Given h0 = 5 cm, u = – 25 cm, f = 10 cm, v = ?
By lens formula
1/v - 1/u = 1/f …. (i)
By putting the given values in eq. (i) with their sign conventions
1/v = 1/10 - 1/25
We get, v = 50/3 cm
image is formed at 50/3 cm away from the mirror on right side i.e. real
m = v/u = h1/h0
-50/3/25 = h1/5
We get, h1 = -10/3
Here, m is less than unity and also negative so image will be diminished, inverted and real in nature.
