Question

an object of 2cm hight is placed 0.4m away from a concave mirror on its principal axis whose radius of curvature is 60cm, find the position , nature and size of image ?​

Answer 1

Given that,

Height of object =2 cm

Distance of object = 0.4 m = 40 cm

Radius of curvature = 60 cm

We need to calculate the focal length

Using formula of radius of curvature

$R=2f$

$f=\dfrac{R}{2}$

Put the value into the formula

$f=\dfrac{60}{2}$

$f=30\ cm$

We need to calculate the image distance

Using formula of mirror

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

Put the value into the formula

$\dfrac{1}{-30}=\dfrac{1}{v}+\dfrac{1}{-40}$

$\dfrac{1}{v}=\dfrac{1}{-30}+\dfrac{1}{40}$

$\dfrac{1}{v}=-\dfrac{1}{120}$

$v=-120\ cm$

The position of image is 120 cm from the concave mirror.

The image obtained is real image.

We need to calculate the height of image

Using formula of magnification

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

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

Put the value into the formula

$\dfrac{h'}{2}=\dfrac{-(-120)}{-40}$

$h'=-30\times2$

$h'=-60\ cm$

The height of image is 60 cm enlarge and inverted.

Hence,  The position of the image is 120 cm from mirror and the image is real.

The size of image is 60 cm  and the image is enlarge and inverted.