An object is moving towards a convergent lens L as shown in figure. If the distance of object from the lens is x, and if ${x=3f, θ=60}^{0}$ , then speed of image is

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$\frac{v \sqrt{13}}{16}$

$\frac{v \sqrt{13}}{4}$

$\frac{v \sqrt{13}}{8}$

answer is D.

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Detailed Solution

$\Rightarrow v_{l y} = \frac{1}{2} v \sin 60^{0} = \frac{v \sqrt{3}}{4}$ $\frac{1}{y} - \frac{1}{x} = \frac{1}{f}$

$\Rightarrow \frac{d y}{d x} = {(\frac{f}{x - f})}^{2} \frac{d x}{d t}$

$\Rightarrow v_{l x} = {(\frac{f}{x - f})}^{2} \times v \cos 60^{0} = \frac{v}{8}$

For lateral magnification,
$\frac{l^{'}}{l} = \frac{y}{x} \Rightarrow \frac{d l^{'}}{d t} = (\frac{y}{x}) \frac{d l}{d t} = (\frac{f}{x - f}) . v \sin θ$
$\Rightarrow v_{l y} = \frac{1}{2} v \sin 60^{0} = \frac{v \sqrt{3}}{4}$

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