JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Geometrical Optics

The critical angle for a denser-rarer interface is

4 5^{\circ}

. The speed of light in rarer medium is

3 \times 1 0^{8}

ms. The speed of light in the denser medium is:

(A) $5 \times 1 0^{7} m / s$
(B) $2.12 \times 1 0^{8} m / s$ ✓ Correct
(C) $3.12 \times 1 0^{7} m / s$
(D) $2 \times 1 0^{8} m / s$

Answer: (B)

Solution

At the critical angle

sin C = \frac{v _{d e n ser}}{v _{r a r er}}

, since the refractive index of each medium is inversely proportional to the speed of light in it. With

C = 4 5^{\circ}

sin 4 5^{\circ} = \frac{1}{2} = \frac{v _{d e n ser}}{v _{r a r er}}

. Hence

v_{d e n ser} = \frac{v _{r a r er}}{2} = \frac{3 \times 1 0 ^{8}}{1.414} \approx 2.12 \times 1 0^{8}

m/s.

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Single correct option

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