Page 413 - Schaum's Outline of Theory and Problems of Applied Physics

398 LENSES [CHAP. 32

The above formula can be rewritten in the more convenient form
f 1 f 2
f =
f 1 + f 2
Since f 1 = 10 cm and f 2 =−20 cm,

(10 cm)(−20 cm)
f = =+20 cm
10 cm − 20 cm
The combination acts as a converging lens of focal length +20 cm.

RAY TRACING
As with a spherical mirror, the position and size of the image of an object formed by a lens can be found by
constructing a scale drawing. Again, what is done is to trace two different light rays from a point of interest in
the object to where they (or their extensions, in the case of a virtual image) intersect after being refracted by the
lens. Three rays especially useful for this purpose are shown in Fig. 32-3; any two of these are sufficient.
1. A ray that leaves the object parallel to the axis of the lens. After refraction, this ray passes through the
far focal point of a converging lens or seems to come from the near focal point of a diverging lens.
2. A ray that passes through the focal point of a converging lens or is directed toward the far focal point of
a diverging lens. After refraction, this ray travels parallel to this axis of the lens.
3. A ray that leaves the object and proceeds toward the center of the lens. This ray is not deviated by
refraction.

SOLVED PROBLEM 32.6

What is the nature of the image of a real object formed by a diverging lens?
It is virtual, erect, and smaller than the object, as in Fig. 32-3(b).

F = near focal point
F′ = far focal point
Object Ray 1 Object Ray 1
Ray 2
Ray 3
Ray 3
F′ F
F Virtual F′
image
Ray 2
Real
image
(a) (b)

Fig. 32-3

SOLVED PROBLEM 32.7
An object is placed at the focal point of a converging lens. Is an image formed?

As Fig. 32-4 shows, the refracted rays are parallel to each other, and so no image is formed.