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JEE Main 2020 (3 Sep, Shift 1) · YDSE: Fringe Width & Fringe Positions

YDSE: Fringe Width & Fringe Positions question from JEE Main 2020 (3 Sep, Shift 1)

Q64JEE Main 2020 · 3 SEP · SHIFT 1Easy

YDSE: Fringe Width & Fringe Positions

In a Young’s double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to

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Answer: (c)

  1. Angular width θ=λd=500×10−90.05×10−3=0.01 rad\theta = \frac{\lambda}{d} = \frac{500\times10^{-9}}{0.05\times10^{-3}} = 0.01\text{ rad}
  2. θ=0.01×180∘π≈0.57∘\theta = 0.01 \times \frac{180^\circ}{\pi} \approx 0.57^\circ
  3. So the answer is (c) 0.57∘0.57^\circ.
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