The speed of light in medium A is 1.5 \times 10^8\text{ m/s} and in medium B is 3.0 \times… — JAMB UTME

The speed of light in medium $A$ is $1.5 \times 10^8\text{ m/s}$ and in medium $B$ is $3.0 \times 10^8\text{ m/s}$ . What is the critical angle for a light ray traveling from medium $A$ towards medium $B$ ?

$30^\circ$ Answer
B
$45^\circ$
C
$60^\circ$
D
$90^\circ$

Answer

The critical angle for light traveling from medium

A

to medium

B

30^\circ

The critical angle

C

is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is

90^\circ

. Using the wave speed form of Snell's law,

\sin C = \frac{v_A}{v_B} = \frac{1.5 \times 10^8}{3.0 \times 10^8} = 0.5

. Taking the inverse sine yields

C = 30^\circ

Step-by-Step Solution

Determine the relationship between critical angle and wave speeds in the two media.

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

By Snell's Law, critical angle occurs when the angle of refraction is

90^\circ

, giving

\sin C = \frac{n_2}{n_1}

. Since refractive index

n = \frac{c}{v}

, the ratio

\frac{n_B}{n_A}

simplifies to

\frac{v_A}{v_B}

Substitute the given wave speeds into the equation.

\sin C = \frac{1.5 \times 10^8\text{ m/s}}{3.0 \times 10^8\text{ m/s}} = 0.5

Light is traveling from the optically denser medium

A

(lower speed) to the less dense medium

B

(higher speed).

Calculate the inverse sine to find the critical angle

C

C = \arcsin(0.5) = 30^\circ

The angle whose sine is

0.5

30^\circ

Key Concept

Critical Angle and Total Internal Reflection in terms of Wave Speed

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic