Question

A ray of light is incident on the surface of a block of clear ice at an angle of 39.0° with the normal. Part of the light is reflected and part is refracted. Find the angle between the reflected and refracted light.

Answer 1

The angle between the reflected and refracted light is 39.0°.

Step-by-step explanation:

When a ray of light passes from one medium to another, it can be either reflected or refracted. The angle of reflection is equal to the angle of incidence, which is the angle between the incident ray and the normal. The angle of refraction is determined by Snell's Law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of light in the two different media.

In the given question, the angle of incidence is 39.0° with respect to the normal of the surface. Since the angle of reflection is equal to the angle of incidence, the angle between the reflected and refracted light would also be 39.0°.

Answer 2

112.76°

Step-by-step explanation:

The angle of the reflected light is the same as the angle of the incident light:

`\theta_r=\theta_i=39.0\°`

The refracted angle is obtained by using the Snell's law:

`n_1sen\theta_1=n_2sin\theta_2`

n1: air index of refraction ≈ 1.00

n2: ice index of refraction = 1.31

By replacing the values for the incident angle and the index of refraction you can calculate the angle of refraction:

`(1.00)sin(39.0\°)=(1.33)sin\theta_2\\\\\theta_2=sin^(-1)(((1.00)sin(39.0\°))/(1.33))=28.24\°`

the angle between the reflected and refracted angle will be:

`\phi=(90\°-39.0\°)+(90\°-28.24\°)=112.76\°`

where you have taken into account that the angles aremeasured according to the horizontal axis.