Example from Physics, 16.3 Lenses

Image Formation by a Concave Lens

Suppose an object, such as a book page, is held 6.50 cm from a concave lens with a focal length of –10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. What magnification is produced?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
This example is identical to the preceding one, except that the focal length is negative for a concave or diverging lens. The method of solution is therefore the same, but the results are different in important ways.
Solution
$d_{i} = \frac{f d_{o}}{d_{o} − f} = \frac{(− 10.0) (6.50)}{6.50 - (− 10.0)} = - 3.94 cm$
16.20

Now the magnification equation can be used to find the magnification, m, because both d_i and d_o are known. Entering their values gives

$m = - \frac{d_{i}}{d_{o}} = - \frac{- 3.94}{6.50} = 0.606.$
16.21

0 of 2 steps open

How did it go?

Strategy

Solution