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 :

1. double

2. half

3. four times

4. one-fourth

Doubtly Asked 30 Oct 2026 NEET UG 2020 Practice test neet Save WhatsApp Report

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Answer: 3. four times

Fringe width $\beta=\dfrac{\lambda D}{d}$.

With $d\to d/2$ and $D\to2D$:

$$\beta'=\frac{\lambda(2D)}{d/2}=4\beta$$

The fringe width becomes four times. Answer: option 3.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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