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JEE Main 2026 · 2 Apr · Shift 2, Physics question 41

One side of an equilateral prism is painted by a transparent material of refractive index n2n_2. The refractive index of prism is 1.6. The minimum value of n2n_2 required for total internal reflection from painted face is ________.

Equilateral prism standing on its base; a ray enters the left face at normal incidence (right-angle mark) and travels to the base, which is coated with paint of refractive index n₂.
Choose one option
Answer and worked solution

Official answer: D

Geometry. The ray enters the left face normally, so it meets the painted base at 60° (equilateral prism).

Total internal reflection at the glass-paint interface needs 1.6sin⁡60∘≥n21.6\sin60^\circ\ge n_2, i.e.
n2≤1.6×32=435.n_2\le1.6\times\frac{\sqrt3}2=\frac{4\sqrt3}5.

The question says "minimum", but the critical value 435\frac{4\sqrt3}5 is the largest n2n_2 that still allows TIR; the key gives this value.

(D)

Topics:Total internal reflectionPrism and dispersion
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