Question

An object is approaching a thin convex lens of focal length $0.3m$ with a speed of $0.01m{s}^{-1}$ . 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
$-\frac{1}{v^2}\cdot \frac{\text{d}\upsilon }{\text{dt}}+\frac{1}{{\left(\text{u}\right)}^2}\cdot \frac{\text{d}\text{u}}{\text{dt}}=0\left(\text{as f = constant}\right)$
$\therefore \left(\frac{\text{d}\upsilon }{\text{dt}}\right)=\left(\frac{v^2}{{\left(\text{u}\right)}^2}\right)\cdot \frac{\text{du}}{\text{dt}}$
Further, substituting proper values in lens formula, we
have
$\frac{1}{v}+\frac{1}{0\cdot 4}=\frac{1}{0\cdot 3}\left(\text{u}=-0\cdot 4\text{m}\text{,}\text{f}=0\cdot 3\text{m}\right)$
or, $v=1\cdot 2\text{m}$
Putting the values in Eq.(i) Magnitude of rate of change of position of image
$=0\cdot 09\text{m}/\text{s}$
Lateral magnification, $\text{m}=\frac{v}{\text{u}}$
$\therefore \frac{\text{dm}}{\text{dt}}=\frac{\text{u}\cdot \frac{\text{dv}}{\text{dt}}-v\frac{du}{dt}}{{\left(\text{u}\right)}^2}=\frac{\left(-0\cdot 4\right)\left(0\cdot 09\right)-\left(1\cdot 2\right)\left(0\cdot 01\right)}{{\left(0\cdot 4\right)}^2}$
$=-0\cdot 3/\text{s}$
$\therefore$ Magnitude of rate of change of lateral magnification $=0\cdot 3/\text{s}$