Solve Antique Telescope Homework: Find Fe Focal Length

[xinlan]

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..

 

[tiny-tim]

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!

I don't know much about optics …

Why have you put a minus there?

 

[andrevdh]

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.

 

[tiny-tim]

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?

 

[xinlan]

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

 

[xinlan]

I got it..
thanks..

 

[The_Hunter]

How'd you find it?

Here's How:

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

Fe= 2.857