IPhO 1968, Теория и експеримент — Задача 3. Problem 3

• The rays emerge on the arc CE, for $75^{\circ} < \varphi < 165^{\circ}$; the arc subtends a central angle of 90°.

Let us use angle $\varphi$ to describe the position of the rays in the glass (Fig. 5). According to the law of refraction $\sin 45^{\circ}/\sin\beta = \sqrt{2}$, $\sin\beta = 0.5$, $\beta = 30^{\circ}$. The refracted angle is 30° for all of the incoming rays. We have to investigate what happens if $\varphi$ changes from 0° to 180°.

It is easy to see that $\varphi$ can not be less than 60° ($AOB\angle = 60^{\circ}$). The critical angle is given by $\sin\beta_{crit} = 1/n = \sqrt{2}/2$; hence $\beta_{crit} = 45^{\circ}$. In the case of total internal reflection $ACO\angle = 45^{\circ}$, hence $\varphi = 180^{\circ} - 60^{\circ} - 45^{\circ} = 75^{\circ}$. If $\varphi$ is more than 75° the rays can emerge the cylinder. Increasing the angle we reach the critical angle again if $OED\angle = 45^{\circ}$. Thus the rays are leaving the glass cylinder if:

CE, arc of the emerging rays, subtends a central angle of 90°.