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. 2f
• D. $\frac { 1 } { 4 }$f

Answer 1

Answer: (B)

$\frac { 1 } { 2 }$f

Answer 2

If end A of rod acts on object for mirrors then its image will be A¢

u = 2f – =

by using mirrors formula

$\frac { 1 } { v } + \frac { 1 } { \frac { - 5 f } { 3 } }$̃

; |v| =

Length of image = $\frac { 5 \mathrm { f } } { 2 } - 2 \mathrm { f } = \frac { \mathrm { f } } { 2 }$