Chapter 44: Problem 7

Question: Fig29 (chap 44) shows the cross section of a hollow glass tube of internal radius r and external radius R, and index of refraction n. Convince yourself that the ray ABC shown defines the apparent internal radius r^* as seen form the side. Show that r^* = nr, independent of R.

Solution: I assume you are convinced. At point B, we have that $\sin\theta_1~=~n\sin\theta_2$, and $R\sin\theta_2~=~r$. Further, we have that $R\sin\theta_1~=~r^*$. Combining these we have that

$${r^*\over R}~=~n{r\over R}$$

and therefore r^* = nr.