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

Hint: 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.