Question

A 6cm tall object is placed perpendicular to the principal axis of a concave mirror of focal length 30 cm the distance of the object from the mirror is 45 CM use mirror formula to determine the position nature and size of the image formed

Answer 1

1/f = 1/v + 1/u

f= -30cm

u= -45cm

1/v = 1/f-1/u == -1/30+1/45 = -1/90

Therefore , V = -90cm

Real image

Size == hi/ho = -v/u == hi = -12cm

Height of image = 12cm below the principal axis

Answer 2

Answer:

The distance of the image is 90 cm, size of the image is 12 cm and enlarged.

Explanation:

h = 6cm

f = -30cm

u = -45

1/f = 1/v + 1/u

1/v = 1/f - 1/u

1/v = 1/ ( -30) - 1/ (-45)

1/v = 1/-90

v = -90cm from the pole of the mirror

v = -90cm from the pole of the mirrorSize of the image

m = -v /u

= -90/45

= -2

ux ☞ m = h'/h = h'

= -2×6 = -12 cm

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