A car is fitted with a convex side-view mirror of focal length 20cm . A second car 2.8m behind the first car is overtaking the first car at a relative speed of 15m/s . The speed of the image of the second car as seen in the mirror of the first one is :

Geometrical Optics PYQ — JEE Main 2011 · Shift 0 · Q56 with Solution

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Correct answer: A

For a spherical mirror,

\frac{1}{f}=\frac{1}{v}+\frac{1}{u}

Here the mirror is convex, so using the Cartesian sign convention:

Thus,

\frac{1}{0.2}=\frac{1}{v}+\frac{1}{-2.8}

5=\frac{1}{v}-\frac{1}{2.8}

\frac{1}{v}=5+\frac{1}{2.8}=5+0.357142857=5.357142857

v\approx 0.1867\text{ m}

So the image is formed $0.1867\,$ m behind the mirror.

Differentiate the mirror formula:

\frac{1}{f}=\frac{1}{v}+\frac{1}{u}

Since $f$ is constant,

0=-\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt}

Hence,

\frac{dv}{dt}=-\frac{v^2}{u^2}\frac{du}{dt}

We need the speed of image, so take magnitude:

\left|\frac{dv}{dt}\right|=\frac{v^2}{u^2}\left|\frac{du}{dt}\right|

Given relative speed of the second car toward the first car is

\left|\frac{du}{dt}\right|=15\text{ m/s}

Now,

\frac{v}{u}=\frac{0.1867}{2.8}=\frac{1}{15}

So,

\frac{v^2}{u^2}=\left(\frac{1}{15}\right)^2=\frac{1}{225}

Therefore,

\left|\frac{dv}{dt}\right|=\frac{1}{225}\times 15=\frac{1}{15}\text{ m/s}

Hence the correct option is A.

Stored correct answer: A

My derived answer: A

They agree.