An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be :

1.16 \times 10–3 m/s towards the lens

2.26 \times 10–3 m/s away from the lens

3.22 × 10–3 m/s towards the lens

0.92 \times 10 - 3 m/s away from the lens

Geometrical Optics PYQ — JEE Main 2019 · 11 Jan · Shift 1 · Q63 with Solution

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Correct answer: A

For a thin convex lens,

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

Using the Cartesian sign convention:

So,

\frac{1}{0.3}=\frac{1}{v}-\frac{1}{-20}

\frac{10}{3}=\frac{1}{v}+\frac{1}{20}

\frac{1}{v}=\frac{10}{3}-\frac{1}{20} = \frac{200-3}{60}=\frac{197}{60}

Hence,

v=\frac{60}{197}\approx 0.3046\,\text{m}

Starting from

\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}

Thus,

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

The object moves away from the lens at $5\,\text{m/s}$ . Since $u$ is negative, moving away means $u$ becomes more negative, so

\frac{du}{dt}=-5\,\text{m/s}

Now,

\frac{dv}{dt}=\left(\frac{0.3046^2}{20^2}\right)(-5)

0.3046^2 \approx 0.0928

\frac{dv}{dt}=\frac{0.0928}{400}(-5)

\frac{dv}{dt}= -1.16\times 10^{-3}\,\text{m/s}

Since $v>0$ and $\frac{dv}{dt}<0$ , the image distance decreases. Therefore the image moves towards the lens.

Magnitude of image speed:

\left|\frac{dv}{dt}\right|=1.16\times 10^{-3}\,\text{m/s}

Direction: towards the lens.

So the correct option is:

\boxed{\text{A: }1.16\times 10^{-3}\,\text{m/s towards the lens}}