Question

A convex driving mirror of focal length 20 cm, is fitted in a motor car. If the second car 2 m broad and 1.6 m high is 6 m away from the first car and overtakes the first car at a relative speed of 15 m/s, then how fast will the image be moving?
A) $0.016m/s$
B) $0.0257m/s$
C) $0.162m/s$
D) $0.0073m/s$

Answer 1

A convex mirror converges the parallel lights at the focus of the mirror. As the mirror moves the focal point also moves and the converging point also moves. But similarly, if the object also moves then the relative distance is also changed eventually with relative speed. So differentiation is an expected operation in this case.

Formulae Used:
If the distance of the mirror from the object is $u$, the distance of the mirror from the image is $v$ and the focal length is $f$, then the mirror formula is
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$

Complete step by step answer:
Given:
The focal length of the convex mirror is $f = 20cm$.
The second car is $2m$ broad and $1.6m$ high.
The second car is at a distance of $6m$ from the first car.
The second car overtakes the first one at a relative speed $15m/s$.
To get: The speed of the image.
Step 1:
The distance of the object from the mirror is $u = - 6m = - 600cm$.
Rewrite the expression eq (1) to calculate the distance of the image from the mirror.
$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} \\\ \Rightarrow \dfrac{1}{v} = \left[ {\dfrac{1}{{20}} - \left( { - \dfrac{1}{{600}}} \right)} \right]c{m^{ - 1}} \\\ \Rightarrow \dfrac{1}{v} = \left( {\dfrac{1}{{20}} + \dfrac{1}{{600}}} \right)c{m^{ - 1}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{{30 + 1}}{{600}}c{m^{ - 1}} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{{31}}{{600}}c{m^{ - 1}} \\\ \Rightarrow v = \dfrac{{600}}{{31}}cm \\\ \therefore v = 19.355cm \\\$

Step 2:
Now, the eq (1) can be written in terms of the velocities by differentiating eq (1):
$\dfrac{d}{{dt}}\left( {\dfrac{1}{v}} \right) = \dfrac{d}{{dt}}\left( {\dfrac{1}{f} - \dfrac{1}{u}} \right) \\\ \Rightarrow - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} = 0 + \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}} \\\ \Rightarrow \dfrac{{dv}}{{dt}} = - \dfrac{{{v^2}}}{{{u^2}}}\dfrac{{du}}{{dt}} \\\$; as the focal length $f$ is constant for a mirror.
Step 3:
You have, by the problem, the velocity of the object $\dfrac{{du}}{{dt}} = 15m/s$.
So, put the values and calculate the speed of the image $\dfrac{{dv}}{{dt}}$
$\dfrac{{dv}}{{dt}} = - \dfrac{{{v^2}}}{{{u^2}}}\dfrac{{du}}{{dt}} \\\ \Rightarrow \dfrac{{dv}}{{dt}} = - {\left( {\dfrac{{19.355}}{{600}}} \right)^2}15m/s \\\ \Rightarrow \dfrac{{dv}}{{dt}} = - {\left( {\dfrac{1}{{31}}} \right)^2}15m/s \\\ \Rightarrow \dfrac{{dv}}{{dt}} = - \dfrac{1}{{961}} \times 15m/s \\\ \Rightarrow \dfrac{{dv}}{{dt}} = - 0.00104 \times 15m/s \\\ \Rightarrow \dfrac{{dv}}{{dt}} = - 0.0156m/s \\\ \therefore \dfrac{{dv}}{{dt}} = - 0.016m/s \\\$

A convex driving mirror of focal length 20 cm, is fitted in a motor car. If the second car 2 m broad and 1.6 m high is 6 m away from the first car and overtakes the first car at a relative speed of 15 m/s, then the image will be moving with $0.016m/s$. Hence, option (A) is correct.

Note:
The distance of the object from the mirror is of opposite signature from the distance of the image and the focal length. This is because the light comes from the object to the mirror and gets reflected in the opposite direction. So, the direction of $u$ should be carefully taken. The relative velocity is the velocity with which the object drifts away and hence you should take that as the velocity of the object itself.