Question

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

• A. 0.09 m/sec
• B. 0.01 m/sec
• C. 0.04 m/sec
• D. 0.02 m/sec

Answer 1

Answer: (A)

0.09 m/sec

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ .....{1}

$f=0.3$

at given point of time given $u=-0.4$

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

$v=1.2$

differentiating $1$ with respect to time

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

given $\frac{du}{dt}=0.01\frac{m}{s}$

putting the values of $u,v,\frac{du}{dt}$ in equation $2$

$\frac{dv}{dt}=0.09\frac{m}{s}$