In the diagram shown, the lens is moving towards the object with a velocity V m/s and the object is also moving towards the lens with the same speed. What speed of the image with respect to earth when the object is at a distance 2f from the lens? (f is the focal length.)

Engineering

Physics

Plane Mirror New

Question

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We analyze the motion of the image using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ . Given object distance $u = - 2 f$ (negative sign by convention), we find image distance $v$ .

Substitute into the formula: $\frac{1}{v} - \frac{1}{- 2 f} = \frac{1}{f}$ → $\frac{1}{v} + \frac{1}{2 f} = \frac{1}{f}$ → $\frac{1}{v} = \frac{1}{f} - \frac{1}{2 f} = \frac{1}{2 f}$ . Therefore, $v = 2 f$ .

To find the image speed, we differentiate the lens formula with respect to time. The relative velocity between the lens and object is 2V (both moving towards each other at speed V). The formula for image velocity is $v_{i} = - {(\frac{v}{u})}^{2} v_{o}$ . Substituting $v = 2 f$ and $u = - 2 f$ , we get $v_{i} = - {(\frac{2 f}{- 2 f})}^{2} (2 V) = - {(- 1)}^{2} (2 V) = - 2 V$ . The negative sign indicates direction. The speed of the image with respect to Earth is 2V.

Final Answer: 2V