Optics Ques 95
- 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: 95.(d)
Solution:
- Lens formula is given as
$ \begin{aligned} \Rightarrow \quad & \frac{1}{f} =\frac{1}{v}-\frac{1}{u} \quad …….(i) \\ \Rightarrow \quad & \frac{1}{v} =\frac{1}{f}+\frac{1}{u} \\ \Rightarrow \quad & \frac{u f}{u+f} =v \\ \Rightarrow \quad & \frac{v}{u} =\frac{f}{u+f} \quad …….(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 } \quad & \frac{d v}{d t} =\frac{v^{2}}{u^{2}} d u / d t \\ \Rightarrow \quad & \frac{d v}{d t} =(\frac{f}{u+f})^2 \cdot \frac{d u}{d t} \end{aligned} $
[using Eq. (ii)]
Given, $f=0.3 m, u=-20 m, d u / d t=5 m / s$
$\therefore \quad \frac{d v}{d t} =(\frac{0.3}{0.3-20})^{2} \times 5=(\frac{3}{197})^{2} \times 5 $
$= \quad 1.16 \times 10^{-3}$ $ m / s$
Thus, the image is moved with a speed of $1.16 \times 10^{-3} $ $m / s$ towards the lens.
BETA
