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A particle revolves in clockwise direction (as seen from point A) in a circle C of radius I cm and completes one revolution in 2 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 cm. Then, the direction of revolution (as seen from A) of the image of the particle and its speed is

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$clockwise, 1.57 {cms}^{- 1}$

$clockwise, 3.14 {cms}^{- 1}$

$anticlockwise, 1.57 {cms}^{- 1}$

$anticlockwise, 3.14 {cms}^{- 1}$

answer is A.

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Detailed Solution

$By mirror formula: \frac{1}{v} + \frac{1}{- 10} = \frac{1}{10} \Rightarrow v = + 5 cm ∴ m = + \frac{1}{2}$

The image revolves in a circle of radius $\frac{r}{2}$ cm. Image of a radius is erect =+ particle will revolve in the same direction as the particle. The image will complete one evolution in the same time 2 s.

$Velocity of image v is ωr = \frac{2 π}{2} \times \frac{1}{2} = \frac{π}{2} {cms}^{- 1} = 1 .57 {cms}^{- 1}$

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