A long horizontal slit is placed 1 mm above a horizontal plane mirror. The interference between the light coming directly from the slit and the after reflection is seen on a screen 1 m away from the slit. If the mirror reflects only 64% of the light falling on it, the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen is

Options

A. `8:1`
B. `3:1`
C. `81:1`
D. `9:1`

Answer

c

Explanation

Intensity of direct ray `= I_(0) = kA_(0)^(2)`
Intensity of reflected ray `= (64)/(100) I_(0) = k ((8 A_(0))/(10))^(2)`
`:. (I_(max))/(I_(min)) = ((A_(0) + 0.8 A_(0))^(2))/((A_(0) - 0.8 A_(0))^(2)) = ((1.85)/(0.8))^(2) = (81)/(1)`