Question

A light beam is traveling through an unknown substance. When it exits that substance and enters into air, the angle of reflection is 32.0° and the angle of refraction is 42.0°. What is the index of refraction of the substance?

Answer 1

Answer:

The index of refraction of the substance is 0.79

Explanation:

According to the second law of refraction which states that the ratio of the sine of angle of incidence (i) to the sine of angle of refraction (r) is a constant for a given pair of media. This constant is known as the refractive index (n). Mathematically;

sin(i)/sin(r) = n

If a light travelling through an unknown substance reflects at an angle of 32°, then the angle of incidence will also be 32° according to the second law of reflection which states that the angle of incidence is equal to the angle of reflection r' i.e i = r' = 32°

Given angle of refraction r = 42°

The index of refraction;

n = sin32°/sin42°

n = 0.5299/0.6691

n = 0.79

Therefore the index of refraction of the substance is 0.79

Answer 2

Answer:

0.79

Explanation:

Using Snell's law, we have that:

n(1) * sin θ1 = n(2) * sinθ2

Where n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection through the unknown substance is the same as the angle of incidence of air. This means that θ1 = 32°

=> 1.0003 * sin32 = n(2) * sin42

n(2) = (1.0003 * sin32) / sin42

n(2) = 0.79