Question

A ray of light propagates in air, entering a rectangular quartz prism with index of refraction 1.72 at an incidence angle of 58⁰. After it passes through the quartz, it enters a fluid with index of refraction 1.27.

(a) What is the refracted angle when the light enters the quartz?
⁰

(b) What is refracted angle of the light when it enters the fluid?
⁰

(c) How much does the light speed up when it enters the fluid?
m/s

Answer 1

Solution:

Given:

Angle of incidence (i) = 58^o

Refractive index of Quartz (n_Quartz) = 1.72

Refractive index of Fluid (n_Fluid) = 1.27

Part (a) Solution:

let the angle of refraction is: r

Then: Using Snell's law: (n_Air) sin(i) = (n_Quartz) sin(r)

(1) sin(58) = (1.72) sin(r)

r = 29.54^o

the angle of refraction is 29.54^o

Part (b) Solution:

in the fluid: let the angle of refraction is: r'

Then: Using Snell's law: (n_Quartz) sin(r) = (n_Fluid) sin(r')

(1.72) sin(29.54) = (1.27) sin(r')

r' = 41.89^o

the refracted angle of light is 41.89^o

Part (c) Solution:

speed of light in fluid = c / n_Quartz = (3 x 10⁸ m/s) / (1.72) = 1.744 x 10⁸ m/s

speed of light in fluid = c / n_fluid = (3 x 10⁸ m/s) / (1.27) = 2.362 x 10⁸ m/s

Therefore: Change in speed = (2.362 x 10⁸ m/s) - ( 1.744 x 10⁸ m/s) = 0.618 x 10⁸ m/s