Question

A thin rod of length f/3  lies along the axis of a concave mirror of focal length f. One end of its magnified image touches an end of the rod. The length of the image is

(a)  f

(b)  1/2 f

(c)  2f

(d)  1/4 f

Answer 1

Correct option  (b)  1/2 f

Explanation :

Since image of B­ - B' comcides

So, B is must be at 2f and to A' and A 

1/v + 1/u = 1/f  ⇒ 1/v - 3/5f = -1/f

v = -f/2

So A 'B' = 5f/2 - 2f = f/2