Answer to Question #263674 in Physics for Karisa

Question

You place an object 22 cm in front of a concave lens with a focal length of -35 cm. Find the image distance and magnification that results from this arrangement.

Answer 1

(a) We can find the image distance from the thin lens equation:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$\dfrac{1}{d_i}=\dfrac{1}{f}-\dfrac{1}{d_o},$$ $$d_i=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{d_o}}=\dfrac{1}{\dfrac{1}{-35\ cm}-\dfrac{1}{22\ cm}}=-13.5\ cm.$$

The sign minus means that the image is virtual (it appears on the same side as the object).

(b) We can find the magnification of the lens as follows:

$$m=-\dfrac{d_i}{d_o}=-\dfrac{-13.5\ cm}{22\ cm}=0.61.$$

The sign plus means that the image is upright.