When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distance of 10 cm from the mirror. If the object is moved with a speed of 9 cms–1, the speed (in cms–1) with which image moves at that instant is \underline\hspace2cm .

Geometrical Optics PYQ — JEE Main 2020 · 3 Sep · Shift 2 · Q44 with Solution

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Correct answer: 1

For a spherical mirror,

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

where $u$ is object distance and $v$ is image distance (with sign convention).

Given magnitudes:

For a concave mirror, at this situation both object and real image are in front of the mirror, so

u=-30\,\text{cm},\qquad v=-10\,\text{cm}

Thus,

\frac{1}{f}=\frac{1}{-30}+\frac{1}{-10}=-\frac{1}{30}-\frac{3}{30}=-\frac{4}{30}=-\frac{2}{15}

f=-\frac{15}{2}=-7.5\,\text{cm}

Since $f$ is constant,

\frac{d}{dt}\left(\frac{1}{u}+\frac{1}{v}\right)=0

Therefore,

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

So,

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

Object speed is $9\,\text{cm s}^{-1}$ . Since the problem asks for speed, we use magnitude.

At the instant:

|u|=30,\qquad |v|=10

Hence,

\left|\frac{dv}{dt}\right|=\frac{10^2}{30^2}\times 9 =\frac{100}{900}\times 9 =1\,\text{cm s}^{-1}

Thus, the speed of the image is

1\,\text{cm s}^{-1}

Derived answer = $1$

Stored correct answer = $1$

They agree.