Question

A converging lens of focal length 3.6 m produces a magnified image 1.8 times the size of the object. What is the object distance if the image is a) real b) virtual?

Answer 1

f = 3.6m
m = 180% = 1.8

a) i > 0
m = -i/o

If i > 0, then (+) = -(+)/o
o must be negative, which means that the object is virtual.

o/i = f/(i - f)
m = -i/o
(o/i)/-1 = -i/o
(f - i)/f = -i/o
(3.6 - i)/3.6 = 1.8
3.6 - i = 6.48
i = -2.88

Thin-lens Eq.) (1/f) = (1/i) + (1/o)
(1/3.6) = (1/-2.88) + (1/o)
o = 1.6

The object distance is -1.6m from the lens.

b) i < 0
m = -i/o

If i < 0, then (+) = -(-)/o
o must be positive, which means that the object is real.

We already did all the work.

o = 1.6

The object distance is 1.6m from the lens.