A motor car is fitted with a convex driving mirror of focal length `20 cm`. A second motor car is `6 m` away from the driving mirror of the first car. Calculate (i) position of second car as seen in the first car mirror.
(ii) if the second car is overtaking the first car at a relative speed of `15 m//s`, how will its image be moving and in what direction ?

Answer

(i) `0.1935 m`
(ii) `- 1.56 cm//s`.

Explanation

(i) Here, `f = + 20 cm = 0.2 m , u = -6 m`
From the mirror formula
`(1)/(v)=(1)/(f)-(1)/(u)=(1)/(0.2) - (1)/(-6)=(31)/(6)`
`v = (6)/(31)m = 0.1935 m`
As `v` is positive, iamge of second car is virtual erect and id formed behind the mirror of first car.
(ii) As is known, the speed of image of second car is
`(dv)/(dt) = -(f^2)/((u - f)^2) xx (du)/(dt)`
where `(du)/(dt) = 15 ms^-1`
`:. (dv)/(dt) = ((0.2)^2)/((-6 - 0.2)^2) xx 15`
= `-1.56 xx 10^-2 m//s`
`(dv)/(dt) = -1.56 cm//s`
Negative sign implies that as `2nd` car approaches the first car, image of `2nd` car moves away from convex driving mirror.