- #1
jias
- 1
- 0
Can anyone explain to me the procedure that one would follow to you find the focal length of a concave lens using a convex lens? Thanks.
Find the focal length of a concave lens using a convex lens?
- Context: Undergrad
- Thread starter jias
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Discussion Overview
The discussion revolves around the procedure for determining the focal length of a concave lens using a convex lens. It includes theoretical aspects and practical considerations related to lens behavior and image formation.
Discussion Character
- Technical explanation
- Homework-related
Main Points Raised
- One participant inquires about the procedure to find the focal length of a concave lens using a convex lens.
- Another participant describes a method involving the formation of images by the convex lens and the subsequent effect of introducing a concave lens, including the use of the lens formula.
- A third participant comments on the phrasing of the explanation, suggesting a more direct approach would be preferable.
- The initial poster acknowledges the feedback and expresses understanding of the concept after the clarification.
Areas of Agreement / Disagreement
Participants generally agree on the method described for finding the focal length, but there is a difference in how the explanation should be framed. The discussion remains somewhat unresolved regarding the best way to communicate the procedure.
Contextual Notes
There are assumptions made about the understanding of lens formulas and image formation that may not be explicitly stated. The discussion does not resolve all potential uncertainties in the method described.
- #2
armis
- 103
- 0
Yeah there is a way
Assume the convex lens creates an image of S at the point S'. Now if you put a concave lens in front of the convex one then the image of S will move further and will be at point S''. Now consider a reverse path of light: S'' is now the object and you'll see it's image at S' which is created by the concave lens. Let the distance from the "center" of the concave lens to S'' be a and the distance from S' to the "center" of the concave lens be b. Then using the lens formula:
[tex]-\frac{1}{f}=\frac{1}{a}-\frac{1}{b}[/tex] notice there is a minus near b and f. It's because the image is imaginary
Thus
[tex]f=\frac{ab}{b-a}[/tex]
I hope I didn't make any mistakes
Assume the convex lens creates an image of S at the point S'. Now if you put a concave lens in front of the convex one then the image of S will move further and will be at point S''. Now consider a reverse path of light: S'' is now the object and you'll see it's image at S' which is created by the concave lens. Let the distance from the "center" of the concave lens to S'' be a and the distance from S' to the "center" of the concave lens be b. Then using the lens formula:
[tex]-\frac{1}{f}=\frac{1}{a}-\frac{1}{b}[/tex] notice there is a minus near b and f. It's because the image is imaginary
Thus
[tex]f=\frac{ab}{b-a}[/tex]
I hope I didn't make any mistakes
- #3
tiny-tim
Science Advisor
Homework Helper
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- 258
erm, armis … you're doing it again!
If you'd changed "Then using the lens formula:" to "Then use the lens formula!
", and stopped there, then that would have been a good hint. 
(The OP can always come back and ask for more help, if necessary, of course)
If you'd changed "Then using the lens formula:" to "Then use the lens formula!
", and stopped there, then that would have been a good hint. (The OP can always come back and ask for more help, if necessary, of course
)- #4
armis
- 103
- 0
Oh... I thought I was supposed to do that only in the homework section, sorry. But it does make a lot more sense your way. I think I finally got the idea
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