Answer to Question #122031 in Physics for Yasien mahmoud

Question

A 0.2 cm tall object lies 10 cm from a 25 cm focal length magnifying glass. Find the location, magnification, and size of the image. Is it erect or inverted? Real or virtual?

Answer 1

The location of the image will be, according to thin lens equation:

$$i=\frac{of}{o-f}=\frac{10\cdot25}{10-25}=-16.7\text{ cm}.$$

Therefore, the image is virtual.

The magnification is:

$$M=-\frac{i}{o}=-\frac{-16.7}{10}=1.67.$$

The height of the erect image is

$$h_i=h_oM=0.2\cdot1.67=0.334\text{ cm}.$$