Question

A car is fitted with a convex side-view mirror of focal length 20cm. A second car 2.8m behind the first car is overtaking the first car at a relative speed of 15m/s. The speed of the image of the second car as seen in the mirror of the first one is:
A. 10m/s
B. 15m/s
C. $\dfrac{1}{{10}}$m/s
D. $\dfrac{1}{{15}}$m/s

Answer 1

Hint: Object distance, image distance and focal length of the mirror are related using the mirror equation. Find the image distance and then differentiate the mirror equation to find the image speed with focal length as constant.

Complete answer:
First list the details provided and to be calculated. Let the object (second car) distance be $o = - 2.8m = - 280cm$ and focal length, $f = 20cm$. The focal length of a convex mirror is positive according to sign convention for mirrors and thus object distance is negative. Relative speed of the car behind can be written as the rate of change of object distance, $= \dfrac{{do}}{{dt}} = 15m{s^{ - 1}} = 1500cm{s^{ - 1}}$.
We know the mirror equation as $\Rightarrow \dfrac{1}{o} + \dfrac{1}{i} = \dfrac{1}{f}$; where $i$ is the image distance. Substituting the values given, we will find the image distance.
$\Rightarrow \dfrac{1}{{ - 280}} + \dfrac{1}{i} = \dfrac{1}{{20}}$
$\Rightarrow \dfrac{1}{i} = \dfrac{1}{{20}} + \dfrac{1}{{280}} = \dfrac{{15}}{{280}}$
Differentiating the mirror equation on both sides with respect to time $t$,
$\Rightarrow - \dfrac{1}{{{o^2}}}\dfrac{{do}}{{dt}} - \dfrac{1}{{{i^2}}}\dfrac{{di}}{{dt}} = 0$. Right hand side is zero as the focal length remains constant in the exercise. So substituting the known values in the above derived equation, we will be calculating the speed of the image in the mirror,
$\Rightarrow \dfrac{{di}}{{dt}} = - {\left( {\dfrac{i}{o}} \right)^2}\dfrac{{do}}{{dt}} = - {\left( {\dfrac{{280}}{{15 \times - 280}}} \right)^2} \times 1500cm{s^{ - 1}}$
$\Rightarrow \dfrac{{di}}{{dt}} = - \dfrac{1}{{15}}m{s^{ - 1}}$
Please note the units in the last two steps. The negative sign shows that the image is moving in the direction opposite to the incident light. Comparing this magnitude with the choices provided, we get that option D is the correct one.

Note: Whenever doing operations using the mirror equation it is preferred to understand the sign convention of measuring distances with respect to the pole of the mirror. The focal length is negative for the concave mirror.