Chapter 18, Problem 86A

To determine

The change in the size of the critical angle as the index of refraction of the medium in which the ray is travelling into increases.

Answer to Problem 86A

Increases

Explanation of Solution

Introduction:

When a ray of light moves from a medium of higher index of refraction in to a medium with a lower index of refraction, then light reflects back to its incident medium for certain angles. This phenomenon is known as total internal reflection. And, the angle of incidence at which the ray of line lines along the boundary of the mediums is called critical angle which is given as,

  $\sin {\theta }_c=\frac{n_2}{n_1}$

Here, $n_1$ is the index of refraction of incident medium and $n_2$ is the index of refraction of refracting medium.

As can be seen from above formula, the critical angle is directly proportional to the index of refraction of the refracting medium. Therefore, if the index of refraction of the medium in which the ray is travelling into increases, the size of critical angle also increases.

Conclusion:

Hence, there will be increase in size of the critical angle as the index of refraction of the medium in which the ray is travelling into increases.