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A converging lens forms a real image of an object on a fixed screen. The lens is then moved at constant speed v0 along its principal axis away from the screen while the object is simultaneously moved to keep the image on the screen at all times. If the image size is twice the object size at some instant, what is the required speed of the object at that instant?
- 3*v0/2
- v0/4
- 4*v0
- 3*v0
Correct answer: 3*v0
Solution
With the image size being half the object size, |m| = 1/2. Differentiating the lens formula gives v_image = m² * v_obj in the lens frame. The screen moves at v0 toward the lens in the lens frame, so v_image = v0 and v_obj(lens frame) = v0 / (1/4) = 4*v0. Converting to ground frame (object moves in same direction as lens): v_obj(ground) = 4*v0 - v0 = 3*v0.
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