Question

A convex mirror of focal length $20\text{cm}$ is fitted to a motor car. If the second car with width $2\text{m}$ and height $1.6\text{m}$ is $6\text{m}$ behind the first car and overtakes the first car at a relative speed of $15\text{m/s}$, then how fast will the image of the second car in the convex mirror of first car be moving?

• A. $0.016\text{m/s}$
• B. $0.026\text{m/s}$
• C. $0.036\text{m/s}$
• D. $0.046\text{m/s}$

Answer 1

The correct option is A $0.016\text{m/s}$

For the given convex mirror of first car the second car is $6\text{m}$ behind,
$u=-6\text{m}=-600\text{cm}$
$f=+20\text{cm}$

Using mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}-\frac{1}{600}=\frac{1}{20}$
$\Rightarrow \frac{1}{v}=\frac{31}{600}$
$\text{or},v=\frac{600}{31}\text{cm}$

Using the relation for magnification produced by mirror,
$m=\frac{-v}{u}=\frac{(-\frac{600}{31})}{-600}=\frac{1}{31}$

Also, velocity of the image when the object is moving along the principal axis is given by,
$v_i=-m^2{v}_o$
Here $v_0=15\text{m/s}$ is the relative velocity of object w.r.t mirror.

$\Rightarrow v_i=-{\left(\frac{1}{31}\right)}^2\times 15\simeq -0.016\text{m/s}$

Hence, the speed of the image will be $0.016\text{m/s}$.

$\begin{array}{c}\text{Why this question ?}\\ \text{Caution: Dimension of object is given to}\\ \text{confuse the students.}\\ \text{Tips: Remember that we consider only point}\\ \text{object in the formula of image velocity}\\ \text{and ignore the dimensions.}\end{array}$