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?
Solution
Here, ${1 \over v} + {1 \over u} = {1 \over f}$ .......(1)
<br/><br/>$\Rightarrow$ ${1 \over v} + {1 \over { - 190}} = {1 \over {10}}$
<br/><br/>$\Rightarrow$ v = ${{19} \over 2}$
<br/><br/>Differentiating equation (1) w.r.t t we get
<br/><br/>$$ - {1 \over {{v^2}}}\left( {{{dv} \over {dt}}} \right) - {1 \over {{u^2}}}\left( {{{du} \over {dt}}} \right) = 0$$
<br/><br/>$\Rightarrow$ $$\left( {{{dv} \over {dt}}} \right) = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)$$
<br/><br/>$\Rightarrow$ $$\left( {{{dv} \over {dt}}} \right) = - {\left( {{{{{19} \over 2}} \over {190}}} \right)^2} \times 40$$
<br/><br/>$= {1 \over {400}} \times 40 = 0.1$ m/s
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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