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

Suppose the amplitude of the light wave coming from the narrower slit is A and that coming from the wider slit is 2A. The maximum intensity occurs at a place where constructive interference takes place. Then the fesultant amplitude is the sum of the individual amplitudes. Thus
`A_("max")=2A + A = 3A`
The minimum intensity occurs at a place where destructive interference takes place. The resultant amplitude is then difference of the individual amplitudes. Thus
`A_("min")= 2A-A=A`
As the intensity is proportional to the square of the amplitude.
`I_(max)/I_(min) = (A_(max))^(2)/(A_(min))^(2) = (3A)^(2)/A^(2) =9`