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JEE Main 20207 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 7 January
Shift: Morning Shift
Subject: Physics
Chapter: Wave Optics

JEE Main 2020 · 7 January · Morning Shift · Physics · Wave Optics

Visible light of wavelength 6000 × 10 ^ - 8 cm falls normally on a...

Visible light of wavelength

6000 \times {10}^{- 8}cm

falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at

{60}^{o}

from the central maximum. If the first minimum is produced at

{\theta}_{1},

then

{\theta}_{1}

is close to

Options

A ${20}^{o}$
B ${30}^{o}$
C ${25}^{o}$
D ${45}^{o}$

Correct answer

C. ${25}^{o}$

Step-by-step solution

For

{2}^{nd}

minima,

dsin \theta = 2\lambda

sin \theta = \dfrac{\sqrt{3}}{2}

, (given)

⇒ \dfrac{\lambda}{d} = \dfrac{\sqrt{3}}{4}

...(1)
So, for

{1}^{st}

minima

dsin\theta = \lambda

⇒ sin\theta = \dfrac{\lambda}{d} = \dfrac{\sqrt{3}}{4}

⇒ \theta = {25.65}^{o}

⇒ \theta \approx {25}^{o}

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