Question

A concave mirror has as a radius of curvature of 0.4m.find the position and size if the image of an object 0.2m high placed 0.8 m in front of the mirror

Answer 1

Explanation:

Radius of curvature, R = 0.4 m

So, focal length, f = 0.4/2 = -0.2 m

Object distance, u = -0.8 m

Height of object, h = 0.2 m

Using mirror's formula :

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

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

$\dfrac{1}{v}=\dfrac{1}{-0.2}-\dfrac{1}{-0.8}$

v = -0.26 m

So, the image distance is 0.26 m

The magnification of mirror is given by :

$m=\dfrac{-v}{u}=\dfrac{h'}{h}$

$h'=\dfrac{-vh}{u}$

$h'=\dfrac{-(-0.26)\times 0.2}{-0.8}$

h' = -0.065 m

h' = -0.07 m

Hence, the size of the image is 0.07 m

Answer 2

thanks for giving me the opportunity