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

JEE Main 2021 — Physics Optics

2021integermedium

In a Young's double slit experiment, the slits are separated by $0.3\mathrm{mm}$ and the screen is $1.5m$ away from the plane of slits. Distance between fourth bright fringes on both sides of central bright fringe is $2.4\mathrm{cm}.$ The frequency of light used is $x\times {10}^{14}\mathrm{Hz}.$

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Held on 31 Aug 2021 · Verified 6 Jul 2026.

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Solution

The distance between ${n}^{\mathrm{th}}$ bright fringe on both side of central maxima is $2.4\mathrm{cm}.$ The $f=?$

$\Delta y=\frac{D\lambda }{d}\times 8$

$\lambda =\frac{d\Delta y}{8D}=\frac{0.3\times {10}^{-3}\times 2.4\times {10}^{-2}}{8\times 1.5}=\frac{0.12\times {10}^{-5}}{2}=0.06\times {10}^{-5}$

$\lambda =6\times {10}^{-7}m$

$f=\frac{C}{6\times {10}^{-7}}=0.5\times {10}^{+15}$

$f=5\times {10}^{14}\mathrm{Hz}$

So, $x=5$

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