Question

A diverging lens of focal length $-10$ cm is moving towards right with a velocity $5$ m $/$ s. An object is placed on the principal axis is moving towards left with a velocity 3 m $/$ s . What is the velocity of the image at the instant when the lateral magnification produced is $\frac{1}{2}$ ? (All velocities are measured with respect to ground)

• A. $3$ m $/$ s towards right
• B. $3$ m $/$ s towards left
• C. $7$ m $/$ s towards right
• D. $7$ m $/$ s towards left

Answer 1

Answer: (A)

Both Object velocity and Image velocity are w.r.t the pole. So, we have to find out velocity of object and image w.r.t pole.
$v_{p/g}=5m/s\hat{i}\left(\right.right\left.\right)v_{o/g}=-3m/s\hat{\overset{\cdot }{i}}\left(\right.left\left.\right)v_{o/p}=v_{o/g}-v_{p/g}=-8m/s\hat{\overset{\cdot }{i}}$
We know the magnification formula :
$v_{i/p}=m^{2}\times v_{o/p}$
Applying this :
$v_{i/p}=\left(\right.\frac{1}{2}\left(\left.\right)^{2}\times -8\hat{i}=-2\hat{i}Findingvelocityofimagewrtgroundv_{i/g}=v_{i/p}+v_{p/g}=-2\hat{i}+5\hat{i}=3\hat{i}=3m/\left(s \right)totheright.$