Question

A layer of liquid B floats on liquid A. A ray of light begins in liquid A and undergoes total internal reflection at the interface between the liquids when the angle of incidence exceeds 38.0°. When liquid B is replaced with liquid C, total internal reflection occurs for angles of incidence greater than 49.5°. Find the ratio nB/nC of the refractive indices of liquids B and C. nB/nC =?

Answer 1

Answer:

The ratio of the refractive indices of the liquids is 1.22

Solution:

Critical angle for A and B interface, $\theta_{AB} = 38.0^{\circ}$

Critical angle for A and B interface, $\theta_{AC} = 49.5^{\circ}$

Now,

From the relation in between the critical angle and refractive index:

$n = \frac{1}{\theta_{c}}$              (1)

where

n = rafractive index

$\theta_{c}$ = critical angle

Thus

For AB interface:

$n_{B} = \frac{1}{sin\theta_{AB}} = \frac{1}{sin(38.0^{\circ})} = 1.606$

For AC interface:

$n_{C} = \frac{1}{sin\theta_{AC}} = \frac{1}{sin(49.5^{\circ})} = 1.315$

Thus the ratio of the refractive indices of these liquids can be given as:

$\frac{n_{B}}{n_{C}} = \frac{1.606}{1.315}$

$\frac{n_{B}}{n_{C}} = 1.22$