Optics 3 Question 35
35. An object is approaching a thin convex lens of focal length $0.3 m$ with a speed of $0.01 m / s$. Find the magnitudes of the rates of change of position and lateral magnification of image when the object is at a distance of $0.4 m$ from the lens.
(2004, 4M)
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Answer:
Correct Answer: 35. $0.09 m / s, 0.3 / s$
Solution:
- Differentiating the lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ with respect to time, we get
$$ \begin{aligned} & -\frac{1}{v^{2}} \cdot \frac{d v}{d t}+\frac{1}{u^{2}} \cdot \frac{d u}{d t}=0 \quad \text { (as } f=\text { constant) } \\ \therefore & \frac{d v}{d t}=\frac{v^{2}}{u^{2}} \cdot \frac{d u}{d t} \end{aligned} $$
Further, substituting proper values in lens formula, we have
or
$$ \begin{aligned} \frac{1}{v}+\frac{1}{0.4} & =\frac{1}{0.3} \quad(u=-0.4 m, f=0.3 m) \\ v & =1.2 m \end{aligned} $$
Putting the values in Eq. (i), we get
Magnitude of rate of change of position of image $=0.09 m / s$
Lateral magnification, $m=\frac{v}{u}$
$$ \begin{aligned} \therefore \quad \frac{d m}{d t} & =\frac{u \cdot \frac{d v}{d t}-v \cdot \frac{d u}{d t}}{u^{2}}=\frac{(-0.4)(0.09)-(1.2)(0.01)}{(0.4)^{2}} \\ & =-0.3 / s \end{aligned} $$
$\therefore$ Magnitude of rate of change of lateral magnification
$$ =0.3 / s $$
