A thief is running away in a car with velocity of `20 m//s`. A police jeep is following him, which is sighted by thief in his rear view mirror, which is a convex mirror of focal length 10 m. He observes that the image of jeep is moving towards him with a velocity of `1 cm//s`. if the magnification of mirror for the jeep at that time is `1/10`. Find
(a) the actual speed of jeep,
(b) rate at which magnification is changing.
Assume the police's jeep is on the axis of the mirror.

Answer

A, B, C

Explanation

(a) Differenting the mirror formula, (with respect to time)

`1/v +1/u =1/f`
We get velocity of image,
`(v_(I))=(m^(2)) (v_(o)) ...(i)`
Here, `(v_(o))=` relative of object with respect to mirror
`v_(I)=` relative velocity of image
`m=`linear magnification
Here, `(v_(0))=(v-20)m//s`
`v_(I)=1 cm//s =0.01 m//s`
`m=(1/10)`
Subsitituting in eQ. (i) we have,
`0.01=(1/10)^(2) (v-20)`
`:. v=21 m//s`.
(b) `1/v+1/u=1/f`
Multiplying with u we get
`(u/v)+1=(u/f)`
or, `(1/m) = (u/f)-1 =(u-f)/(f)`
`:. m=(f)/(u-f)`
Differentiating we have,
`((dm)/(dt))=-(f)/((u-f)^(2)).(du)/(dt)` ...(ii)
Using mirror formula to find u with magnification
`m=(1/10)` we get
`(1)/(u//10)-(1/u)=(1/10)`
or, `u=90 m` (with sign `u=-90m`)
Subsitituting in Eq. (ii) we have,
`(dm)/(dt)=-(10)/((-90-10)^(2)) (-1)`
`=10^(-3)` per second.