JEE Main 2026 — Jan 23, Shift 1 — Physics solution

JEE Main2026Jan 23, Shift 1Physics

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
$tan^{- 1} (\frac{8}{17})$
D
$cos^{- 1} (\frac{8}{9})$

Solution

At the critical angle, $μ_{A} sin c = μ_{B} sin 9 0^{\circ}$ , so:

$sin c = \frac{μ _{B}}{μ _{A}} = \frac{v _{A}}{v _{B}} = \frac{2.4 \times 1 0 ^{8}}{2.7 \times 1 0 ^{8}} = \frac{8}{9}$

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

$tan c = \frac{sin c}{cos c} = \frac{8/9}{17 /9} = \frac{8}{17}$

$∴ c = tan^{- 1} (\frac{8}{17})$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →