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In a single slit diffraction pattern, a light of wavelength 6000 Ao is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from slits. The width of the slit is ____×10-4 m.
PhysicsWave OpticsJEE MainJEE Main 2024 (29 Jan Shift 2)
Solution:
2367 Upvotes Verified Answer
The correct answer is: 2

For nth minima:

bsinθ=nλ

(λ is very small, so sinθ is very small, hence sinθtanθ.

btanθ=nλ

byD=nλ

yn=nλDb (Position of nth minima)

B1st  minima and A3rd :

y3=3λDb,y1=λDb

Δy=y3-y1=2λDb

3×10-3=2×6000×10-10×0.5b

b=2×6000×10-10×0.53×10-3

b=2×10-4 m

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