Question

A diverging lens of focal length - 10 cm is moving towards right with a velocity 5 m/s. An object, placed on the Principal axis, is moving towards the left with a velocity 3 m/s. The velocity of the image at the instant when the lateral magnification produced is 1/2 is: (All velocities are with respect to ground)
(A) 3 m/s towards right
(B) 3m/s towards left
(C) 7m/s towards right
(D) 7m/s towards left

Answer 1

Hint: The lens and the object are moving in different directions. Lateral magnification of the image is the ratio between the image distance and the object distance.

Formula Used: The following formulas are used to solve this question.
Lateral Magnification $ M = \left( {\dfrac{v}{u}} \right) $ , $ u $ being the object distance and $ v $ being the image distance.
 $ {v_I} - {v_M} = \dfrac{1}{4}\left( {{v_O} - {v_M}} \right) $ where velocity of object is $ {v_o} $ , $ {v_I} $ is the velocity of the image and $ {v_M} $ is the velocity of the image.

Complete step by step solution:
Diverging lenses are biconcave lenses. Any incident ray travelling parallel to the principal axis of a diverging lens will refract through the lens and travel in line with the focal point (i.e., in a direction such that its extension will pass through the focal point).
Given that, a diverging lens having focal length $ - 10cm $ is moving towards right with a velocity $ 5m/s $ .
Thus focal length $ f = - 10cm $ and velocity of lens $ {v_m} = 5m/s $ .
 An object, placed on the principal axis is moving towards left with a velocity $ 3m/s $ . Thus velocity of object $ {v_o} = 3m/s $
Let lateral magnification be represented by $ M $ where $ M = \left( {\dfrac{v}{u}} \right) $ , $ u $ being the object distance and $ v $ being the image distance.
Given that, $ M = \dfrac{1}{2} = \left( {\dfrac{v}{u}} \right) $ .
Now, we know that, $ {v_{IM}} = \left( {\dfrac{{{v^2}}}{{{u^2}}}} \right){v_{OM}} $
 $ \Rightarrow {v_I} - {v_M} = \dfrac{1}{4}\left( {{v_O} - {v_M}} \right) $ where $ {v_I} $ is the velocity of the image.
Assigning the values given in the question,
 $ {v_I} - 5 = \dfrac{1}{4}\left( { - 3 - 5} \right) $ , the velocity of the object is negative because the object is moving towards left, opposite to the direction of movement of the lens.
 $ {v_I} - 5 = - 2 $
Thus, the velocity of the image is, $ {v_I} = 3m/s $ . Since it is not a negative value, it is moving towards the right.
The correct answer is Option A.

Note:
If the value of image velocity were negative, the image would be moving towards the opposite direction with respect to the lens. Then the object shall move to the left. Object distances are positive to the left, negative to the right. Image distances are negative to the left, positive to the right.