In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern

Options

A. The intensities of both maxima and the minima increase
B. The intensity of the maxima increases and the minima has zero intensity
C. The intensity of the maxima decreases and that of the minima increases
D. The intensity of the maxima decreases and the minima has zero intensity

Answer

A

Explanation

In interference we know that
`I_("max") = (sqrt(I_(1)) + sqrt(I_(2)))^(2)` and `I_("min") = (sqrt(I_(1)) - sqrt(I_(2)))^(2)`
Under normal conditions (when the widths of both the slits are equal)
`I_(1) ~~ I_(2) = I` (say)
`therefore I_("max") = 4I` and `I_("min") = 0`
When the width of one of te slits is increased. Intensity due to that slit would increase, while that of the other will remain same. So let :
`I_(1) = I` and `I_(2) = eta I " "(eta gt 1)`
Then `I_("max") = I(1+sqrt(eta))^(2) gt 4 I` and `I_(min") = I (sqrt(eta) - 1)^(2) lt 0`
`therefore` Intensity of both maximum and minima is increased.