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Question

A double slit experiment is performed with sodium light of wavelength 600 nm and interference pattern is observed on a screen 100 cm away from the slits. The 4th dark fringe is at a distance of 7 mm from the central maximum. Find the separation between the slits.

A
0.15 mm
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B
0.30 mm
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C
0.60 mm
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D
1.2 mm
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Solution

The correct option is B 0.30 mm
Given:
λ=600 nm=6×107 m
D=100 cm=1 m
4th dark fringe is formed at y=7 mm=7×103 m

Path difference, Δx=ydD
For the 4th dark fringe,
Δx=7λ2
Δx=(n12)λ, for a dark fringe.

So, ydD=7λ2

d=7λD2y=7×6×107×12×7×103

d=3×104 m=0.3 mm

Hence, (B) is the correct answer.

Why this question :Concept : For dark fringes to form, path difference :Δx=(2n1)λ2Where,n=1,2,3,.....

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