Answer to Question #343817 in Optics for Joe

Question

An object is placed 10 cm infront of a diverging lens of focal length 11cm. where is the image located?Is the image real or virtual? What is the magnification?

Answer 1

The diverging thin lens equation says

$$\frac{1}{d_o}+\frac{1}{d_i}=-\frac{1}{f}$$

Hence

$$\frac{1}{d_i}=-\frac{1}{f}-\frac{1}{d_o}$$

$$\frac{1}{d_i}=-\frac{1}{11}-\frac{1}{10}$$

$$d_i=-5.2\:\rm cm$$

Image is virtual.

The magnification

$$m=-\frac{d_i}{d_o}=-\frac{-5.2}{10}=0.52$$