Exit pupil and AFOV

[miguel87]

5 minutes ago, vlaiv said:

Great

Next thing to realize, and here moon will be great help - is that different points in infinity arrive at different angles.

Do this thought experiment.

Take a ruler and point it at the center of the moon. When you look at one edge of the moon - it is at slight angle to that ruler. When you look at the other side of the moon - it is at slight angle again but to the other side.

Angle at which parallel rays arrive at aperture is related to where in the sky point of origin is.

If scope is aimed directly at a star - parallel rays will arrive at 90 degrees to aperture.

If scope is not aimed directly at a star, this will happen:

Rays will arrive at an angle to front of the lens, but they will also converge not directly behind lens - but a bit "lower" - also on a focal plane but some distance to center.

This is why image forms at focal plane of telescope - star in the center of the FOV is one scope is aiming at while star at the edge is at an angle to telescope tube.

Ok I think I am with you.

So the FOV of the created image, in a newtonian reflector is limited by the aperture of the OTA and very slightly reduced by a dew shield for example.

The maximum angle would be greater for a refractor than a reflector?

Hope I am not over complicating, that's just the way I picture different points in infinity.

[miguel87]

So every star is brought to focus at a different point on the focal plane.

I now suddenly understand (I think) why a faster scope (steeper angles from infinity) would have a curved 'perfect' focal plane and as the angles from infinity decrease the focal plane will get flatter?

Thanks for this by the way!

[vlaiv]

4 minutes ago, miguel87 said:

Ok I think I am with you.

So the FOV of the created image, in a newtonian reflector is limited by the aperture of the OTA and very slightly reduced by a dew shield for example.

The maximum angle would be greater for a refractor than a reflector?

Hope I am not over complicating, that's just the way I picture different points in infinity.

No.

FOV depends on focal length, or how much bent light rays after they arrive.

If you block portion of rays while they are still parallel - it is just as using smaller aperture.

 

 

[miguel87]

7 minutes ago, vlaiv said:

No.

FOV depends on focal length, or how much bent light rays after they arrive.

If you block portion of rays while they are still parallel - it is just as using smaller aperture.

 

 

Ok,

I'm just imagining positioning my eye at the edge of a primary mirror compared the edge of a primary lens, and I can see stars at much greater angles from the lens?

Anyway, I get your first two points.

Single point at infinity=parallel photons

Different infinity points on = different angles of photons striking the primary

Each star covers the entire primary in parallel photons, each stars angle onto the primary is different and therefore creates an image at the focal plane.

The focal POINT being the area of the focal plane that is on the optical axis?

Edited May 8, 2020 by miguel87

[vlaiv]

1 minute ago, miguel87 said:

So every star is brought to focus at a different point on the focal plane.

I now suddenly understand (I think) why a faster scope (steeper angles from infinity) would have a curved 'perfect' focal plane and as the angles from infinity decrease the focal plane will get flatter?

Thanks for this by the way!

Yes, in simple design curvature of focal plane has to do with focal length (and hence to some degree F/ratio - but aperture is not that important).

Focal plain is curved because rays that converge further away from center of focal plane, converge "closer".

This is in fact not true - they converge at the "same distance" - but since distance is measured at an angle - it look shorter. Look at central rays - both blue and red - those two that go thru exact center of the lens. Points where rays converge are at the same distance to center of the lens and these two rays traveled same distance to their respective converging points.

This is field curvature and it depends on focal length of lens.

 

 

[vlaiv]

1 minute ago, miguel87 said:

Ok,

I'm just imagining positioning my eye at the edge of a primary mirror compared the edge of a primary lens, and I can see stars at much greater angles from the mirror?

What you have here is aperture obstruction or aperture mask. It just makes less light reach objective / mirror. Here is an example:

Some of parallel rays lend outside of tube - they miss telescope.

Some of parallel rays lend on "inside" of the tube but don't reach mirror

Some enter tube and hit main mirror

Some hit tube on the outside.

We don't know nor don't care about all those parallel rays that did not make it - mirror will collect all photons that made it and converge those onto a star.

What you see here is form of vignetting - image is a bit fainter further away from principle axis you get because less photons made it at a larger angle.

[vlaiv]

6 minutes ago, miguel87 said:

Single point at infinity=parallel photons

Different infinity points on = different angles of photons striking the primary

Each star covers the entire primary in parallel photons, each stars angle onto the primary is different and therefore creates an image at the focal plane.

