Question

A small piece of wire bent into an L shape, with upright and horizontal portions of equal length, is placed with the horizontal portion along the axis of the concave mirror whose radius of curvature is $10\text{ cm}$ . If the bend is $20\text{ cm}$ from the pole of the mirror, then the ratio of the lengths of the image of the uprights and horizontal portions of the wire is
A. \[1:2\]
B. \[3:1\]
C. \[1:3\]
D. \[2:1\]

Answer 1

Hint: The upright part of the wire laterally magnified and the horizontal part is longitudinally magnified. The ratio of the lengths of the image of the uprights and horizontal portions of the wire can be obtained by finding the ratio of lateral magnification of the wire to the longitudinal magnification.
Formula used:
The mirror equation:
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
Where v is the image distance from the mirror, $u$ is the object distance and $f$ is focal length of the mirror.
Lateral magnification m is given by:
m = $-\dfrac{v}{u}$
Longitudinal magnification l is given by:
 l = \[{{m}^{2}}\]

Complete answer:
Let's discuss and find out the correct answer.
Given data: object distance, u =\[-20\text{ cm}\][ negative sign implies the image is on the left side of the mirror]
Radius of curvature, R = \[10\text{ cm}\]
Therefore, Focal length, $f=-\dfrac{10}{2}=-5$ cm [ focal length is half of radius of curvature]
Now, from mirror formula, we get ,
$\begin{align}
  & \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} \\
 & \Rightarrow \dfrac{1}{v}+\dfrac{1}{-20}=\dfrac{1}{-5} \\
 & \Rightarrow \dfrac{1}{v}-\dfrac{1}{20}=-\dfrac{1}{5} \\
 & \Rightarrow \dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{5} \\
 & \Rightarrow \dfrac{1}{v}=\dfrac{5-20}{100} \\
 & \Rightarrow \dfrac{1}{v}=-\dfrac{15}{100} \\
 & \Rightarrow \dfrac{1}{v}=-\dfrac{3}{20} \\
 & \Rightarrow v=-\dfrac{20}{3} \\
\end{align}$
Therefore, image distance, $v=-\dfrac{20}{3}$ cm.
Now, substituting the value of v and u in lateral magnification formula, we get
$\begin{align}
  & m=-\dfrac{v}{u} \\
 & \Rightarrow m=-\dfrac{(\dfrac{-20}{3})}{-20} \\
 & \Rightarrow m=-\dfrac{-20}{3\times (-20)} \\
 & \Rightarrow m=-\dfrac{1}{3} \\
\end{align}$
And from longitudinal magnification formula ,we get ,
\[l={{m}^{2}}={{\left( \dfrac{1}{3} \right)}^{2}}=\dfrac{1}{9}\]
$\therefore $ Ratio of up-right portion of the image to the lateral portion of the image is$\dfrac{m}{l}=\dfrac{\dfrac{1}{3}}{\dfrac{1}{9}}=\dfrac{1}{3}\times \dfrac{9}{1}=\dfrac{3}{1}=3:1$

Hence, option B is the correct answer.

Note:
Sign convention is a very important aspect of solving mirror problems. ‘Coordinate geometry sign convention’ is used to solve the problem. By this convention,
1. Light rays fall on the mirror surface from the left side
2. All distances measured in the direction of incident light are taken with positive sign and those measured in the direction opposite to the direction of incident light are taken with negative sign.
3. The lengths of the object and the image measured upwards perpendicular to the principal axis are taken as positive, while those measured downwards are taken as negative.