Question

An antelope is at a distance of 20.0 m from a converging lens of focal length 27.0 cm. The lens forms an image of the animal.
(a) If the antelope runs away from the lens at a speed of 4.10 m/s, how fast does the image move? mm/s

Answer 1

Solution: From the question we have

An antelope is at a distance of 20.0 m from a converging lens of focal length 27.0 cm

The lens forms an image of the animal.

To find: If the antelope runs away from the lens at a speed of 4.10 m/s, how fast does the image move? mm/s

Let:
u be the object distance,
v be the image distance,
f be the focal length.

With Real is Positive sign convention:
1 / u + 1 / v = 1 / f ...(1)
1 / 20.0 + 1 / v = 1 / 0.27
1 / v = 1 / 0.27 - 1 / 20.0
1 / v = 3.65 m^(- 1)

That's a real image on the opposite side of the lens from the antelope.

Differentiating (1) with respect to time:
- (1 / u^2)(du / dt) - (1 / v^2)(dv / dt) = 0
dv / dt = - (v^2 / u^2)(du / dt)

Putting du / dt = - 4.10 m/s, v = 3.65 m, u = 20.0 m:
dv / dt = - [ (3.65)^2 / 20^2 ](- 4.10)
= 0.136 m/sec.=136.55mm/sec to 3 sig. fig.

As dv / dt > 0, v is increasing, and the image is moving away from the lens.