Solve Antique Telescope Homework: Find Fe Focal Length

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The discussion centers on calculating the focal length of the eyepiece (Fe) for an antique telescope with a total length of 60 cm and a magnification of 20. The initial attempt incorrectly used a negative focal length due to confusion about the sign convention for magnification. Clarification was provided that the angular magnification should be treated as -20, leading to a positive focal length for the eyepiece. The correct formula derived from the relationship between focal lengths and magnification resulted in Fe being calculated as approximately 2.857 cm. This highlights the importance of correctly applying sign conventions in optics calculations.
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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 throught 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 throught 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 throught 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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