Figure shows a concave mirror of focal length 20 cm. A point object P is oscillating along te principle axis of the mirror. Position co-ordinates of image at any time `t` is given by:
`x_(i)=40+20sin(2t+(5pi)/(6))`
Here x is in cm and t is in positive.

Options

A. Speed of the object at `t=0` is `40sqrt(3)cm//s`
B. speed of the object at `t=0` is `(80sqrt(3))/(9)cm//s`
C. if we start observing at `t=0` minimum time `t` at which object & image will cross each other is `(pi)/(12)` s
D. if we start observing at `t=0`, minimum time `t` at which object & imega will cross each other is `(pi)/(6)s`

Answer

B, C

Explanation

At `t=0` image distance `v=-50cm`
velocity of image `V_(I)=-20sqrt(3)cm//s`
`V_(I)=-m^(2).v_(0)`
`20sqrt(3)=((f-v)/(f))^(2)V_(0)`
`20sqrt(3)=-((-20+50)/(-20))^(2)V_(0)`
`20sqrt(3)=-(9)/(4)V_(0)`
`V_(0)=80(sqrt(3))/(9)cm//s`
Object & image will cross each other at centre of curvature i.e., when `x_(i)=40`
`2t+(5pi)/(6)=pi`
`2t+(pi)/(6)impliest=(pi)/(42)`s