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JEE Main 2026 · 5 Apr · Shift 1, Physics question 40

A ray of light passing through an equilateral prism is having velocity 2.12×1082.12\times10^8 m/s in the prism material, then the minimum angle of deviation is ________ degrees.

Choose one option
Answer and worked solution

Official answer: B

Refractive index. μ=3×1082.12×108≈1.414=2\mu=\dfrac{3\times10^8}{2.12\times10^8}\approx1.414=\sqrt2.

Minimum deviation (A=60∘A=60^\circ):
sin⁡60∘+δm2=2sin⁡30∘=12,\sin\frac{60^\circ+\delta_m}2=\sqrt2\sin30^\circ=\frac1{\sqrt2},
so 60∘+δm2=45∘\dfrac{60^\circ+\delta_m}{2}=45^\circ and δm=30∘\delta_m=30^\circ.

(B)

Topics:Prism and dispersion
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