JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Geometrical Optics

Car B overtakes another car A at a relative speed of

40 ms^{- 1}

. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car

A

, when the car

B

is 1.9 m away from the car A ?

Answer: (A)

Solution

The mirror fitted in a car is convex, so

f = + 0.1

m, and the other car is a real object at

u = - 1.9

m. From the mirror formula,

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{0.1} + \frac{1}{1.9} = \frac{200}{19} \Rightarrow v = \frac{19}{200} = 0.095 m

The transverse magnification is

m = - \frac{v}{u} = \frac{0.095}{1.9} = 0.05

For motion along the axis the image speed is

m^{2}

times the object speed, so

v_{image} = (0.05)^{2} \times 40 = 0.1 m s^{- 1}

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option