Question

You place an object 24.0 cm in front of a diverging lens which has a focal length with a magnitude of 14.6 cm. Determine how far in front of the lens the object should be placed in order to produce a new image that is 4.25 times smaller than the original image.

Answer 1

1/u + 1/v = 1/f ? u/u + u/v = u/f .. ..( u/v = 1/magnification)

1/m = u/f - 1 = (u - f) / f

m1 = f / (u1 - f) ---- (1)

By comparison with derivation for m1 above .. m2 = f / (u2 - f)

But .. m2 = m1/4.25 = 0.235m1

0.235m1 = f / (u2 -f) ----- (2)

combining (1) and (2) ..

0.235 {f / (u1 - f)} = f / (u2 -f) .. .. top-line f cancels out, u1 = 24cm, f = -14.6cm**

0.235 / {24 - [-14.6]} = 1 / (u2 - [-14.6])

0.0060957 (u2 + 14.6) = 1

u2 = 149.45cm ?

** concave lens has virtual focal points, hence f = -14.6cm by convention