Yes, yes, yes

6 minutes ago, miguel87 said:

The focal POINT being the area of the focal plane that is on the optical axis?

Yes - that is what we sometimes call focal point - but we also call focal point - any point on focus plane that is of interest. Principal focal point shall we say.

When we talk about lens then focus point / focal point of that lens is the principal focal point - places where rays parallel to principal optical axis converge.

If we are talking about focused star that is off axis and want to refer to place on focal plain where all those rays converge - we will say - it's the focal point of that star.

[vlaiv]

Important point for further understanding:

Distance of star image (focal point for that particular star) - depends on angle but it also depends on focal length of that telescope.

Short focal length telescopes are "wide field" and long focal length telescopes are "narrow field" because of this.

If we have a star at a same angle and we have two telescopes one with 500mm FL and other with 1000mm FL  - second telescope will form image of the star at twice the distance to center of the frame compared to first telescope.

Note - this is not magnification (although it looks like it) but it is related to magnification - this is why it is easier for longer FL telescopes to magnify more.

[miguel87]

4 minutes ago, vlaiv said:

Yes, yes, yes

Yes - that is what we sometimes call focal point - but we also call focal point - any point on focus plane that is of interest. Principal focal point shall we say.

When we talk about lens then focus point / focal point of that lens is the principal focal point - places where rays parallel to principal optical axis converge.

If we are talking about focused star that is off axis and want to refer to place on focal plain where all those rays converge - we will say - it's the focal point of that star.

Understood 👍

So what exactly IS the exit pupil

3 minutes ago, vlaiv said:

Important point for further understanding:

Distance of star image (focal point for that particular star) - depends on angle but it also depends on focal length of that telescope.

Short focal length telescopes are "wide field" and long focal length telescopes are "narrow field" because of this.

If we have a star at a same angle and we have two telescopes one with 500mm FL and other with 1000mm FL  - second telescope will form image of the star at twice the distance to center of the frame compared to first telescope.

Note - this is not magnification (although it looks like it) but it is related to magnification - this is why it is easier for longer FL telescopes to magnify more.

Ok,

So this would give a better resolution, per degree of night sky, as the focal length increases?

[vlaiv]

10 minutes ago, miguel87 said:

So this would give a better resolution, per degree of night sky, as the focal length increases?

In a sense yes - for same "grid" (like camera sensor with pixels) - longer focal length will increase resolution in terms of arc seconds per pixel.

We have to be careful when using word resolution as it has so many meanings

10 minutes ago, miguel87 said:

So what exactly IS the exit pupil

Final piece of puzzle.

First here is this:

Which stands for - converging rays will come to one point at focal plane but will continue to diverge if nothing stops them. Simple as that. It is important to note that angle of convergence is the same as angle of divergence for a ray - same thing as saying light rays are straight lines.

Now we take a small telescope and run things in "reverse".

Here in this diagram arrows are put on rays and objective is marked as objective and eyepiece is marked as eyepiece. In reality - this diagram can be read in reverse - it can go from right to left and things would remain the same. If we remove labels and change arrow directions - it will still be valid diagram.

Eyepiece is just a small telescope, or rather telescope with short FL where light is "running" in reverse" - or rather light does what it does usually - move in straight line.

Same as rays arrive parallel to entrance pupil - same way they leave at exit pupil. Difference being that entrance pupil is larger because focal length of front scope is larger and we call it aperture. Exit pupil is smaller because focal length of second "telescope" is smaller. Rays diverge at same angle - they just don't have enough room to spread as much since focal length is shorter - that is all.

In fact aperture:fl = aperture:fl

where left side is one telescope and right side is other "telescope" (or eyepiece).

Only thing that we did not see in above diagram is magnification. We have seen how parallel rays become parallel rays again at exit. How their pupil decreases (or increases if we swap telescopes around - its up to focal lengths).

Last piece of puzzle has to do with focal lengths and angles and distance from optical axis that we talked about.

We said distance of a point in focal plane depends on

1) angle of parallel rays

2) focal length

If point on focal plane is the same for two scopes, and one scope has smaller focal length than other - then one scope will have angles smaller than the other. In fact - scope with larger focal length has smaller angles.

