Question

A 8.0-cm-diameter disk emits light uniformly from its surface. 20 cm from this disk, along its axis, is an 9.0-cm-diameter opaque black disk; the faces of the two disks are parallel. 20 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

To solve this problem draw the following diagram, sketched to an appropriate scale:

Draw a horizontal line of length 40cm representing the central axis of the system, marked with points A, B, C representing the positions of the 2 discs and the screen.

Represent the radius of the light source as a vertical line Aa of length 4cm, and the radius of the opaque disc as a vertical line Bb of length 4.5cm. Erect a vertical line at C representing the screen. Project line ab to the position where the screen is, and mark the point of intersection of this line with the screen as c. Cc is the radius of the completely dark part of the shadow of the opaque disc.

Line ab declines by 2cm in the 20cm distance between A and B, and will decline a further distance of 20*2/20 = 2cm over the span BC. The length of Cc is therefore 6 - (2 + 2) = 2cm. This is the radius of the completely dark part of the shadow. The required diameter is therefore 4cm

To convince yourself that c is the limit of the completely dark shadow, chose any point on Aa and project a line from the chosen point via b to the screen. You will see that the intersection of this line with the screen is at a greater radius than c, so points at greater radius than c will be illuminated by the light source and will therefore not be in complete shadow.