Question

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

• A. clockwise, $1.57{\text{cms}}^{-1}$
• B. clockwise, $3.14{\text{cms}}^{-1}$
• C. anticlockwise, $1.57{\text{cms}}^{-1}$
• D. anticlockwise, $3.14{\text{cms}}^{-1}$

Answer 1

The correct option is A clockwise, $1.57{\text{cms}}^{-1}$

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

Why this question? Tip: The one common thing for object and image motion is time- so time period for both is same.