In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is do - Physics MCQ NEET 2020 E1 2026 | ExamTest.live | ExamTest.live

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In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:

A $double$
B $half$
C $four times$
D $one-fourth$

Solution & Step-by-step Explanation

Fringe width β = \frac{λ D}{d} .New distance D^{'} = 2 D .New separation d^{'} = d /2 . β^{'} = \frac{λ ( 2 D )}{( d /2 )} = 4 (\frac{λ D}{d}) = 4 β The fringe width becomes four times.

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In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes:

A

double

B

half

C

four times

D

one-fourth

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Wave Optics Young's Double Slit

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