Question

Car B overtakes 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 car B is 1.9 m away from car A?

Options

• 0.1 ms^-1

• 0.2 ms^-1

• 4 ms^-1

• 40 ms^-1

Answer 1

0.1 ms^-1

Explanation:

As, `1/f = 1/u + 1/v`

On differentiating w.r.t. time 't', we get

`-(du)/(dt) . 1/u^2 - (dv)/(dt) . 1/v^2 = 0`

⇒ `(du)/(dt)` = object speed, (V₀)

`(dv)/(dt)` = image speed, (V₁)

`V_1 = -m^2V_0` ....(i) `{{:(∵u = -1.9  cm),(∵f = 10  cm):}}`

Magnification, m = `v/u = f/(f - u)`

m = `10/(10 - (-190))`

⇒ m = `10/200 = 1/20`

Putting numbers into an equation (i),

`V_1 = -(1/20)^2 xx 40`

V₁ = -0.1 m/s

Hence, the can will appear to move with a speed of 0.1 m/s.