In a YDSE experiment two slits `S_(1)` and `S_(2)` have separation of d=2mm the distance of the screen is `D=(8)/(5)` m source S starts moving from a very large distance towards `S_(2)` perpendicular to `S_(1)S_(2)` as shown in figure the wavelength of monochromatic light is 500 nm. The number of maximas observed on the screen at point P as the moves towards `S_(2)` is

Options

A. 4001
B. 3999
C. 3998
D. 4000

Answer

4

Explanation

`S_(1)O -S_(2)P =d^(2)/(2D) = (2xx10^(-3)xx2xx10^(-3))/(2xx 8/5 ) 5/2 lambda(lambda=500nm)`
So when S is at `infty` there is `I^(st)` minima and when S is at `S_(2)` there is last minima because `d/lambda=4000`
So the number of minima.s will be 4001 and number of maxima.s will be 4000.