Magnification vs. Exit Pupil

[Utwo]

Hi guys, sorry to bug you with another question about this.

Logically, it seems that magnification works by reducing the field of view angle, so that you're looking at a smaller part of the sky. Now, does this smaller view of the sky get projected over your entire vision? Or is it just enough to fill the exit pupil?

I guess what I mean to say is, since the size of the exit pupil is equal to the aperture divided by the magnification, then the higher the power, the lower the exit pupil for any given aperture. Now, one thing I've noticed, using my tiny 70mm refractor (350mm focal length), is that the image of Saturn with the 6mm eyepiece PLUS 2x Barlow is not that much larger than the image of Saturn with just a 25mm eyepiece. You'd think that, since the former is 116x and the latter is 14x, it seems like Saturn should be more than 8 times larger in the former case, but this is not my experience.

I'm wondering if this is because of the exit pupil. At 116x the size of my exit pupil is about 0.6mm, whereas at 14x it is about 5mm. I'm wondering if the reason that Saturn doesn't look much larger is that, at 116x, the image is magnified at over 8 times the 14x, but that image is displayed on a pupil that is about 8 times smaller.

And I guess that, if this is true, a corralary to it would be that increasing the aperture without increasing the magnification would result in an increase in the exit pupil size, and therefore, an increase in apparent size of the objects being viewed.

I guess that doesn't make sense.

Edit: In other words, I guess what I'm asking is, does a larger exit pupil, given the same magnification, show a wider piece of the sky? Or does it show the same piece of the sky stretched across a wider area? I guess the former would have to be true, or there would be no point to magnification, because it seem that the math works out such that, as magnification increases, all objects increase in size to the same proportion that the exit pupil decreases in size. So, if the latter were true, there would be no point to magnification at all.

[yuzuha]

the exit pupil is the diameter of the beam from the eyepiece as it enters your eye. Smaller than about 1/2mm and things look very dim and can be obscured by junk floating in your eyeball. Larger than about 6 (depending on age... youngsters can take in about 7.7mm) and you are wasting light since the iris in your eye can't expand far enough to take it all in.

Here is a nice scope calculator that tells you all kinds of things about various scope and ep combinations http://www.stargazing.net/naa/scopemath.htm

[Hornblower]

First things first. If your 14x image of Saturn looks about as large as the 116x image, it must be hideously out of focus. Try to sharpen the focus by carefully moving the eyepiece in and out. If that fails, it might mean that the telescope, or at least one of the eyepieces, is a piece of junk.

Let's try again to explain what the exit pupil is and what it is not. This is difficult to explain verbally, and I continue to be P.O.ed by the difficulty of finding good ray tracing diagrams online. Here goes.

The light rays from all points in the field of view of the eyepiece converge until they pass the plane of the exit pupil and then diverge. If you have placed the pupil of your eye in that same plane and have it properly centered, all of the light gets into the eye and is projected onto the retina. The result is an enlarged image on your retina.

The exit pupil is not a place on which a field of view is projected. It is merely the narrowest place in the beam that comes out of the telescope. Its diameter becomes an issue only at extremely low magnification, in which case it might be larger than your eye pupil with the result that some light is cut off.

Clear as mud? If so, don't try to make sense of it without a good ray tracing diagram and someone to walk you through it visually. Just use the telescope at various amounts of magnification and enjoy the view, provided the optical quality is good enough to make that possible.

[Dave Mitsky]

There's a ray tracing in this article on exit pupil.

Dave Mitsky

[Hornblower]

There's a ray tracing in this article on exit pupil.

Dave Mitsky

Thanks for the link. Your searching endurance was better than mine.

[Utwo]

I'm sorry, I am still not sure if I fully understand. Let me ask this: if my pupil is 5mm, and the exit pupil of the telescope I'm looking through is 2.5mm, will the image coming through the telescope appear to take up about half of my own field of view?

