Answer to Question #140588 in Optics for sharfa

Question

A screen is placed 90 cm from an object. When a converging lens is moved between the screen and the object , at two different positions images can be produced on the screen. If the distance between the positions of the lens is 20 cm , find the focal length of the lens

Answer 1

d = 20 cm

u + v = 90 cm

v – u = 20 cm

v = 55 cm

u = 35 cm

Lens formula:

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{55} - \frac{1}{-35} = \frac{1}{f}$

$\frac{1}{55} + \frac{1}{35} = \frac{1}{f}$

$\frac{35 + 55}{1925} = \frac{1}{f}$

f = 21.38 cm

Answer: 21.38 cm