Question

A virtual image is formed 17.0 cm from a concave mirror having a radius of curvature of 39.0 cm. (a) Find the position of the object. cm in front of the mirror
(b) What is the magnification of the mirror?

Answer 1

Explanation:

Image distance, v = -17 cm (-ve for virtual image)

Radius of curvature of concave mirror, R = 39 cm

Focal length, f = -19.5 cm (-ve for a concave mirror)

(a) Using mirror's formula as :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

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

$\dfrac{1}{u}=\dfrac{1}{-19.5}-\dfrac{1}{(-17)}$  

u = 132.6 cm    

So, the object is placed 132.6 cm in front of the mirror.

(b) Magnification of the  mirror, $m=\dfrac{-v}{u}$

$m=\dfrac{-17}{132.6}$

m = -0.128

Hence, this is the required solution.