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. $\frac { 1 } { 2 } f$
• C. 2 f
• D. $\frac { 1 } { 4 } f$

Answer 1

Answer: (B)

$\frac { 1 } { 2 } f$

Answer 2

If end A of rod acts an object for mirror then it's image will be A' and if $u = 2 f - \frac { f } { 3 } = \frac { 5 f } { 3 }$

So by using $\Rightarrow \frac { 1 } { - f } = \frac { 1 } { v } + \frac { 1 } { \frac { - 5 f } { 3 } }$ $\Rightarrow v = - \frac { 5 } { 2 } f$

$= \frac { 5 } { 2 } f - 2 f = \frac { f } { 2 }$