Here,

h = 5cm = Height of image, h′ = Height of image = ?, u = – 25 cm, f = + 25 cm, v = ?

Using Lens formula :

1/v - 1/u = 1/f

1/v = 1/f + 1/u

1/v = 1/10 + 1/(-25)

1/v = 1/10 - 1/25

1/v = 5-2/50

1/v = 3/50

v = 50/3

∴ v = 16.6 cm

v = positive. This implies that the image formed at a distance of 16.6 cm behind the mirror is real and inverted.

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Power of lens

m = h'/h = v/u

h' = v/u x h

h' = 50/3 x 1/(-25) x 5

h' = -3.33

h′ = negative. Thus, the image formed is inverted.