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 mmGiven: λ=600 nm=6×10−7 m D=100 cm=1 m 4th dark fringe is formed at y=7 mm=7×10−3 m Path difference, Δx=ydD For the 4th dark fringe, Δx=7λ2 ∵Δx=(n−12)λ, for a dark fringe. So, ydD=7λ2 ⇒d=7λD2y=7×6×10−7×12×7×10−3 d=3×10−4 m=0.3 mm Hence, (B) is the correct answer. Why this question :Concept : For dark fringes to form, path difference :Δx=(2n−1)λ2Where,n=1,2,3,.....

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