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

Answer

A, C

Explanation

`f=0.3m` `u=-0.4m`
Using lens formula
`(1)/(v)-(1)/(-0.4)=(1)/(0.3)impliesv=1.2m`
Now we have `(1)/(v)-(1)/(u)=(1)/(f)`, differentiationg w.r.t.t
we have `-(1)/(v^2)(dv)/(dt)+(1)/(u^2)(du)/(dt)=0` given `(du)/(dt)=0.01`m/s`
`implies((dv)/(dt))=(1.20)^2/((0.4)^2)xx0.01=0.09m/s`
So, rate of separation of the image (w.r.t. the lens)
Now, `m=(v)/(u)implies(dm)/(dt)=((udv)/(dt)-(vdu)/(dt))/(u^2)=(-(0.4)(0.09)-(1.2)(0.1))/((0.4)^2)=-0.35s^(-1)
Magnitude of rate of change of lateral magnification=0.35`s^(-1)`.