A point object is moving with speed of 10 cm/s infront of converging lens along the principal axis as shown in figure. At this instant, the speed of the image is

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$30 c m s^{- 1}$

$32 c m s^{- 1}$

$44 c m s^{- 1}$

$40 c m s^{- 1}$

answer is C.

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Detailed Solution

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

$\Rightarrow \frac{1}{v} - \frac{1}{- 30} = \frac{1}{20}$

$\Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$

$\Rightarrow v = 60 {cm}^{- 1}$

$V_{i} = {(\frac{v}{u})}^{2} V_{0} = {(\frac{60}{30})}^{2} \times 10 = 40 {cms}^{- 1}$

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