Question

A man running at a speed 5 m/s is viewed in the side view mirror of the radius of curvature $R = 2m$ of a stationary car. Calculate the speed of image when the man is at a distance of 9m from the mirror
(A) $0.3m/s$
(B) $0.2m/s$
(C) $0.1m/s$
(D) $0.05m/s$

Answer 1

The side view mirror of a car is a convex lens, so the image distance and focal length are negative. We need to investigate the change in distance after one second as seen in the mirror.

Formula used: In this solution we will be using the following formula;
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ where $f$ is the focal length, $v$ is the image distance in the mirror, $u$ is the object distance.
$f = \dfrac{R}{2}$ where $R$ is the radius of curvature.
$s = \dfrac{d}{t}$ where $s$ is the speed of an object, $d$ is the distance covered by the object, and $t$ is the time taken to cover the distance.

Complete step by step solution:
A man is said to be moving at 5 m/s in real life, we are to calculate his speed as observed in the mirror at a distance of 9m from the mirror.
The mirror equation given by ; $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ where $f$ is the focal length, $v$ is the image distance in the mirror, $u$ is the object distance.
First the focal length is $f = \dfrac{R}{2}$ where $R$ is the radius of curvature. Then,
$f = \dfrac{2}{2} = 1m$
At the object distance of 9m, the mirror equation can be written as,
$- \dfrac{1}{1} = - \dfrac{1}{v} + \dfrac{1}{9}$ (because the image distance and focal length can be considered negative for a convex mirror which is the mirror used as side view mirror).
Hence, by making $\dfrac{1}{v}$ subject and calculating, we have
$\dfrac{1}{v} = \dfrac{1}{1} + \dfrac{1}{9} = \dfrac{{10}}{9}$
Inverting, we have
$v = \dfrac{9}{{10}} = 0.9m$
Now since speed is given by
$s = \dfrac{d}{t}$ where $d$ is the distance covered by the object, and $t$ is the time taken to cover the distance.
Hence, after 1 second, the distance covered would be
$d = 5 \times 1 = 5m$
Then the new object distance would be
$u = 9 - 5 = 4m$
Then the image distance at this point would be
$\dfrac{1}{v} = \dfrac{1}{1} + \dfrac{1}{4} = \dfrac{5}{4}$
$\Rightarrow v = \dfrac{4}{5} = 0.8m$
Hence, in the mirror the distance travelled in one second is
${d_i} = 0.9 - 0.8 = 0.1m$ which means speed is $0.1m/s$
Hence, the correct option is C.

Note:
Alternatively, we could decide to not put the negative before the image distance, but our answer will be negative to show that the image is virtual, as in;
$- \dfrac{1}{1} = \dfrac{1}{v} + \dfrac{1}{9}$
$\Rightarrow \dfrac{1}{v} = - \dfrac{1}{1} - \dfrac{1}{9} = - \dfrac{{10}}{9}$ .
The negative can simply be discarded.