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JEE Main 2024 (6 Apr, Shift 1) · Prism, Dispersion & Scattering

Prism, Dispersion & Scattering question from JEE Main 2024 (6 Apr, Shift 1)

Q36JEE Main 2024 · 6 APR · SHIFT 1Medium

Prism, Dispersion & Scattering

The refractive index of prism is μ=3\mu = \sqrt{3} and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is _________∘{}^\circ.

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

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Answer: 60

  1. With δm=A\delta_m = A: μ=sin⁡Asin⁡A2=2cos⁡A2\mu = \dfrac{\sin A}{\sin\frac{A}{2}} = 2\cos\frac{A}{2}
  2. 3=2cos⁡A2⇒A2=30∘\sqrt{3} = 2\cos\frac{A}{2} \Rightarrow \frac{A}{2} = 30^\circ
  3. A=60∘A = 60^\circ, so the answer is 60.
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