Focal length of the mirror shown in figure is 20cm. Find the image position and its velocity.

Options

A. ` sqrt(272)) m//s` and `theta=tan^(-1)(5/8)`
B. ` sqrt(292)) m//s` and `theta=sin^(-1)(3/8)`
C. ` sqrt(292)) m//s` and `theta=tan^(-1)(3/8)`
D. none

Answer

C

Explanation

Substuting the values,
` u=-30 cm` and` f=-20 cm`
in the mirror formula, we have `(/v)+(1/(-30))+(1/(-20))`
Solving the equation , we get
`v=-60cm`
Further, `m=-(v/u)=-((-60))/((-30))=-2`
and `m^2=4`
Object velocity along the axis is `5 cos 37^(@)=4 min//s`(towards OP).Therefore image velocity the axis should be `m^@` times or `16 mm//s` in the opposite direction of object velocity. Object velocity perpendicular to axis is `5 sin 37^(@)=3 mm//s`(upward).Therefore ,image velocity will be m times or -`6mm//s` downwards .The position and velocity of image is shown below.

` v_(1)=(sqrt((16^2)+(6^2)))=( sqrt(292)) mm/s.`
` tan theta =(6/16)=(3/8)` or `theta=tan^(-1)(3/8)`.