Question

A light ray enters a solid glass sphere of refractive index \(\mu = \sqrt{3}\) at an angle of incidence \(60^\circ\). The ray is both reflected and refracted at the farther surface of the sphere. The angle (in degrees) between the reflected and refracted rays at this surface is ________.

Hint:

Find the refraction angle at the first surface; by symmetry of the chord it equals the angle of incidence at the farther surface. Apply Snell's law again and use the geometry of reflected and refracted rays.

Answer 1

Correct Answer: 90

At the first surface: \(\sin 60^\circ = \sqrt{3}\,\sin r\) so \(\sin r = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}\), hence \(r = 30^\circ\).
The two radii to the entry and exit points and the chord form an isosceles triangle, so the angle of incidence at the farther surface is also \(i' = r = 30^\circ\).
Refraction at the farther surface (glass to air): \(\sqrt{3}\sin 30^\circ = \sin e\) so \(\sin e = \frac{\sqrt{3}}{2}\), hence \(e = 60^\circ\).
The reflected ray makes \(30^\circ\) with the normal (inside the sphere) and the refracted ray makes \(60^\circ\) with the normal (outside, on the other side of the surface), both lying on the same side of the normal line as seen along the surface.
Angle between them \(= 180^\circ - 30^\circ - 60^\circ = 90^\circ\).
Answer: 90