Question

determine how far in front of the lens the object should be placed..

You place an object 19.6 cm in front of a diverging lens which has a focal length with a magnitude of 14.2 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 4.35.

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 = m1/4.35 = 0.247.m1
0.2298.m1 = f / (u2 -f) ----- (2)

combining (1) and (2) ..
0.2298 {f / (u1 - f)} = f / (u2 -f) .. .. top-line f cancels out, u1 = 19.6cm, f = -14.2cm**

0.2298 / {19.6 - [-14.2]} = 1 / (u2 - [-14.2])

0.006798 (u2 + 14.2) = 1

u2 = 1/0.006798 - 14.2 .. .. ?132.90 cm

Answer 2

Start with the Magnification

M = -q/p

1/4.35 = -q/p

p = -4.35q

Then, 1/f = 1/p + 1/q

1/-14.2 = 1/-4.35q + 1/q

1/-14.2 = 3.35/4.35q

4.35q = -47.57

q = -10.9 cm

p = -4.35(-10.9)

p = 47.6 cm

Thus the object should be 47.6 cm in front of the lens