Question

An object is approaching a thin convex lens of focal length $0.3\text{m}$ with a speed of $0.01\text{m/s}$. The magnitude of the rate of change of position of image when the object is at a distance of $0.4\text{m}$ from the lens is

• A. $0.09\text{m/sec}$
• B. $0.01\text{m/sec}$
• C. $0.04\text{m/sec}$
• D. $0.02\text{m/sec}$

Answer 1

The correct option is A $0.09\text{m/sec}$

From lens formula,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}.....\left(1\right)$

Given , $f=0.3,u=-0.4\text{m}$

$\therefore \frac{1}{v}=\frac{1}{0.3}-\frac{1}{0.4}=\frac{1}{1.2}$

$\Rightarrow v=1.2\text{m}$

Differentiating $\left(1\right)$ with respect to time on both sides we get,

$\frac{-1}{v^2}\times \frac{dv}{dt}+\frac{1}{u^2}\times \frac{du}{dt}=0....\left(2\right)$

Given , $\frac{du}{dt}=0.01\text{m/s}$

Substituting the value of $u,v,\frac{du}{dt}$ in equation $\left(2\right)$

$\frac{1}{(1.2{)}^2}\times \frac{dv}{dt}=\frac{1}{(-0.4{)}^2}\times 0.01$

$\Rightarrow \left|\frac{dv}{dt}\right|=0.09\text{m/s}$

Hence, option (a) is the correct answer.