Question

An object is approaching a thin convex lens of focal length 0.3 m with a speed of 0.01 m/s. Find the magnitudes of the rate of change of position and rate of change of lateral magnification, of image when the object is at a distance of 0.4 m from the lens.

Answer 1

from lens maker formula,

1/v - 1/u = 1/f

given, f = 0.3 m, u = -0.4m

so, 1/v - 1/-0.4 = 1/0.3

or, 1/v = 1/0.3 - 1/0.4 = 1/1.2

v = 1.2 m

now, differentiating 1/v - 1/u = 1/f with respect to t,

-1/v² dv/dt + 1/u² du/dt = 0

or, dv/dt = (v/u)² du/dt

putting, du/dt = 0.01 m/s , v = 1.2 m and u = -0.4 m

so, dv/dt = (1.2/-0.4)² × 0.01

= 0.09 m/s

hence, speed of image with respect to lens is 0.09 m/s .

from formula of magnification

magnification, m = v/u

differentiating with respect to time both sides,

dm/dt = (u dv/dt - vdu/dt)/u²

= (-0.4 × 0.09 - 1.2 × 0.01)/(-0.4)²

= (-0.036 - 0.012)/0.16

= -0.048/0.16

= -0.3 m/s

hence, magnitude of rate of change of lateral magnification is 0.3 m/s