Question

A man running at a speed 5 m/s is viewed in the
side view mirror of radius of curvature R = 2 m of a
stationary car. Calculate the speed of image when the
man is at a distance of 9 m from the mirror.
(a) 0.3 m/s
(b) 0.2 m/s
(c) 0.1 m/s s
(d) 0.05 m/s
(AMU (Med.))​

Answer 1

Answer:

Explanation:

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;

1f=1v+1u where f is the focal length, v is the image distance in the mirror, u is the object distance.

f=R2 where R is the radius of curvature.

s=dt 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.

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 ; 1f=1v+1u 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=R2 where R is the radius of curvature. Then,

f=22=1m

At the object distance of 9m, the mirror equation can be written as,

−11=−1v+19 (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 1v subject and calculating, we have

1v=11+19=109

Inverting, we have

v=910=0.9m

Now since speed is given by

s=dt 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=5×1=5m

Then the new object distance would be

u=9−5=4m

Then the image distance at this point would be

1v=11+14=54

⇒v=45=0.8m

Hence, in the mirror the distance travelled in one second is

di=0.9−0.8=0.1m which means speed is 0.1m/s