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NEET 2020 (Phase 1) · YDSE: Fringe Width & Fringe Positions

YDSE: Fringe Width & Fringe Positions question from NEET 2020 (Phase 1)

Q61NEET 2020 · PHASE 1Easy

YDSE: Fringe Width & Fringe Positions

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:

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Answer: (b)

$\beta = \frac{\lambda D}{d}$
$\beta' = \frac{\lambda (2D)}{d/2} = 4\beta$
So the answer is (b) four times.

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