View Full Version : Ocular Question

ArgoNavis

2005-Sep-12, 09:34 AM

dear all

sometimes beginners come out with the most interesting questions.....

had a few people over on saturday night and we were explaining how to caculate magnifications: divide the telescope focal length by the eyepeice focal length. I was then asked why? This is a question I cannot explain. Why does the magnification depend on how many times the focal length of the eyepeice (which I understand does the actual magnification) divides into the telescope focal length?

can anyone with more knowledge of optics than I please explain?

ta

Ricimer

2005-Sep-12, 11:12 AM

After taking two classes in electrodynamics/Optics and lasers...I don't know. Well...I feel silly, as I used to know it, and I've even derived the equations. I'll do what I can to find out exactly whats up.

Ricimer

2005-Sep-12, 11:31 AM

Okay, I'm starting to remember. Answer without math (that part of my brain isn't working just yet):

The magnification of an object, for a single lense, is a ratio of the object size, to the image size. These sizes are directly related to the distance between the object and the image locations, and this distance is controled by the lenses focal length. Magnification is generally proportional to the inverse of the focal length. Long focal lengths are smaller magifications (an infinitely long focal length, or a flat mirror or flat piece of glass, produces no magnification).

So the focal length of a lense controls the ratio between object and image sizes. Put two lenses together, and you've got to factor their magnifications together. So magnification of lense one * magnification of lens two gives the total magnification. However, the magnification is the inverse of the focal length...so you need the focal lengths divided by eachother.

I think. That's what you'll get from me at 4 am without pulling out an optics textbook.

Dave Mitsky

2005-Sep-12, 12:18 PM

Think of the ocular as a magnifying glass that enlarges the prime focus image. Its focal length corresponds to the distance from that image. A 5mm eyepiece produces a large, up-close image since it "places" your eye a short distance, namely 5mm, from prime focus. A 55mm eyepiece gives one a correspondingly long view at low magnification.

Dave Mitsky

Jeff Root

2005-Sep-12, 03:50 PM

Argo,

Any good answer to your question will include the words

"refraction" and "angle". It is about the behavior of light,

and geometry.

-- Jeff, in Minneapolis

Powered by vBulletin® Version 4.2.2 Copyright © 2014 vBulletin Solutions, Inc. All rights reserved.