JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Geometrical Optics

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Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in media A and B are

2.4 \times 1 0^{8} m/s

and

2.7 \times 1 0^{8} m/s

respectively, then the value of critical angle is:

(A) $cot^{- 1} (\frac{3}{13})$
(B) $sin^{- 1} (\frac{9}{8})$
(C) $tan^{- 1} (\frac{8}{17})$ ✓ Correct
(D) $cos^{- 1} (\frac{8}{9})$

Answer: (C)

Solution

At the critical angle

μ_{A} sin c = μ_{B}

. Since

μ = \frac{c}{v}

sin c = \frac{μ _{B}}{μ _{A}} = \frac{v _{A}}{v _{B}} = \frac{2.4}{2.7} = \frac{8}{9}

. Then

cos c = 1 - \frac{64}{81} = \frac{17}{9}

and

tan c = \frac{sin c}{cos c} = \frac{8}{17}

. The critical angle is

tan^{- 1} (\frac{8}{17})

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Geometrical Optics

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