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JEE Main 2026 (2 Apr, Shift 1) · Prism, Dispersion & Scattering

Prism, Dispersion & Scattering question from JEE Main 2026 (2 Apr, Shift 1)

Q11JEE Main 2026 · 2 APR · SHIFT 1Medium

Prism, Dispersion & Scattering

For a thin symmetric prism made of glass (refractive index 1.5), the ratio of incident angle and minimum deviation will be _______.

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Answer: (b)

  1. At minimum deviation r=A2r = \frac{A}{2}; small angles give i=μr=μA2=0.75Ai = \mu r = \frac{\mu A}{2} = 0.75A.
  2. Thin prism: δm=(μ−1)A=0.5A\delta_m = (\mu - 1)A = 0.5A
  3. i:δm=0.75:0.5=3:2i : \delta_m = 0.75 : 0.5 = 3 : 2, so the answer is (b).
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