A particle is dropped along the axis from a height $\frac{f}{2}$ on a concave mirror of focal length f as shown in the figure. The acceleration due to gravity is g . What is the maximum speed of the image.

$\frac{1}{2} \sqrt{2 f g}$

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$\frac{3}{8} \sqrt{2 f g}$

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$\frac{3}{4} \sqrt{3 f g}$

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$\sqrt{f g}$

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Solution

The correct option is C
$\frac{3}{4} \sqrt{3 f g}$

If the distance of the vertical image from the pole is ‘y’ and ‘x’ is distance of the object from O then

$\frac{1}{y} - \frac{1}{x} = - \frac{1}{f} y = \frac{f x}{f - x} \frac{d y}{d x} = {(\frac{f}{f - x})}^{2} \frac{d x}{d y} V_{1} = {(\frac{f}{f - x})}^{2} \sqrt{2 g (\frac{f}{2} - x)}$
For $V_{1}$ to be maximum
$\frac{d V_{1}}{d T} = 0 x = \frac{f}{3} (V_{1})_{m a x i m u m} = \frac{3}{4} \sqrt{3 f g}$

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A particle is dropped along the axis from a height f2 on a concave mirror of focal length f as shown in the figure. The acceleration due to gravity is g . What is the maximum speed of the image.

The correct option is C 34√3fg If the distance of the vertical image from the pole is ‘y’ and ‘x’ is distance of the object from O then 1y−1x=−1fy=fxf−xdydx=(ff−x)2dxdyV1=(ff−x)2√2g(f2−x) For V1 to be maximum dV1dT=0x=f3(V1)maximum=34√3fg

A particle is dropped along the axis from a height $\frac{f}{2}$ on a concave mirror of focal length f as shown in the figure. The acceleration due to gravity is g . What is the maximum speed of the image.