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JEE Main 2020 (5 Sep, Shift 2) · Prism, Dispersion & Scattering

Prism, Dispersion & Scattering question from JEE Main 2020 (5 Sep, Shift 2)

Q68JEE Main 2020 · 5 SEP · SHIFT 2Easy

Prism, Dispersion & Scattering

A prism of angle A = 1° has a refractive index μ\mu = 1.5. A good estimate for the minimum angle of deviation (in degrees) is close to N10\frac{N}{10}.

Value of N is ____.

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

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Answer: 5

  1. Thin prism: δm=(μ−1)A\delta_m = (\mu - 1)A.
  2. δm=(1.5−1)×1∘=0.5∘=510\delta_m = (1.5 - 1) \times 1^\circ = 0.5^\circ = \frac{5}{10}
  3. So N=5N = 5.
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