Aptitude Test Problems in Physics Science for Everyone by S Krotov ( PDFDrive.com )

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 hole and undergoing re-
fraction at the left base of the cylinder will prop-
agate 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 hole the rays within
a solid angle of 2n sr.
In the absence of the cylinder, the luminous
flux confined in a solid angle of nd 4 /(41) 1 gets into
the hole in the screen. Therefore, in the presence
of the transparent cylinder, the luminous flux
through the hole will increase by a factor of

= 8 X 10 4.

ad 2 /(40 1

4.9. The thickness of the objective lens can be found

from geometrical considerations (Fig. 235). Indeed,

r2

ri = (MI —h) h pe, 211^1 h, h.

2Ri

where R 1 is the radius of curvature of the objective.

Fig. 235

Let us write the condition of equality of optical

paths A BF and CD F for the case when the telescope

90*