Converging and Diverging lenses?

Question

I got half of this problem correct so far
. A diverging lens (f = -10 cm) is located 20.0 cm to the left of a converging lens (f = 31.5 cm). A 3.00 cm tall object stands to the left of the diverging lens, exactly at its focal point.
(a) Determine the distance of the final image relative to the converging lens.
I got this answer correct, -121.15 cm
(b) What is the height of the final image (including proper algebraic sign)? in cm
I know that hi/ho = -di/do but I'm not getting the correct answer. Any help will be appreciated.

Answer 1

Ok, this is a very interesting problem, because is a system of lenses, so, first of all make a graphic, then you will have :

The distance from the object to the diverging lens is 10 cm

the object distance is positive, the focal distance of a diverging lens is negative, so let's find the image made by the diverging lens :

Diverging lens :

Using : 1 / p + 1/q = 1 / f

p = object

q = image

1 / 10 + 1 / q = -1 / 10

q = - 5 cm

amplification one : m1 = -(-5) / 10 = 1/2

Second lend : Converging lens :

p = 5 + 20 cm = 25 cm

1 / 25 + 1 / q = 1 / 31.5

25 - 31.5 / 31.5*25 = 1 / q

-6.5 / 31.5*25 = 1 / q >>>> q = -121.15 cm

Second amplification = -( -121.15 / 25) = 4.8

It's correct

To find the altitude of the object :

FINAL AMPLIFICATION : m1*m2 = 4.8*1/2 = 2.4

Let "x" be the final altitude :

x / 3 = 2.4

x = 7.2 cm

The final image is 7.2 cm tall.

Hope that helped