Understanding the Magnification of a Telescope: Why is M = f_o / f_e?

In summary, the 2 converging lens telescope's magnification is given by M = f_o / f_e, which is the angular magnification formula. This is different from the linear magnification formula, M = image distance / object distance. The angular magnification formula can also be expressed as M = tan E / tan E_0, where E_0 is the angle subtended by the object at the front focal point of the objective and E is the angle subtended by the image at the rear focal point of the eyepiece. While these two formulas may seem different, they are equivalent and in direct proportion.
  • #1
quietrain
655
2
hi, i don't understand why the 2 converging lens telescope's magnification is given by M = f_o / f_e

where f_o = focal length of objective, and f_e = focal length of eyepiece...

issn't M = image distance / object distance...

in the case of a infinity object, the image is also infinity for the eye to view it without straining.

so why is there still a magnification ? infinity over infinity?

need help thanks a lot!
 
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  • #2
quietrain said:
the 2 converging lens telescope's magnification is given by M = f_o / f_e

This is angular magnification.

issn't M = image distance / object distance...

This is linear magnification.

They're different things.
 
  • #3
oh. i see.

i look up the angular magnification and the formula was

M = tan E / tan E_0 (wikipedia)

where E_0 = the angle subtended by the object at the front focal point of the objective and E =the angle subtended by the image at the rear focal point of the eyepiece.

so how does this become the formula

M = f_o / f_e ?

even with small angle approximation the formula M = tan E / tan E_0 is still not M = f_o / f_e ?
 
  • #4
The two formulas for angular magnificaion are equivalent by their very nature and you don't need to prove equality by making arithmetic substitutions
They are both ratios (the focal length equation is simply a ratio of spatial dimensions, where as the angle equations is a ratio of the angles subtended by the magnitude of these dimensions)
Both ratios are in direct proportion.
 

1. What is magnification in a telescope?

Magnification in a telescope is the degree to which the image of an object is enlarged when viewed through the telescope. It is determined by the ratio of the focal length of the telescope's objective lens or primary mirror and the focal length of the eyepiece.

2. How is magnification calculated in a telescope?

Magnification is calculated by dividing the focal length of the objective lens or primary mirror by the focal length of the eyepiece. For example, if the focal length of the objective lens is 1000mm and the focal length of the eyepiece is 10mm, the magnification would be 100x (1000/10=100).

3. Is higher magnification always better in a telescope?

No, higher magnification does not always mean better in a telescope. While higher magnification can make objects appear larger, it can also decrease the field of view and make the image appear dimmer or blurrier. The optimal magnification for a telescope will depend on the aperture and atmospheric conditions.

4. Can a telescope have too much magnification?

Yes, a telescope can have too much magnification. Using too much magnification can result in a blurry or distorted image due to the limitations of the telescope's optics. It is important to find the right balance of magnification for the telescope's aperture and atmospheric conditions.

5. How does the type of telescope affect magnification?

The type of telescope can affect magnification in several ways. Refracting telescopes, which use lenses, have a maximum practical magnification of around 50 times the aperture in inches. Reflecting telescopes, which use mirrors, can typically handle higher magnifications. Additionally, the quality of the telescope's optics and its focal length will also impact the achievable magnification.

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