Solve Antique Telescope Homework: Find Fe Focal Length

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Homework Help Overview

The problem involves an antique telescope with a specified magnification and a fixed tube length, requiring the determination of the focal length of the eyepiece lens. The context is within the subject area of optics, specifically dealing with lens systems and magnification concepts.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between the focal lengths of the objective and eyepiece lenses, referencing the magnification formula. There are questions about the sign convention used in the magnification equation and the implications of the telescope's configuration on the focal lengths.

Discussion Status

Some participants have offered clarifications regarding the sign of the magnification and its effect on the focal length of the eyepiece. There is ongoing exploration of the implications of using converging lenses and the resulting focal lengths, with no explicit consensus reached on the correct approach.

Contextual Notes

Participants are navigating potential misunderstandings regarding the sign conventions in optics and the characteristics of converging lenses. The original poster's calculations and assumptions are under scrutiny, with some noting discrepancies in the expected outcomes.

xinlan
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Homework Statement



On one of the shelves in your physics lab is displayed an antique telescope. A sign underneath the instrument says that the telescope has a magnification of 20 and consists of two converging lenses, the objective and the eyepiece, fixed at either end of a tube 60.0 cm long. Assuming that this telescope would allow an observer to view a lunar crater in focus with a completely relaxed eye, what is the focal length Fe of the eyepiece?

Homework Equations



Length (L) = Fo+Fe
Which Fo = Focal length of the objective lens
Fe = Focal length of the eyepiece

M = -Fo/Fe

The Attempt at a Solution



L = Fo+Fe
60 = Fo+Fe
Fe = 60-Fo

M = -Fo/Fe
20 = -(60-Fe) / Fe
Which Fe would be -3.16

But I got wrong..
Please help me..

thank you..
 
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xinlan said:
A sign underneath the instrument says that the telescope has a magnification of 20 and consists of two converging lenses
…
M = -Fo/Fe

Hi xinlan! :smile:

I don't know much about optics …

Why have you put a minus there? :confused:
 
Make the angular magnification -20, that is the final image is inverted when viewed through the eyepiece. Which would give you a positive focal length for the eyepiece as required for a convex lens.
 
Last edited:
andrevdh said:
Make the angular magnification -20, that is the final image is inverted when viewed through the eyepiece. Which would give you a positive focal length for the eyepiece as required for a convex lens.

But the question says "two converging lenses" - wouldn't that make both negative? :confused:
 
andrevdh said:
Make the angular magnification -20, that is the final image is inverted when viewed through the eyepiece. Which would give you a positive focal length for the eyepiece as required for a convex lens.

Even though the M is -20, but the answer would still be the same..
it just change the minus sign into positive sign..
I put the positive focal length and I still got wrong
 
I got it..
thanks..
 
How'd you find it?

Here's How:

M=Fo/Fe
20= (60-Fe)/Fe
Fe = (60/21)

Fe= 2.857
 
Last edited:

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