In the Young's double slit experiment, the… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Optics

2021mcqhard

In the Young's double slit experiment, the distance between the slits varies in time as $d(t)={d}_{0}+{a}_{0}\mathrm{sin}\omega t$ ; where ${d}_{0},\omega$ and ${a}_{0}$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as:

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Held on 25 Jul 2021 · Verified 6 Jul 2026.

Options

A
$\frac{2\lambda D({d}_{0})}{({d}_{0}^{2}-{a}_{0}^{2})}$
B
$\frac{2\lambda D{a}_{0}}{({d}_{0}^{2}-{a}_{0}^{2})}$
C
$\frac{\lambda D}{{d}_{0}^{2}}{a}_{0}$
D
$\frac{\lambda D}{{d}_{0}+{a}_{0}}$

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Solution

Fringe width,

$\beta =\frac{\lambda D}{d}$

${\beta }_{\mathrm{max}}\Rightarrow {d}_{\mathrm{min}}$ and ${\beta }_{\mathrm{min}}\Rightarrow {d}_{\mathrm{max}}$

$d={d}_{0}+{a}_{0}\mathrm{sin}\omega t$

${d}_{\mathrm{max}}={d}_{0}+{a}_{0}$ and ${d}_{\mathrm{min}}={d}_{0}-{a}_{0}$

$\therefore {\beta }_{\mathrm{min}}=\frac{\lambda D}{{d}_{0}+{a}_{0}}$ and $\therefore {\beta }_{\mathrm{max}}=\frac{\lambda D}{{d}_{0}-{a}_{0}}$

${\beta }_{\mathrm{max}}-{\beta }_{\mathrm{min}}=\frac{\lambda D}{{d}_{0}-{a}_{0}}-\frac{\lambda D}{{d}_{0}+{a}_{0}}=\frac{2\lambda D{a}_{0}}{{d}_{0}^{2}-{a}_{0}^{2}}$

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