(a) A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole. What is the size of the image? (b) If the object begins to move with speed `V_(0)` , what will be the speed of its image?

Explanation

a. According to mirror formula,
`(1)/(v)+(1)/(u)=(1)/(f)` (i)
For a given mirror, focal length f= constant. Therefore, on differentiating, Eq. (i) gives
`-(1)/(v^(2))dv-(1)/(u^(2))du=0rArr dv=-((v)/(u))^(2)du` (ii)
Multiplying Eq. (i) throughout by u, we get
`(u)/(v)+1=(u)/(f)rArr (u)/(v)=(u)/(f)-1=(u-f)/(f)`
`:. (v)/(u)=(f)/(u-f)` (iii)
Differential length of the object on the axis, `du=b `(given).
Therefore, Eq. (ii) gives `dv=-((f)/(u-f))^(2) b` .
The negative sign shows that the image is longitudinally inverted.
`:.` Size of image `|dv|= b((f)/(u-f))^(2)`.
b. From Eqs. (ii) and (iii), `dv =-f((f)/(u-f))^(2)du`
Dividing both the sides by differential time dt, we get
`(dv)/(dt)=-((f)/(u-f))^(2)(du)/(dt)`
Given `(du)/(dt)=v_(0)` adn speed of image `(dv)/(dt)=v_(1)` (say)
`rArr v_(i) =-((f)/(u-f))^(2)v_(0)`