Question

You send a beam of light from a material with index of refraction 1.17 into an unknown material. In order to help identify this material, you determine its index of refraction by measuring the angles of incidence and refraction for which you find the values 40.1 ° and 36.3 ° , respectively. What is the index of refraction of the unknown material?

Answer 1

Answer:

The index refraction of the unknown material is 1.08

Explanation:

Light has characteristics of a wave. When light travels or propagates through two media with different index of refraction, the light will bend or will change its speed. 

Using Snells' law which is stated mathematically as:

n1Sintheta1 = n2Sinthetha2

Given:

n1 = 1.17

Theta1 = 40.1° = angle of incidence

Theta2 = 36.3° = angle of refraction

n2 = unknown

Substituting into the equation

n1Sin40.1 = 1.17Sin36.3°

0.6441n1 = 0.6937

n1 = 0.6937/0.6441

n1 = 1.08

Answer 2

Answer:

1.09

Explanation:

From Snell's law, the refractive index of the unknown material is the ratio of the sine of the angle of incidence in the unknown material to the sine of the angle of refraction in the unknown material.

n= sini/ sinr

i= 40.1°

r= 36.3°

n= sin40.1°/sin36.3°

n= 0.6441/0.5920

n= 1.09