A light ray travels from Medium X into Medium Y. The speed of light in Medium X is 1.50 \t… — JAMB UTME

A light ray travels from Medium X into Medium Y. The speed of light in Medium X is $1.50 \times 10^8\text{ m/s}$ and the speed of light in Medium Y is $2.50 \times 10^8\text{ m/s}$ . What is the sine of the critical angle for total internal reflection at the boundary between these two media?

$0.60$ Answer
B
$1.67$
C
$0.50$
D
$0.40$

Answer

The sine of the critical angle for total internal reflection at the boundary is

0.60

Total internal reflection occurs when light moves from an optically denser medium to an optically less dense medium. By Snell's law,

n_X \sin\theta_c = n_Y \sin 90^\circ

, which gives

\sin\theta_c = n_Y / n_X

. Since refractive index is inversely proportional to speed (

n = c/v

\sin\theta_c = v_X / v_Y = (1.50 \times 10^8) / (2.50 \times 10^8) = 0.60

Step-by-Step Solution

Relate speed of light in each medium to their respective refractive indices

n_X = \frac{c}{v_X}

and

n_Y = \frac{c}{v_Y}

Refractive index is defined as the ratio of light speed in vacuum to light speed in the medium.

Apply Snell's Law for the critical angle condition

\sin\theta_c = \frac{n_Y}{n_X}

At the critical angle, the angle of refraction in the less dense medium is

90^\circ

(

\sin 90^\circ = 1

Substitute light speeds into the critical angle formula and calculate

\sin\theta_c = \frac{v_X}{v_Y} = \frac{1.50 \times 10^8\text{ m/s}}{2.50 \times 10^8\text{ m/s}} = 0.60

Substituting the expressions for

n_X

and

n_Y

simplifies the refractive index ratio to the direct ratio of speeds

v_X / v_Y

Key Concept

Critical angle and total internal reflection relation to speed of light in media

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