How do I determine the distance between two lenses in a telephoto lens system?

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Homework Help Overview

The discussion revolves around a telephoto lens system consisting of an objective lens and a concave lens, focusing on determining the distance between the two lenses and the placement of film in relation to these lenses. The problem involves understanding focal lengths, image distances, and magnification in optics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between image distance and object distance for the two lenses, with one participant attempting to calculate the necessary distances and expressing confusion over the signs used in the lens formula. Others question the understanding of image formation in the context of the system.

Discussion Status

The discussion includes attempts to clarify the calculations related to the distances and magnifications involved. Some participants have provided insights into potential mistakes in the calculations, while others express uncertainty about the conceptual understanding of image formation in this lens system.

Contextual Notes

Participants note the importance of sign conventions in lens equations and the implications of virtual images in the context of the problem. There is also a mention of the need for clarity in the calculations to avoid confusion.

creative_wind
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Homework Statement



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?

Homework Equations



1/f=1/di + 1/do

M=-di/do

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!
 
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creative_wind said:
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!

Hi creative_wind! :smile:

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? :smile:
 
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!
 
… don't talk to me about reality …

creative_wind said:
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 …

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

Forget reality!

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

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!
 

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