[Hornblower]

I'm sorry, I am still not sure if I fully understand. Let me ask this: if my pupil is 5mm, and the exit pupil of the telescope I'm looking through is 2.5mm, will the image coming through the telescope appear to take up about half of my own field of view?

As long as the edge of your eye's pupil does not encroach on any of the light that is coming out of the eyepiece, you will see the entire field of view that the telescope is capable of providing with that particular eyepiece.

Reminder: The telescope's exit pupil has to fit inside your eye's pupil, not vice versa. It is the narrowest part of the bundle of light rays that emerge from the eyepiece. For any given objective its diameter is determined concurrently with the magnification by the focal length of the eyepiece. When the exit pupil is small, as is the case with high magnification, you have a greater margin for error in your eye position. You can be slightly off center or a bit too far in front of or behind the plane of the exit pupil, and you can still get all of the light into your eye cleanly and have an unimpaired view. At low power, with the exit pupil just as large as the eye's pupil, your eye position must be exact. If you are off center you will lose some brightness, and if you are too far forward or back you will lose the outer parts of the field of view.

[Utwo]

Thanks. I guess what I'm wondering is, what is the FOV that a telescope is capable of showing with a particular eyepiece? Let me use an example with a hypothetical telescope so that we can use numbers.

Suppose I have a telescope with a 50mm aperture. I have 2 eyepieces, one that allows me to view the sky at 10x power and another that allows me to view the sky at 40x power. The top row shows the AFOV for each magnification, pointed at some unnamed planet-like object.

Now, at 10x, my exit pupil is 5mm (50mm aperture / 10x power = 5mm exit pupil), and at 40x, my exit pupil is 1.25mm.

So I guess my question is, does each respective field of view get "squeezed" to fit the exit pupil, as in the second row? Or does it cut "clipped" to fit the exit pupil, as in the third row?

It seems that what is happening in the second row couldn't be true, because if the "exit pupil = diameter of aperture / power", then any time you increase the power, you exactly undo what you just magnified by squeezing it into a proportionally smaller exit pupil.

But the second option, in the third row, seems perhaps just as strange, because in addition to the AFOV being smaller with the higher magnification, the image is further "clipped" so as to fit the smaller exit pupil.

Then again, perhaps the first option (second row) is correct, and the lense of the eye projects this small exit pupil across the entire retina, thereby blowing the image back up.

Here's the image I was referring to:

http://i67.photobucket.com/albums/h2.../exitpupil.jpg

[Hornblower]

I have concluded that any further attempts at a verbal explanation from afar are futile. I would recommend finding an astronomy club in your area, where someone could walk you visually through a ray trace diagram to show how the optical geometry really adds up.

In the meantime, just enjoy the view without worrying about a technical explanation of how it works. Point your telescope at the Moon, so you will have something bright filling up the field of view. Then practice moving your eye around behind the eyepiece so you can see what it looks like when you are in the sweet spot, and how it deteriorates when you are out of position.

[hhEb09'1]

It seems that what is happening in the second row couldn't be true, because if the "exit pupil = diameter of aperture / power", then any time you increase the power, you exactly undo what you just magnified by squeezing it into a proportionally smaller exit pupil.

I agree with the "couldn't be true" part , and also with Hornblower's suggestion in their first post that you must be doing something wrong. I would go back and check to make sure that you've used the lenses in the sequence that you report in the OP. Check, for instance, that the image looks twice as large using the 6mm eyepiece PLUS 2x Barlow, as it does with the 6mm alone.

[Dave Mitsky]

Thanks. I guess what I'm wondering is, what is the FOV that a telescope is capable of showing with a particular eyepiece?

Edit

The answer to your question depends entirely on the focal length of the telescope, the focal length of the eyepiece, and the apparent field of view of the eyepiece.

See the first two formulae at http://www.astro-tom.com/technical_d...l_formulas.htm

The formula given for true field of view is an approximation. A more accurate version is TFOV = (eyepiece field stop/objective focal length) x 57.3 degrees/radian.

Dave Mitsky