Calculating the Final Image Location in a Compound Lens System

[jcvince17]

Homework Statement

A compound lens system consists of two converging lenses, one at x= -20cm with focal length f₁ = +10cm, and the other at x= +20cm with focal length f₂ = +8cm . An object 1 centimeter tall is placed at x = -50cm.

What is the location of the final image produced by the compound lens system? Give the x coordinate of the image.

Homework Equations

1/f = 1/s' + 1/s

m = y'/y = -(s'/s)

The Attempt at a Solution

I am not sure. I believe I need to do the first equation I posted twice. Using the s' from the first lense as the s (object) for the second lens.

for lens 1 i get:

1/f = 1/s' + 1's

1/10 = 1/s' + 1/-30 (used negative since it is on oppostie side of mirror?)

s' = 7.5 cm (is that telling me it is 7.5 from the lens correct? so in between the lens and the f.)

for lens 2 i get:

1/f = 1/s' + 1/s

1/8 = 1/s' + 1/-7.5 (used negative since it is on oppostie side of mirror?)

s' = 3.87 cm (is that telling me it is 3.87 from the lens correct? so in between the lens and the f.)


if these are corect it is telling me the final image is located at x = + 23.87 but that is wrong.

so where did i go wrong in my calculations?

 

[jcvince17]

well, I have gotten several wrong answers and have only two chances left.


edit - still working on it

i found something about solving for a resultant focal point.

F = 1 / f1 + 1 / f2 - d / f1f2
F = (1/10 + 1/8) - (40/10*8)
F = -.275

if i then take this into my original equation
1/f = 1/s' + 1/s
1/-.275 = 1/s' + 1/30
s' = -3.67

if this is correct then s' is located at what?

 

[jcvince17]

tried it again

lens 1:

1/f = 1/s' + 1/s
1/10 = 1/s' + 1/30
s' = 7.5 = @ x = -12.5


lens 2:
1/f = 1/s' + 1/s
1/8 = 1/s' + 1/32.5
s' = 10.6 = final image is at x = + 30.6 ?

 

[jcvince17]

This is due at 11pm tonight, if anyone can please help me. I don't know where I am going wrong here.

Thanks.