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Ray Optics

JEE Main level, single correct, medium

53 of 295

Question 53 of 295

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 $2.4 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and $2.7 \times 10^{8} \mathrm{~m} / \mathrm{s}$ , respectively, then the value of critical angle is :

Show answer and solution

Answer: (D) $\tan ^{-1}\left(\frac{8}{\sqrt{17}}\right)$

The refractive index of a medium is inversely proportional to the speed of light in that medium, given by

n = \frac{c}{v}

.
For light travelling from medium

A

to medium

B

, the critical angle

\theta_c

is defined when light travels from a denser to a rarer medium.
Given speeds are

v_A = 2.4 \times 10^8

m/s and

v_B = 2.7 \times 10^8

m/s.
The refractive indices are

n_A = \frac{c}{v_A}

and

n_B = \frac{c}{v_B}

.
The critical angle

\theta_c

is given by

\sin \theta_c = \frac{n_B}{n_A} = \frac{v_A}{v_B}

.
Substituting the values:

\sin \theta_c = \frac{2.4 \times 10^8}{2.7 \times 10^8} = \frac{24}{27} = \frac{8}{9}

.
To find the value in terms of

\tan^{-1}

, we use the trigonometric identity for a right-angled triangle where perpendicular

p = 8

and hypotenuse

h = 9

.
The base

b = \sqrt{h^2 - p^2} = \sqrt{9^2 - 8^2} = \sqrt{81 - 64} = \sqrt{17}

.
Thus,

\tan \theta_c = \frac{p}{b} = \frac{8}{\sqrt{17}}

.
Therefore,

\theta_c = \tan^{-1}\left(\frac{8}{\sqrt{17}}\right)

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JEE Main 2025 Apr 2 Shift 1 question paper

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