Question 179: Ray Optics

Question

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?

Accepted Answer

Here, $${1 \over v} + {1 \over u} = {1 \over f}$$ .......(1)
$$ \Rightarrow $$ $${1 \over v} + {1 \over { - 190}} = {1 \over {10}}$$
$$ \Rightarrow $$ v = $${{19} \over 2}$$
Differentiating equation (1) w.r.t t we get
$$ - {1 \over {{v^2}}}\left( {{{dv} \over {dt}}} ight) - {1 \over {{u^2}}}\left( {{{du} \over {dt}}} ight) = 0$$
$$ \Rightarrow $$ $$\left( {{{dv} \over {dt}}} ight) =  - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} ight)$$
$$ \Rightarrow $$ $$\left( {{{dv} \over {dt}}} ight) =  - {\left( {{{{{19} \over 2}} \over {190}}} ight)^2} 	imes 40$$
$$ = {1 \over {400}} 	imes 40 = 0.1$$ m/s

Answer

4 ms$$-$$1

Answer

0.2 ms$$-$$1

Answer

40 ms$$-$$1

Answer

0.1 ms$$-$$1

Ray 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?

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