A rod `AB = 10 cm` in length is placed along the principal axis of a concave mirror having focal length equal to `10 cm` as shown in Fig. The distance `PB = 20 cm`. What is the length of the image of the rod `AB` ?

Answer

5 cm

Explanation

As is clear from Fig., end `B` lies at centre of curvature `C` of the mirror `(because PC = R = 2f = 20 cm)`. Therefore, its image is formed at `B'` itself (i.e., at centre of curvature C only).
For end `A, u = -(20 + 10) = -30 cm`,
`f = -10 cm`
As `(1)/(v)+(1)/(u)=(1)/(f)`
`:. (1)/(v)=(1)/(f)-(1)/(u)=(1)/(-10)+(1)/(3)=(-3 + 1)/(30)`
`:. v = -15 cm`. `:. PA' = 15 cm`
Size of image `=A'B' = 20 - 15 = 5 cm`.