Let us prove that after refraction all the rays emerging from the point source towards the screen will lie within the cylinder, and after multiple reflections from the lateral surface of the cylinder, will ultimately pass through the **** in the screen.
Indeed, the extreme ray from the point source , whose angle of incidence on the left base of the cylinder is pi/2, after refraction will form an angle α with the cylinder axis such that sin α = 1/n (according to Snell' law). The angle of incidence φ of such a ray on the lateral surface of the cylinder satisfies the condition α + φ = pi/2 (Fig. 234).
Since
α < pi/4 and φ > pi/4, i.e. the angle of incidence on the lateral surface of the cylinder will be larger than the critical angle of total internal reflection. Therefore, this ray cannot emerge from the cylinder at any point other than that lying on the right base. Any other ray emerging from the source towards the screen with a and undergoing refraction at the left base of the cylinder will propagate at a smaller angle to the axis, and hence will be incident on the lateral surface at an angle exceeding the critical angle. Thus, the transparent cylinder will "converge" to the the rays within a solid angle of 2pi sr.
In the absence of the cylinder, the luminous flux confined in a solid angle of pid2/(4l)2 gets into the in the screen. Therefore, in the presence of the transparent cylinder, the luminous flux through the will increase by a factor of
