Question

A slide with an image 4 cm $\times$ 2 cm is placed at a distance of 10 cm behind a converging lens and a clear image is formed on a screen 1.1 m from the slide. What is the size of the image on the screen?

Answer 1

Given u = -10cm

v = 1.1 m

So magnification $m=\frac{v}{u}=\frac{110}{-10cm}$

$m=-11$

hence, size of image will be $44cm\times 22cm$.