Question

A motor car is fitted with a convex driving mirror of focal length 20 cm. A second motor car 2 m broad and 1.6 m high is 6 m away from the first car.

(i) Calculate the position and size of the image of the second car seen in the mirror of first.

(ii) If the second car is overtaking at a relative speed of 14 ms^-1, how fast will the image be moving and in what diection?

Answer 1

(i) Here u = - 6m, f = 20 cm = 0.2 m

From mirror formula

\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) ...........(i)

∴ \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{0.2} - \frac{1}{6}\)

or v = \(\frac{6}{31}\) = 0.194 m = 19.4 cm

Linear magnification

m = \(\frac{I}{O} = \frac{v}{u} \) or I = \(\frac{v}{u}\) O

Breadth b = \(\frac{0.194}{6}\) x 2 = 0.064 m = 6.4 cm

Height h = \(\frac{0.194}{6} \) x 1.6 = 0.052 m = 5.2 cm

(ii) Differentiating both sides of eq.(i), we get