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JEE Main 20188 April
Exam
JEE Main
Year
2018
Date
8 April
Subject
Physics
Chapter
Wave Optics

JEE Main 2018 · 8 April · Physics · Wave Optics

The angular width of the central maximum in a single slit...

The angular width of the central maximum in a single slit diffraction pattern is 60°60 ° . The width of the slit is 1μm1\mu m. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young's fringes can be observed on a screen placed at a distance 50cm50cm from the slits. If the observed fringe width is 1cm1cm, what is slit separation distance? (i.e., the distance between the centres of each slit.)The angular width of the central maximum in a single slit diffraction pattern is 60 ° . The width of the slit is 1μ m .The angular width of the central maximum in a single slit diffraction pattern is 60 ° . The width of the slit is 1μ m .

Options

  1. A100μm100\mu m
  2. B25μm25\mu m
  3. C50μm50\mu m
  4. D75μm75\mu m

Correct answer

B. 25μm25\mu m

Step-by-step solution

In diffraction,& dsin30°=λdsin30 ° = \lambda,~λ=d2\lambda = \dfrac{d}{2}

The angular width of the central maximum in a single slit diffraction pattern is 60 ° . The width of the slit is 1μ m .

Young’s fringe width [~d′d^{'}- the separation between two slits]~

β=λ×Dd′⇒10−2=d2×50×10−2d′\beta = \dfrac{\lambda \times D}{d'}⇒ {10}^{- 2} = \dfrac{d}{2} \times \dfrac{50 \times {10}^{- 2}}{d'}
⇒10−2=10−6×50×10−22×d′⇒d′=25μm⇒ {10}^{- 2} = \dfrac{{10}^{- 6} \times 50 \times {10}^{- 2}}{2 \times d^{'}}⇒ d^{'} = 25\mu m.
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