Question

You place an object 23.6 cm in front of a diverging lens which has a focal length with a magnitude of 10.0 cm. Determine how far in front of the lens the object should be placed in order to produce an image that is reduced by a factor of 3.75.

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 = m1/3.75 = 0.267.m1

0.267.m1 = f / (u2 -f) ----- (2)

combining (1) and (2) ..

0.267 {f / (u1 - f)} = f / (u2 -f) .. .. top-line f cancels out, u1 = 23.6cm, f = -10cm**

0.267 / {23.6 - [-10]} = 1 / (u2 - [-10])

0.00794 (u2 + 10) = 1

u2 = 1/0.00794 – 10

=> u2 = 116 cm (ans)

** concave lens has virtual focal points, hence f = -10cm by convention