Question

An object of height 10cm is kept at a distance of 150cm in front of a concave mirror having radius of curvature 2m. What is the nature,height and distance of image formed by the mirror ?

Answer 1

Step 1: Given

Mirror is concave.

Height of the object, $h_o=10cm$

Object distance, $u=150cm$

Radius of curvature, $R=2m=200cm$

Focal length, $f=\frac{R}{2}=100cm$

Image distance,$v$

Image height, $h_i$

Step 2: Formula used

Mirror formula

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

Magnification

$m=\frac{h_i}{h_o}=-\frac{v}{u}$

Step 3: Calculation

Using mirror formula

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

$\frac{-1}{100}=\frac{1}{v}+\frac{-1}{150}$

$\frac{1}{v}=\frac{-1}{100}+\frac{1}{150}$

$\frac{1}{v}=\frac{-1}{300}$

$v=-300cm$

Now, using magnification formula

$m=-\frac{v}{u}$

$m=-\frac{-300}{-150}$

$\frac{h_i}{h_o}=-3$

$h_i=-3\times 10$

$h_i=-30cm$

So,

Image height is $30cm$.

Image is formed at $300$ cm in front of mirror.

Image is real and inverted.