Q. A small piece of wire bent into an $L$ shape with upright and horizontal portions of equal lengths, is placed with the horizontal portion along the axis of the concave mirror whose radius of curvature is $10\ cm$. If the bend is $20\ cm$ from the pole of the mirror, then the ratio of the lengths of the images of the upright and horizontal portions of the wire is
A) $1\ :2$
B) $3\ :1$
C) $1\ :3$
D) $2\ :1$
Explanation
(b) Focal length of mirror $f = \frac{R}{2} = \frac{10}{2} = 5cm$
For part $PQ:$ transverse magnification
length of image $L_{1} = \left( \frac{f}{f - u} \right) \times L_{0}$
$= \left( \frac{- 5}{- 5 - ( - 20)} \right) \times L_{0} = \frac{- L_{0}}{3}$
For part $QR:$ longitudinal magnification
Length of image $L_{2} = \left( \frac{f}{f - u} \right)^{2}L_{0}$
$= \left( \frac{- 5}{- 5 - ( - 20)} \right)^{2} \times L_{0} = \frac{L_{0}}{9} \Rightarrow \frac{L_{1}}{L_{2}} = \frac{3}{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.