Question

3. An object of height 2 cm is placed at a distance
20 cm in front of a concave mirror of focal length
12 cm. Find the position, size and nature of the
image.

How find the image height.
​

Answer 1

Explanation:

h1 = 2 CM

f = -12 CM

u = - 20 CM

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

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

$\frac{1}{v \:} = \frac{1}{ - 20} + \frac{1}{ - 12}$

$\frac{1}{v} = \frac{ - 32}{240}$

$v = \frac{ 240}{ - 32} = - 7.5$

v is negative image formed is real, behind the mirror and inverted

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

$= - \frac{ - 7.5}{20} = 0.375$

m = magnification

$m \: = - \frac{h2}{h1}$

$0.375 = \frac{h2}{2}$

h 2 = - 0.75

Answer 2

Given ,

focal length (f ) = - 12 cm

object distance ( u ) = -20 cm

use formula ,

1/u + 1/ v = 1/ f

1/(-20) + 1/v = 1/(-12)

1/v = 1/20 - 1/12

=-8/20 x 12 = - 1/30

v = - 30 cm

hence image form 30 cm distance from the pole of mirror from left side .

m = -v/u = height of image / height of object

-(-30)/(-20) = height of image /2 cm

height of image = -3 cm

hence image form left side from the pole below the x -axis and image is real inverted and larger then object .

☺️☺️

#bebrainly

#spamfreebrainly