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

0.09 m/s , 0.3 /s

Explanation

Differentiating the lens formula `(1)/(v) - (1)/(u) = (1)/(f)`
with respect to line .
we get `-(1)/(v^(2)) , ("dv")/("dt") + (1)/(u_(2)) , (du)/(dt) = 0 ` (as f = constant )
`therefore ((dv)/(dt)) = ((v^(2))/(u^(2))) (du)/(dt) … (i)`
Further , substituting proper values in lens formula , we have
`(1)/(v) + (1)/(0.4) = (1)/(0.3)(u= -0.4 m , f= 0.3 m )` or v = 1.2 m
Putting the values in equation (i)
Magnitude of rate of change of position of image = `0.9 m//s`
Lateral magnification
`m = (v)/(u) therefore (dm)/(dt) = (u.("dv")/("dt") - v ("du")/("dt"))/(u^(2))`
`=((-0.4)(0.09)-(1.2)(0.01))/((0.4)^(2)) = -0.3//s`
`therefore ` Magnitude of rate of change of lateral magnification = 0.3 /s.