Question

An Object is placed at a distance of 20 cm of a concave mieror of 5 cm focal length if height of the object is 3 cm then what will be the height of the inmage of the object

Answer 1

Given:

Distance of the object (u) = 20 cm

Focal length of the mirror (f) = 5 cm.

The height of the object (O) = 3 cm.

To find:

The height of the image(I)

Solution:

Distance of the object (u) = 20 cm.

Focal length of the mirror (f) = 5 cm.

Distance of the image(v) = ?

From the formula we get that,

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

$or, \dfrac{1}{5} = ( \dfrac{1}{20} + \dfrac{1}{v})$

$or, ( \dfrac{1}{20} + \dfrac{1}{v} ) = \dfrac{1}{5}$

$or, \dfrac{1}v} = ( \dfrac{1}{5} - \dfrac{1}{20} )$

$or, \dfrac{1}{v} = ( \dfrac{4-1}{20})$ [ ∵ The L.C.M. of the denominators is 20.]

$or, \dfrac{1}{v} = \dfrac{3}{20}\\ \\or, v = \dfrac{20}{3}$

∴ The distance of image is (20/3) cm in front of the mirror.

We know that,

$Magnificent(m) =( \dfrac{-v}{u})$

$= \dfrac{\dfrac{-20}{3} }{20}$

= (-20 ×20)/3

$=( \dfrac{-400}{3})$

[ Since, magnificent is a ratio so, it has no unit.]

Magnificent(m) = (-400/3)

Height of the object(O) = 3 cm.

Height of the image(I) = ?

From the formula we get that,

$m = \dfrac{I}{O} \\\\or, (\dfrac{-400}{3} ) = \dfrac{I}{3}\\ \\or, \dfrac{I}{3} =( \dfrac{-400}{3})$

or, I = (-400 ×3) /3

or, I = -400 cm.

∴ The image is inverted and the size of the image is 400 cm.

Answer:

           The size of the image is 400 cm and it is inverted.