Interference effects are produced at point P on a screen as a result of direct rays from a 500 nm source and reflected rays from a mirror, as shown in figure. If the sources is 100m to the left of the screen and 1.00 cm above the mirror, find the distance y (in milimeters) to the first dark band above the mirror.

Answer

B

Explanation

`:.d = 2cm`
` omega = (lambdaD)/d = ((500 xx 10^-9)(100))/(2 xx 10^-2)`
` = 2.5 xx 10^-3 = 2.5mm`

In the figure, P is in a dark fringe, as conditions of maxima
and minima are interchanged.
Hence, next dark fringe will be obtained at a
distance omega or 2.5 mm from P.