Optics 3 Question 7
7. 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
(2019 Main, 11 Jan I)
(a) $3.22 \times 10^{-3} m / s$ towards the lens
(b) $0.92 \times 10^{-3} m / s$ away from the lens
(c) $2.26 \times 10^{-3} m / s$ away from the lens
(d) $1.16 \times 10^{-3} m / s$ towards the lens
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Answer:
Correct Answer: 7. (d)
Solution:
- Lens formula is given as
$$ \begin{aligned} \Rightarrow & \frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \cdots(i)\\ \Rightarrow & \frac{1}{v} & =\frac{1}{f}+\frac{1}{u} \\ \Rightarrow & \frac{u f}{u+f} & =v \\ \Rightarrow & \frac{v}{u} & =\frac{f}{u+f} \cdots(ii) \end{aligned} $$
Now, by differentiating Eq. (i), we get
$$ 0=-\frac{1}{v^{2}} \cdot \frac{d v}{d t}+\frac{1}{u^{2}} \cdot \frac{d u}{d t} $$
$[\because f$ (focal length of a lens is constant)]
$$ \begin{aligned} \text { or } & \frac{d v}{d t} & =\frac{v^{2}}{u^{2}} d u / d t \\ \Rightarrow & \frac{d v}{d t} & =\frac{f}{u+f} \cdot \frac{d u}{d t} [using Eq. (ii)] \end{aligned} $$
Given, $f=0.3 m, u=-20 m, d u / d t=5 m / s$
$$ \begin{aligned} \therefore \quad \frac{d v}{d t} & =\Big(\frac{0.3}{0.3-20}\Big)^{2} \times 5=\Big(\frac{3}{197}\Big)^{2} \times 5 \\ & =1.16 \times 10^{-3} m / s \end{aligned} $$
Thus, the image is moved with a speed of $1.16 \times 10^{-3} m / s$ towards the lens.
