# 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?

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### 2 Answers

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.

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`

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