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JEE Main 201815 AprilMorning ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Morning Shift
Mode
Online
Subject
Physics
Chapter
Wave Optics

JEE Main 2018 · 15 April · Morning Shift · Online · Physics · Wave Optics

Light of wavelength 550 ~nm falls normally on a slit of width 22.0...

Light of wavelength 550 nm550 \mathrm{~nm} falls normally on a slit of width 22.0×10−5 cm22.0 \times 10^{-5} \mathrm{~cm}. The angular position of the second minima from the central maximum will be (in radians)

Options

  1. Aπ8\frac{\pi}{8}
  2. Bπ12\frac{\pi}{12}
  3. Cπ4\frac{\pi}{4}
  4. Dπ6\frac{\pi}{6}

Correct answer

A. π8\frac{\pi}{8}

Step-by-step solution

If angular position of 2nd 2^{\text{nd }} maxima from central maxima is θ\theta then sin⁡θ=(2n−1)λ2a=3λ2a=3×550×10−92×22×10−7∴θ=π8rad\begin{aligned}&\sin \theta=\frac{(2 \mathrm{n}-1) \lambda}{2 \mathrm{a}}=\frac{3 \lambda}{2 \mathrm{a}}=\frac{3 \times 550 \times 10^{-9}}{2 \times 22 \times 10^{-7}}\\ &\therefore \quad \theta=\frac{\pi}{8} \mathrm{rad} \end{aligned}
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