Question

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.

• A. $\frac{1}{2}\sqrt{2fg}$
• B. $\frac{3}{8}\sqrt{2fg}$
• C. $\frac{3}{4}\sqrt{3fg}$
• D. $\sqrt{fg}$

Answer 1

The correct option is C

$\frac{3}{4}\sqrt{3fg}$

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

$\frac{1}{y}-\frac{1}{x}=-\frac{1}{f}y=\frac{fx}{f-x}\frac{dy}{dx}={\left(\frac{f}{f-x}\right)}^2\frac{dx}{dy}{V}_1={\left(\frac{f}{f-x}\right)}^2\sqrt{2g(\frac{f}{2}-x)}$
For $V_1$ to be maximum
$\frac{d{V}_1}{dT}=0x=\frac{f}{3}(V_1{)}_{maximum}=\frac{3}{4}\sqrt{3fg}$