# Light is passing through a transparent material at an angle of 63 degrees.  The material has an index of refraction of 1.6.  At what angle will the light pass into the air?

justaguide | Certified Educator

When a beam of light passes from one medium to another the beam bends if the two mediums do not have the same refractive index.

If the angle of incidence is A and the angle of refraction is B, the two are related as (sin A)/(sin B) = (n2)/(n1) where n1 is the refractive index of the first medium and n2 is the refractive index of the second.

In the problem, the angle of incidence A is 63 degrees and the refractive index of the material is 1.6. The refractive index of air is 1. If B is the angle of refraction, sin 63/sin B = 1/1.6

=> sin B = 1.6*sin 63

=> sin B = 1.425

Is is seen that sin B > 1. This indicates that the beam of light does not pass into the air. The beam of light undergoes total internal reflection.

quirozd | Certified Educator

Snell's Law:

n_1sin(\theta_1)=n_2sin(\theta_2)

let n_1 = 1.0 (air) and n_2 = 1.6

So, \theta_1 = 63^o

Solve for \theta_2

\theta_2 = sin^(-1)(n_1/n_2*sin(\theta_1))

\theta_2 =sin^(-1)(1.0/1.6*sin(63^o))

:. \theta_2 = 33.8^o