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JEE Main 2022 (30 Jun, Shift 1) · Prism, Dispersion & Scattering

Prism, Dispersion & Scattering question from JEE Main 2022 (30 Jun, Shift 1)

Q50JEE Main 2022 · 30 JUN · SHIFT 1Easy

Prism, Dispersion & Scattering

The refractive index of an equilateral prism is 2\sqrt{2}. The angle of emergence under minimum deviation position of prism, in degree, is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 45

  1. At minimum deviation r1=r2=A/2=30∘r_1 = r_2 = A/2 = 30^\circ.
  2. sin⁡e=μsin⁡r2=2×12=12\sin e = \mu \sin r_2 = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}
  3. e=45∘e = 45^\circ
  4. So the angle of emergence is 45∘45^\circ.
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