A ray of light traveling inside a glass prism of refractive index 1.60 strikes the boundar… — JAMB UTME

A ray of light traveling inside a glass prism of refractive index $1.60$ strikes the boundary with an adjacent oil medium of refractive index $1.20$ . What is the sine of the critical angle for total internal reflection at this glass-oil interface?

$0.75$ Answer
B
$1.33$
C
$0.625$
D
$0.80$

Answer

The sine of the critical angle at the glass-oil interface is

0.75

For light traveling from a medium of refractive index

n_1

into a medium of refractive index

n_2

(where

n_1 > n_2

), the critical angle

C

satisfies

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

. Substituting

n_1 = 1.60

and

n_2 = 1.20

gives

\sin C = \frac{1.20}{1.60} = 0.75

Step-by-Step Solution

Identify the refractive indices of the two media.

Denser medium (glass):

n_1 = 1.60

; less dense medium (oil):

n_2 = 1.20

Total internal reflection occurs when light originates in the optically denser medium and strikes the boundary with a less dense medium.

Apply Snell's law at the critical angle

C

n_1 \sin C = n_2 \sin 90^\circ \implies \sin C = \frac{n_2}{n_1}

At the critical angle of incidence, the angle of refraction in the second medium is

90^\circ

, so

\sin 90^\circ = 1

Substitute the given numerical values to compute

\sin C

\sin C = \frac{1.20}{1.60} = \frac{3}{4} = 0.75

Dividing the refractive index of the oil by that of the glass gives the exact sine of the critical angle.

Key Concept

Critical Angle between Two Media

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic