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 32.0°. When liquid B is replaced with liquid C, total internal reflection occurs for angles of incidence greater than 46.0°. Find the ratio nB/nC of the refractive indices of liquids B and C.

Answer 1

Answer:

0.737

Explanation:

$n_{A}$ = Refractive indices of liquid A

$n_{B}$ = Refractive indices of liquid B

$n_{C}$ = Refractive indices of liquid C

Consider the total internal reflection at interface of liquid A and liquid B

$\theta_{i}$ = Angle of incidence = 32.0

Using Snell's law for total internal reflection

$n_{A} Sin\theta_{i} = n_{B} \\n_{B} = n_{A} Sin32$

Consider the total internal reflection at interface of liquid A and liquid C

$\theta_{i}$ = Angle of incidence = 46

Using Snell's law for total internal reflection

$n_{A} Sin\theta_{i} = n_{C} \\n_{C} = n_{A} Sin46$

Ratio is hence given as

$Ratio = \frac{n_{B}}{n_{C}} = \frac{n_{A} Sin32}{n_{A} Sin46} = \frac{Sin32}{Sin46}\\Ratio = 0.737$