Question

A particle revolves in clockwise direction (as seen from point $A$) in a circle $C$ of radius $1cm$ and completes one revolution in $2\sec$. The axis of the circle and the principal axis of the mirror $M$ coincide, a call it $AB$. The radius of curvature of the mirror is $20cm$. Then, the direction of revolution (as seen from $A$) of the image of the particle and its speed is

• A. Clockwise, $1.57c{m}^{-1}$
• B. Clockwise, $3.14c{m}^{-1}$
• C. Anticlockwise, $1.57c{m}^{-1}$
• D. Anticlockwise, $3.14c{m}^{-1}$

Answer 1

The correct option is A Clockwise, $1.57c{m}^{-1}$

By mirror formula : $\frac{1}{v}+\frac{1}{-10}=\frac{1}{10}$
$\Rightarrow v=+5cm$
$\therefore m=+\frac{1}{2}$
The image revolves in a circle of radius $\frac{1}{2}cm$. Image of a radius is erect $\Rightarrow$ particle will revolve in the same direction as the particle. The image will complete one revolution in the same time $2s$.
Velocity of image $v$ is
$\omega r=\frac{2\pi }{2}\times \frac{1}{2}cm{s}^{-1}=1.57cm{s}^{-1}$.