Question

1. A 20-cm-diameter disk emits light uniformly from its surface. 40 cm from this disk, along its axis, is a 16.0-cm-diameter opaque black disk; the faces of the two disks are parallel. 40 cm beyond the black disk is a white viewing screen. The lighted disk illuminates the screen, but there's a shadow in the center due to the black disk. What is the diameter of the completely dark part of this shadow?

Answer 1

In triangle ABC tan theta = AB/AC tan theta = 2/40 equation - 1 In triangle CDE CD = 8 - r DE = 40 tan theta = CD/DE tan theta = (8 - r)/40 equation - 2 using equation - 1 and equation - 2 2/40 = (8 - r)/40 r = 6 cm d = diameter = 2 r = 2 times 6 = 12 cm

In triangle ABC

tan$\theta$ = AB/AC

tan$\theta$ = 2/40                                                eq-1

In triangle CDE

CD = 8 - r

DE = 40

tan$\theta$ = CD/DE

tan$\theta$ = (8 - r)/40                                        eq-2

using eq-1 and eq-2

2/40 = (8 - r)/40

r = 6 cm

d = diameter = 2 r = 2 x 6 = 12 cm