An object is approaching a convex lens of focal length 0.3m with a speed of `0.01m s^(-1)`. Find the magnitudes of the ratio of change of position and lateral magnification of image when the object is at a distance of 0.4m from the lens

Explanation

Given `f=0.3m , u=-0.4m`
Form lens formula,
`(1)/(f)=(1)/(v)-(1)/(u)` (i)
We have,
`(1)/(0.3)=(1)/(v)+(1)/(0.4)rArr (1)/(v)=(1)/(0.3)-(1)/(0.4) or v=1.2m`
Differentiating Eq.(i) with respect to time t, we get
`0=-(1)/(v^(2)) (dv)/(dt)+(1)/(u^(2))(du)/(dt)rArr (dv)/(dt)=(v^(2))/(u^(2))(du)/(dv)` (ii)
Given `(du)/(dt)=0.01ms^(-1)`
`rArr (dv)/(dt)=((1.2)/(0.4))^(2)xx0.01=0.09ms^(-1)`.
Magnitication , `m=(v)/(u)`
Differenting with respect to time t,
`(dm)/(dt) =(u(dv)/(dt)-v(du)/(dt))/(u^(2))=((0.4)xx0.009-1.2xx(0.01))/((-0.4)^(2))`
`=(-0.036-0.012)/(0.16)=-(0.048)/(0.016)=-0.3s^(-1)`