Question

A particle is in uniform circular motion in a vertical plane with its centre at the centre of curvature of a concave mirror of focal length $10\text{m}$ kept in the same plane. The radius of the circle is $5\text{m}$. If the particle moves in clockwise sense, then its image will move in:

• A. anticlockwise path
• B. clockwise path
• C. straight line
• D. Image will not move

Answer 1

The correct option is B clockwise path

At any instant, on breaking velocity into components along and perpendicular to the principal axis.


For motion along the principal axis, velocities of image and object are opposite to each other (in opposite directions) because
$\frac{d(v_i{)}_x}{dt}=\frac{-v^2}{u^2}\frac{d(v_o{)}_x}{dt}$

Similarly, for the motion perpendicular to the principal axis,
$\frac{d(v_i{)}_y}{dt}=\frac{-v}{u}\frac{d(v_o{)}_y}{dt}$

Also, we know that, for object at/beyond the centre of curvature of concave mirror, image is formed between the focus and centre of curvature and vice-versa.

Thus, qualitative ray diagram of the image at that instant will be


The above ray diagram shows that image will also move in clockwise direction.

NOTE: It should be noted that the image will not be performing circular motion in this case.