Question

an object of height 10 cm placed at 30cm in front of concave mirror of focal length 25cm find location and size of image

Answer 1

Given :

In concave mirror,

Object height = 10cm

Object distance = - 30cm

Focal length = - 25cm

To find :

The location and size of image

Solution :

using mirror formula that is,

The relationship between image distance, object distance and focal length of a spherical mirror is known as Mirror formula.

The mirror formula can be written as :

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

where,

• v denotes image distance.
• u denotes object distance.
• f denotes focal length.

substituting all the given values in the formula,

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

$\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{ - 30} = - \dfrac{1}{25} }$

$\leadsto{ \sf \dfrac{1}{v} - \dfrac{1}{30} = - \dfrac{1}{25} }$

$\leadsto{ \sf \dfrac{1}{v} = - \dfrac{1}{25} + \dfrac{1}{30} }$

$\leadsto{ \sf \dfrac{1}{v} = \dfrac{ - 6 + 5}{150} }$

$\leadsto{ \sf \dfrac{1}{v} = - \dfrac{1}{150} }$

$\leadsto{ \sf v = - 150 \: cm }$

thus, the position of image is -150 cm.

we know,

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

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

$\leadsto \sf \dfrac{h'}{10} = \dfrac{ - ( - 150)}{30}$

$\leadsto \sf \dfrac{h'}{10} = 5$

$\leadsto \sf h' = 50 \: cm$

thus size of image is 50cm.