Question

If an object of 10 CM high is placed at a distance of36 CM from concave mirror of focal length 12 CM find position, nature and hight of the image

Answer 1

1/v = 1/f - 1/u

=1/(-12) - 1/(-36)

= -2/36 = -1/18

therefore , v = -18

now magnification M = image height / object height = -v/u

i.e, image height = 10 x -(-18)/-36

= 10 x (-1/2)

= -5

Therefore, image would be real and inverted

Answer 2

height of the object, h1 = 10cm

object distance,u =- 36 cm

focal length= -12 cm

image distance, v =?

$mirror \: formula \: = \frac{1}{f} = \frac{1}{v} \ + \frac{1}{u}$

1/-12= 1/-36+ 1/v

1/v= 1/-12 - 1/-36 = 1/-12 + 1/36

1/v= -3+1/36

1/v= -2/36

1/v=1/-18

v = -18 cm

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

m = -(-18)/-36

m = 18/-36

m = 1/-2

m = -0.5

As magnification is negative the nature of the image is real and inverted