Car B overtakes another car A at a relative speed of $40 \mathrm{~ms}^{-1}$. How fast will the image of car B appear to move in the mirror of focal length $10 \mathrm{~cm}$ fitted in car $\mathrm{A}$, when the car $\mathrm{B}$ is $1.9 \mathrm{~m}$ away from the car A?
Correct Option: , 4
Mirror used is convex mirror (rear-view mirror)
$\therefore \mathrm{V}_{\mathrm{L} / \mathrm{m}}=-\mathrm{m}^{2} \mathrm{~V}_{\mathrm{O} / \mathrm{m}}$
Given,
$\mathrm{V}_{\mathrm{O} / \mathrm{m}}=40 \mathrm{~m} / \mathrm{s}$
$m=\frac{f}{f-u}=\frac{10}{10+190}=\frac{10}{200}$
$\therefore \mathrm{V}_{1 / \mathrm{m}}=-\frac{1}{400} \times 40=-0.1 \mathrm{~m} / \mathrm{s}$
$\therefore$ Car will appear to move with speed $0.1 \mathrm{~m} / \mathrm{s}$.
Hence option (4)
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