IPhO 1968 · Theory problem 3

Parallel rays refracted by a glass semi-cylinder

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Parallel light rays are falling on the plane surface of a semi-cylinder made of glass, at an angle of \(45^\circ\), in such a plane which is perpendicular to the axis of the semi-cylinder (Fig. 4). (Index of refraction is \(\sqrt{2}\).) Where are the rays emerging out of the cylindrical surface?

Figure 1

Figure 4

Official English text of the problem.

Answers

  • Problem 3. Measure the central angle \(\phi\) clockwise from the right-hand end of the horizontal diameter. The rays emerge through the arc bounded by \[ \boxed{\phi=15^\circ\quad\text{and}\quad\phi=105^\circ.} \] This arc subtends \(90^\circ\) and has length \(\pi R/2\), where \(R\) is the cylinder’s radius. The limiting rays emerge tangentially.

Solution

Problem 3. We use geometrical optics, take the surrounding air’s refractive index as \(1\), and neglect partial reflections when transmission is possible. Total internal reflection must, however, be included.

At the flat surface, Snell’s law gives the refracted angle \(r\), measured from the normal: \[ \sin45^\circ=\sqrt2\sin r. \] Hence \[ \boxed{r=30^\circ}. \] The rays inside the glass therefore travel downwards and to the right, at \(60^\circ\) below the horizontal.

Let \(O\) be the centre of the semicircular section. At a point \(P\) on the curved surface, its outward normal is the radius \(OP\). Define \(\phi\) as the angle of this radius measured clockwise from the horizontal direction to the right. For a ray reaching this surface directly, its incidence angle is \[ i=\left|60^\circ-\phi\right|. \]

The critical angle \(i_{\mathrm c}\) for passage from glass to air satisfies \[ \sin i_{\mathrm c}=\frac{1}{\sqrt2}, \qquad \boxed{i_{\mathrm c}=45^\circ}. \] Consequently, emergence is possible when \[ |60^\circ-\phi|\le45^\circ, \] or \[ \boxed{15^\circ\le\phi\le105^\circ}. \]

Thus the emitting arc begins \(15^\circ\) below the right-hand end of the diameter and ends \(15^\circ\) past the lowest point, towards the left. Its angular extent and length are \[ \Delta\phi=90^\circ=\frac{\pi}{2}, \qquad \boxed{s=\frac{\pi R}{2}}. \]

At either boundary, Snell’s law gives an emergence angle of \(90^\circ\) relative to the normal, so the limiting outgoing rays are tangent to the circle. At \(\phi=60^\circ\), incidence is normal and the ray emerges without changing direction.

We must also check rays that initially undergo total internal reflection. Let \(a\) be a ray’s entry coordinate along the diameter, measured from \(O\), positive to the right. Its perpendicular distance from \(O\) is \[ b=|a|\cos30^\circ=\frac{\sqrt3}{2}|a|. \] At a circular-surface encounter, \[ b=R\sin i. \] Specular reflection at a circle preserves this distance and hence the incidence angle at subsequent circular encounters. Any reflection at the flat diameter also preserves \(b\), because that diameter passes through \(O\). Therefore a ray initially totally reflected at the curved surface cannot subsequently escape through another part of that surface.

The rays contributing to the emergent arc are precisely those entering within \[ \frac{\sqrt3}{2}|a|\le\frac{R}{\sqrt2}, \qquad \boxed{|a|\le R\sqrt{\frac23}\approx0.816R}. \] The remaining incident rays add no further emitting region on the curved surface.

The solution was checked against the official solution.