Focal length, power and magnification

In refracting (lens not mirror based) telescopes, to have a large magnification the objective lens needs to have a large focal length.

if $\text{Power} = \frac{1}{\text{focal length}}$ then that means telescopes that have a large magnication have a low powered objective lens?

Also if a large focal length provides a greater magnification then a lens which refracts (bends) the light the least is better, so surely that's no lens at all (or just a pane of glass)?

How does a longer focal length (and so low power lens) provide a greater magnitude? It seems to my common sense (obviously wrong) that a lens that bends the light more would provide a greater magnification? (As the rays would be spread over a greater area for a shorter length)

Larger focal length lenses do have less optical power, as you have stated.

A single lens, however, does not form a telescope. You need two lenses, e.g. an objective lens and an eye lens with respective focal lengths $f_\text{eye}$ and $f_\text{obj}$. The magnification of a telescope constructed with these two lenses will have a magnification

$$M = \left|\frac{f_\text{obj}}{f_\text{eye}}\right|$$

A larger objective focal length does result in larger magnification -- however, the individually higher optical power element in this telescope is actually the eye lens.

To achieve large magnification, you need either a really short focal length eye lens or a really long focal length objective. It is much more difficult to manufacture high quality short focal length lenses than it is to manufacture high quality long focal length lenses. This is because short focal lenses are more curved. In the thin lens approximation, the focal length of a lens of index $n$ in air with radii of curvature $R_1,R_2$ is given by:

$$\frac{1}{f} = (n-1)\left( \frac{1}{R_1} - \frac{1}{R_2}\right)$$

As a lens focal length gets longer and longer, the lens becomes less and less curved. An infinite focal length lens would be a pane of glass, or a window.

Can you use a "window" as your objective lens to achieve infinite magnification? In principle, yes, however, this is not a practical solution primarily because the length of a two-lens telescope is $L = f_\text{obj} + f_\text{eye}$. In other words, such a telescope would need to be infinite in length.

I also have been intrigued by this apparent dilemma. I think I have it worked out, maybe only half way. This is all my conjecture, since most people just are not thinking down these lines.

The longer the focal length of the objective lens the narrower the field of view, right? So the longer the focal length, the more the lens is gathering data from a narrower space.. So for best magnification you would want to gather the most data from the smallest area. Thus long focal length.

Now you might also notice that when you use a shorter focal length eyepiece, the field of view is also narrowed. This is because while light beams that are outside of that field of view are still hitting the eyepiece they are being refracted off to useless places if not aberrated altogether with.

So now, lets take for example a 120cm objective, and couple if with a 25mm eyepiece. That is a 48 power telescope. Photons from outer limits of your field of view are coming at your cornea at particularly steep angles. The closer to the limits of the field of view, the sharper the angle it hits your cornea. Beyond that field of view, light beams are still hitting the objective, but are not refracted down to the eyepiece..

Now let's put on a 6mm eyepiece.. Your field of view is now reduced.. What has happened? All the same information is coming down to the eyepiece from the objective. But outside your new field of view the eyepiece is not sending this information to the cornea. It is refracting it or aberrating it too much to be of use.

So let's see what happens when you shorten the objective's focal length. We'll say you have a 120cm Objective and the 6mm eyepiece... With the 6mm eyepiece you are already missing most of the information that the objective is sending the eyepiece... While I can see that I am still missing part of the full conceptual understanding, I don't see using an increased field of view of a shorter lens getting more magnification.

Perhaps magnification is a balance between limiting your field of view with the objective and limiting your field of view with the eyepiece...... Anytime you increase your field of view you are reducing magnification. Decrease field of view, increase magnification. So what you need to do with a telescope is magnify (with the eyepiece) a smaller field of view (from the objective).... I'll be darned, that's what we do......

I am not crazy about guys using mathematics to explain things .. so I will try to be more graphical than theirs so here you go..

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down vote

It is easy to understand that the higher the focal length is the narrower the angle of the image from the objective lens or mirror to the eyepiece. So the narrower the angle (higher focal length that is) the deeper the image can go into the eyepiece without losing focus. The lower the focal length is the shorter the depth of focus. It means that you have less play with your focuser with shorter focal length than longer focal length. In fact you can still get same magnification for both focal lengths but the one with shorter focal length will run out of its focus long before the one with a longer focal length. This is why you notice the expansion of the "star" as it goes out of focus in either direction. The longer focal length is more forgiving with the depth of focus thus giving you more play with your focuser. The longer focal length affords you to move more forward toward the objective lens or mirror to get larger image while still in focus than with shorter focal length which expands the image beyond the field of your eyepiece in use. It has a lot to do with the eyepiece ability to eat the whole image in one gulp.

• This is a copy-paste of your other answer. Please don't do that. – Kyle Kanos Feb 1 '16 at 2:50