# Compound lenses final image

1. Nov 8, 2008

### creative_wind

1. The problem statement, all variables and given/known data

A simple camera telephoto lens consists of two lenses. The objective lens has a focal length f1 = +36.0 cm. Precisely 33.0 cm behind this lens is a concave lens with a focal length f2 = -10.0 cm. The object to be photographed is 3.67 m in front of the objective lens.

(a) How far behind the concave lens should the film be placed?

(b) What is the lateral magnification (ratio of image height to object height) of this lens combination?

2. Relevant equations

1/f=1/di + 1/do

M=-di/do

3. The attempt at a solution

My understanding is that the di1 (the objective lens) is the do2 (for the second/concave lens). Calculating di1 gives me 0.399 m which is greater than the distance between the two lenses, so that would mean do2 on the same side that the film would be on. I tried using 0.399m - 0.33 m = 0.069m = do2 to calculate di2 anyways, but that answer is wrong. I'm stuck as to what value to use for do2. As for part b) I know to find the magnification of each lens and then multiply the two magnification values to get the final magnification. I just need to figure out part a). Thanks so much for any help!

Last edited: Nov 8, 2008
2. Nov 9, 2008

### tiny-tim

Hi creative_wind!

You really need to show us what you did if we're going to see where the mistake is.

I'll guess … did you remember to make the 1/f for the second lens negative?

3. Nov 9, 2008

### creative_wind

Okay so as I wrote above I was using 0.399 m -0.33 m = 0.069 m. It should be 0.33 m - 0.399 m = -0.069 m, which gives me the correct answer. I can't say I really understand the concept of locating the position of the image of the first lens since it never really forms (hits second lens first) but at least I understand how to answer the question and could do it on an exam now. Thanks for your help!

4. Nov 10, 2008

### tiny-tim

… don't talk to me about reality …

You're really into reality aren't you? :rofl:

Forget reality!

phsyics is reality … maths (including geometry) isn't

and this is a geometrical trick

the rays behave as if they did come from the virtual image …

just like the image in an ordinary flat mirror, which saves you the effort of having to draw incidence and reflection angles carefully!