Question

At what distance from a concave mirror of focal length 10 cm should an object 2 cm long be placed in order to get an exerct of 6 cm tall

Answer 1

f = -10 cm

h = 2cm

h' = 6cm

1/f = 1/v + 1/u

h'/h=-v/u

6/2 = -v/u

3 = -v/u

-3u = v

-1/10 = -1/3u + 1/u

-1/10 = (-1 + 3)/3u

-1/10 = 2 / 3u

3u = -20

u = -20/3 cm

u = -6.67 cm