Q.

In an interference arrangement similar to Young’s double slit experiment, but having four slits S1, S2, S3 and S4 are illuminated with coherent light wave of wavelength 400 nm. Sources are synchronized in way such that S1, S2 and S4 have zero phase difference but S3 is out of phase with others. In central wide part of screen intensity due to all individual slits is I0. The consecutive slits are separated by a distance d = 1 mm. Distance between plane of slits and screen is D = 10 m

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Point O is a point on screen lie on perpendicular bisector of S1S2. ‘P’ is the point on the screen closest to point ‘O’ having intensity 2I0 when only S1 and S2 are illuminated. Now if all slits are illuminated, what will be new intensity at point ‘P’?

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a

4I0

b

2I0

c

3I0

d

Zero

answer is B.

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Detailed Solution

Let distance OP = y
Path difference of S1 and S3 at P=ydD=λ4
y=λD4d..(i)y=d
Now path difference of S1 and S3
=3d2(2d)D=3d2D=3λ4
And path difference of S1 and S4 is =2dD(3d)=32λ
Phasor diagram

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AR=2A          IR=4I0

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