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JEE Main 2021 (25 Jul, Shift 2) · Prism, Dispersion & Scattering

Prism, Dispersion & Scattering question from JEE Main 2021 (25 Jul, Shift 2)

Q60JEE Main 2021 · 25 JUL · SHIFT 2Medium

Prism, Dispersion & Scattering

A prism of refractive index μ\mu and angle of prism A is placed in the position of minimum angle of deviation. If minimum angle of deviation is also A, then the value of A in terms of refractive index is

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Answer: (a)

  1. With δm=A\delta_m = A: μ=sin⁡Asin⁡A2=2sin⁡A2cos⁡A2sin⁡A2=2cos⁡A2\mu = \dfrac{\sin A}{\sin\frac{A}{2}} = \dfrac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}} = 2\cos\frac{A}{2}
  2. cos⁡A2=μ2⇒A=2cos⁡−1(μ2)\cos\frac{A}{2} = \frac{\mu}{2} \Rightarrow A = 2\cos^{-1}\left(\frac{\mu}{2}\right)
  3. So the answer is (a).
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