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JEE Main 2018 · Diffraction at a Single Slit

Diffraction at a Single Slit question from JEE Main 2018

Q73JEE Main 2018Difficult

Diffraction at a Single Slit

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

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Answer: (b)

  1. Central maximum 60∘60^\circ, so first minimum at 30∘30^\circ: asin⁡30∘=λ⇒λ=1×12=0.5 μa\sin 30^\circ = \lambda \Rightarrow \lambda = 1 \times \dfrac{1}{2} = 0.5\ \mum
  2. Fringe width β=λDd⇒d=λDβ\beta = \dfrac{\lambda D}{d} \Rightarrow d = \dfrac{\lambda D}{\beta}
  3. d=0.5×10−6×0.50.01=25×10−6d = \dfrac{0.5 \times 10^{-6} \times 0.5}{0.01} = 25 \times 10^{-6} m
  4. So the answer is (b) 25 μ25\ \mum.
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