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

`5cm`

Explanation

As is clear from Fig. end `B` lies at centre of curvature `C` of the mirror (`because PC = R = 2 f = 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)/(30) = (-3 + 1)/(30)`
`:. v = - 15 cm`. `:. PA' = 15 cm`.
Size of image `= AB' = 20 - 15 = 5 cm`