Question

A ray of light is moving from a material having a high indexof refraction into a material with a lower index of refraction.

Answer 1

(a) Away from the normal

We can find the direction of bending of the ray of light by using Snell's equation:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where we have:

n1, n2: index of refraction of the first and second medium

$\theta_1, \theta_2$; angle that the incident and the refracted ray form with the normal to the surface

Here, the light ray moves from a material with high index of refraction to a material with lower index, so we have

$n_1 > n_2$

Re-arranging Snell's law we find

$sin \theta_2 = \frac{n_1}{n_2} sin \theta_1$

since we have

$\frac{n_1}{n_2}>1$

this implies

$sin \theta_2 > sin \theta_1\\\theta_2 > \theta_1$

so the ray of light bends away from the normal.

(b) The wavelength is greater in the second material (the one with lower index of refraction)

The wavelength of the light in a medium is given by

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0$ is the wavelength of the light in a vacuum

n is the refractive index

The equation can be rewritten as

$\lambda_0 = \lambda_1 n_1 = \lambda_2 n_2$

and again it can be rewritten as

$\lambda_2 = \frac{n_1}{n_2} \lambda_1$

where

$\lambda_1 = 600 nm\\\frac{n_1}{n_2}>1$

Therefore, we have that the wavelength in the second medium (the one with lower index of refraction) is longer than the wavelength in the first medium.

(c) The frequency remains the same

Wavelength and speed of a light ray depend on the medium in which the wave is travelling through, however the frequency does not depend on that, so it remains the same in the two mediums.