The width of one of the two slits in a Young's double slit experiment is double of the other slit. Assuming that the amplitude of the light coming from a slit is proportional to the slit width, find the ratio of the maximum to the minimum intensity in the interference pattern.

Explanation

Amplitudes from the slits are A and 2A.
`A_(max) = A + 2A = 3A` and `A_(min) = 2A - A`
`implies(I_(max))/(I_(min)) = (A_(max)^(2))/(A_(min)^(2)) = 9`