Question

An object is approaching a thin convex lens of focal length $0.3m$ with a speed of $0.01ms^{- 1}$ . The magnitude of the rate of change of lateral magnification (in per second) of image when the object is at a distance of $0.4m$ from the lens is

• A. $0.3$
• B. $0.6$
• C. $0.15$
• D. $-0.3$

Answer 1

Answer: (A)

Differentiating the lens formula $\frac{1}{v}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$ with respect to time, we get
$-\frac{1}{v^{2}}\cdot \frac{\text{d} \upsilon}{\text{dt}}+\frac{1}{\left(\text{u}\right)^{2}}\cdot \frac{\text{d} \text{u}}{\text{dt}}=0 \, \, \, \left(\text{as f = constant}\right)$
$ \, ∴ \, \left(\frac{\text{d} \upsilon}{\text{dt}}\right)=\left(\frac{v^{2}}{\left(\text{u}\right)^{2}}\right)\cdot \frac{\text{du}}{\text{dt}}$
Further, substituting proper values in lens formula, we
have
$\frac{1}{v}+\frac{1}{0 \cdot 4}=\frac{1}{0 \cdot 3}\left(\text{u} = - 0 \cdot 4 \text{m} \text{,} \textit{f} = 0 \cdot 3 \text{m}\right)$
or, $v=1\cdot 2\text{m}$
Putting the values in Eq.(i) Magnitude of rate of change of position of image
$= 0 \cdot 0 9 \text{m} / \text{s}$
Lateral magnification, $\textit{m}=\frac{v}{\textit{u}}$
$ \, ∴ \, \frac{\textit{dm}}{\textit{dt}}=\frac{\textit{u} \cdot \frac{\textit{dv}}{\textit{dt}} - v \frac{d u}{d t}}{\left(\textit{u}\right)^{2}}=\frac{\left(- 0 \cdot 4\right) \left(0 \cdot 09\right) - \left(1 \cdot 2\right) \left(0 \cdot 01\right)}{\left(0 \cdot 4\right)^{2}}$
$= - 0 \cdot 3 / \text{s}$
$\therefore $ Magnitude of rate of change of lateral magnification $= 0 \cdot 3 / \text{s}$