A point sourcs `S` is moving with a speed of `10 m//s` in `x-y` plane as shown in the figure. The radius of curvature of the concave mirror is `4m`. Determine the velocity vector of the image formed by paraxial rays.

Answer

`vec(V_(i))=V_(ix) hat(i) + V_(jy) hat(j) =- 2 hat(i) -4 hat(j)`

Explanation

`u=-6m,f=-2m rArr v=-3m`
let the height of object `Y_(@)`
then height of image is `Y_(i)=-((v))/((u))Y_(@)`
`therefore y`- component of velocity of image `V_(iy)=(dY_(i))/(dt)`
or `V_(iy)=(v.Y_(@))/(u^(2)) ((du)/(dt))-(Y_(0))/(u) ((dv)/(dt))-(v)/(u)((dY_(@))/(dt))` (differentiatiating both sides) .......`(A)`
`(du)/(dt)=x-` component of velocity of objcet `=10 cos37^(@)=8m//s`
`(dv)/(dt)=x` - component of velocity of image `=V_(ix)=-((v)/(u))^(2)(du)/(dt)=-2m//s`
`(dY_(@))/(dt)=Y`- component of velocity of object `=10 sin 37^(@)=6m//s`
(from equation `A`), `V_(iy)==(dY_(i))/(dt)=((-3)(1))/((-6)^(2)).8-(1)/((-6))(-2)-((-3))/((-6))6=-4m//s`
`therefore V_(i)=V_(ix)i+V_(iy)j=-2i-4j`