Wave OpticsHard

Question

In a YDSE merriment two slits S1 and S2 have separation of d = 2 mm. The distance of the screen is D = m. Source S stars moving from a very large distance towards S2 perpendicular to S1S2 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 source moves towards S2 is

Options

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

Solution

S1P - S2P = λ         (λ = 500 nm)
So when S is at ∞ there is Ist minima and when S is at S2 there las t minima because d/λ = 4000
So the number of minima′s will be 4001 and number of maxima′s will be 4000.

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