Chapter 43: Problem 46

Question: A light ray falls on a square glass slab as in Fig40, chap 43. What must be the minimum index of refraction of the glass if total internal reflection occurs at the vertical face?

Solution: We have that for total internal reflection to occur at the vertical face

$$\sin\theta_1~=~n\sin\theta_2$$

$$n\sin (90-\theta_2)~=~1$$

Now, the second relation gives us $\cos\theta_2~=~{1\over n}$ and hence $\sin\theta_1~=~\sqrt{n^2-1}$, which gives $n=\sqrt{1+\sin^2\theta_1}~=~1.18$.