Question

A police inspector is chasing a thief who is running away in a car. The speed of police jeep is 12m/s and the speed of thief car is 3m/s, then speed of image of police jeep as seen by thief in the rear-view mirror when police jeep is at a distance of 30m is (Given that focal length of rear-view mirror is 15m)
A. 1m/s
B. 2m/s
C. 3m/s
D. None of these

Answer 1

The question talks about an object placed on a mirror which will reflect back or be thrown back from the point of projection, the distance of the object from the mirror or the surface is the object distance and the distance of the image after it has been thrown back is known as the image distance.

Complete step by step answer:
We are given that a police inspector is chasing a thief who is running away in a car and the speed of the police jeep is 12m/s and the speed of the thief car is 3m/s and the police jeep is at a distance of 30m from the thief.
We have to find the speed of the police jeep as seen by the thief in the rear-view mirror when the police jeep is at a distance of 30m.
The mirror equation connects the object distance, image distance and the focal length together
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$, where f is the focal length, u is the object distance and v is the image distance.
The focal length (f) of the rear-view mirror is equal to 15m and the object distance (u) is equal 30m. The object distance of a rear-view mirror must be taken negative.
$\dfrac{1}{{15}} = \dfrac{1}{{ - 30}} + \dfrac{1}{v} \\\ \implies \dfrac{1}{v} = \dfrac{1}{{15}} + \dfrac{1}{{30}} \\\ \implies \dfrac{1}{v} = \dfrac{{2 + 1}}{{30}} = \dfrac{3}{{30}} \\\ \therefore v = \dfrac{{30}}{3} = 10m \\\$
The image distance would then be equal to 10m.
The relative speed (object speed) of jeep and car is ${V_{OM}} = 12 - 3 - 9m/s$
Differentiate the mirror equation with respect to time.
$\dfrac{d}{{dt}}\left( {\dfrac{1}{f}} \right) = \dfrac{d}{{dt}}\left( {\dfrac{1}{v} + \dfrac{1}{{ - u}}} \right) \\\ \implies 0 = \dfrac{d}{{dt}}\left( {\dfrac{1}{v}} \right) + \dfrac{d}{{dt}}\left( {\dfrac{1}{{ - u}}} \right) \\\ \left( {\because \dfrac{d}{{dx}}{x^n} = n.{x^{n - 1}}} \right) \\\ \implies \dfrac{{ - 1}}{{{v^2}}}.\dfrac{{dv}}{{dt}} + \dfrac{1}{{{u^2}}}.\dfrac{{du}}{{dt}} = 0 \\\ \implies \dfrac{1}{{{v^2}}}.\dfrac{{dv}}{{dt}} = \dfrac{1}{{{u^2}}}.\dfrac{{du}}{{dt}} \\\ \implies \dfrac{{dv}}{{dt}} = \dfrac{{{v^2}}}{{{u^2}}}.\dfrac{{du}}{{dt}} \\\ \implies {V_{IM}} = {m^2}.{V_{OM}} \implies eq\left( 1 \right) \\\$
$m$ is the magnification of the mirror which can be found by $m = \dfrac{{ - v}}{u}$
$m = \dfrac{{ - v}}{u} \\\ v = 10m,u = - 30m \\\ \implies m = - \left( {\dfrac{{10}}{{ - 30}}} \right) = \dfrac{{10}}{{30}} = \dfrac{1}{3} \\\$
Substituting the value of m in equation 1
${V_{IM}} = {m^2} \times {V_{OM}} \\\ m = \dfrac{1}{3},{V_{OM}} = 9m/s \\\ \implies {V_{IM}} = {\left( {\dfrac{1}{3}} \right)^2} \times 9 = \dfrac{9}{9} = 1m/s \\\$

So, the correct answer is “Option A”.

Note:
The rear-view mirror is a convex mirror, which makes a driver see other vehicles behind it. In a convex mirror, the object is placed in front of the mirror that is why we assume it as negative because the centre of curvature and focal point (which are assumed positive) lies behind the convex mirror.