (2)/(1)`
`(I_(max))/(I_(min)) = ((a + b)^(2))/((a - b)^(2)) = ((2 + 1)^(2))/((2 - 1)^(2)) = 9 : 1`