Fisika Study Center

This problem is taken from the second IphO Budapest, Hungary, Theoretical problems, conducted in 1968.

Problem 3
Parallel light rays are falling on the plane surface of a semi-cylinder made of glass, at an angle of 45°, in such a plane which is perpendicular to the axis of the semi-cylinder (Fig. 4). (Index of refraction is √2.) Where are the rays emerging out of the cylindrical surface?

Solution
Let us use angle φ to describe the position of the rays in the glass (Fig. 5). According to the law of refraction sin 45°/sin β = √2 , sin β = 0.5 , β = 30° . The refracted angle is 30° for all of the incoming rays. We have to investigate what happens if φ changes from 0° to 180°.

It is easy to see that φ can not be less than 60° (AOB ∠ = 60°). The critical angle is given by sin β_crit = 1/n = √2/2; hence β_crit = 45°. In the case of total internal reflection ACO ∠ = 45° , hence φ = 180° − 60° − 45° = 75° . If φ is more than 75° the rays can emerge the cylinder. Increasing the angle we reach the critical angle again if OED ∠ = 45° . Thus the rays are leaving the glass cylinder if:

75°<φ<165°,

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

Sources/Literatures :
-Waldemar Gorzkowski, Institute of Physics, Polish Academy of Sciences, Warsaw, Poland
-Péter Vankó, Institute of Physics, Budapest University of Technical Engineering, Budapest, Hungary, Problems of the 2nd and 9th International Physics Olympiads (Budapest, Hungary, 1968 and 1976)
-The Olympic home page www.jyu.fi/ipho