It is this angle amplification that is actual amplification of image that telescope + eyepiece (or two telescopes, or two lenses) provide.

That is why we see larger image - because for our eye it is like there is no telescope only parallel rays coma at larger angles - and they will come at larger angles if thing is indeed larger -  we see it as enlarged.

Makes sense?

[miguel87]

11 minutes ago, vlaiv said:

In a sense yes - for same "grid" (like camera sensor with pixels) - longer focal length will increase resolution in terms of arc seconds per pixel.

We have to be careful when using word resolution as it has so many meanings

Final piece of puzzle.

First here is this:

Which stands for - converging rays will come to one point at focal plane but will continue to diverge if nothing stops them. Simple as that. It is important to note that angle of convergence is the same as angle of divergence for a ray - same thing as saying light rays are straight lines.

Now we take a small telescope and run things in "reverse".

Here in this diagram arrows are put on rays and objective is marked as objective and eyepiece is marked as eyepiece. In reality - this diagram can be read in reverse - it can go from right to left and things would remain the same. If we remove labels and change arrow directions - it will still be valid diagram.

Eyepiece is just a small telescope, or rather telescope with short FL where light is "running" in reverse" - or rather light does what it does usually - move in straight line.

Same as rays arrive parallel to entrance pupil - same way they leave at exit pupil. Difference being that entrance pupil is larger because focal length of front scope is larger and we call it aperture. Exit pupil is smaller because focal length of second "telescope" is smaller. Rays diverge at same angle - they just don't have enough room to spread as much since focal length is shorter - that is all.

In fact aperture:fl = aperture:fl

where left side is one telescope and right side is other "telescope" (or eyepiece).

Only thing that we did not see in above diagram is magnification. We have seen how parallel rays become parallel rays again at exit. How their pupil decreases (or increases if we swap telescopes around - its up to focal lengths).

Last piece of puzzle has to do with focal lengths and angles and distance from optical axis that we talked about.

We said distance of a point in focal plane depends on

1) angle of parallel rays

2) focal length

If point on focal plane is the same for two scopes, and one scope has smaller focal length than other - then one scope will have angles smaller than the other. In fact - scope with larger focal length has smaller angles.

It is this angle amplification that is actual amplification of image that telescope + eyepiece (or two telescopes, or two lenses) provide.

That is why we see larger image - because for our eye it is like there is no telescope only parallel rays coma at larger angles - and they will come at larger angles if thing is indeed larger -  we see it as enlarged.

Makes sense?

Yeah I think so, the two 'telescopes' sort of cancel out. I.e. you could place a flat photograph perfectly at the focal plane of the telescope and use eyepieces to view the photo in varying magnifications depending on the 'grasp' of the eye piece's lens(es)?

[vlaiv]

1 minute ago, miguel87 said:

Yeah I think so, the two 'telescopes' sort of cancel out. I.e. you could place a flat photograph perfectly at the focal plane of the telescope and use eyepieces to view the photo in varying magnifications depending on the 'grasp' of the eye piece's lens(es)?

Yes - you can use eyepiece as magnifying glass - screw off barrel - put it close to your smartphone and enjoy large pixels

Or you can do following to turn around "diagram" - take binoculars and look at the front - things will look smaller and more distant than larger and close.

[miguel87]

Ok yes I have done that with Binos before. All understood.

 

[miguel87]

If an eyepiece 'outputs' parallel lines, I dont understand how the exit pupil comes about, hovering above the eyepiece?

[jetstream]

5 hours ago, vlaiv said:

We don't know nor don't care about all those parallel rays that did not make it

Your image showed me perfectly the effect of stray light in the scope, which can bounce around and is something I take great care in controlling.

[vlaiv]

8 hours ago, miguel87 said:

If an eyepiece 'outputs' parallel lines, I dont understand how the exit pupil comes about, hovering above the eyepiece?

Ah, ok - exit pupil is imaginary circle all parallel lines from a single star pass thru.

There is special "exit pupil" - or rather same thing but at a special place - imagine that each set of parallel lines forms a circle.

This circle is not at any particular place - it can be anywhere along those parallel lines.

However - if you take all parallel sets of lines - they all intersect in one place (or at least majority of them) - this is exit pupil that we talk about when we talk about eye relief and thing that hovers.

Here we have another diagram and exit pupil and eye relief marked.

Note that place of exit pupil is where all parallel light rays intersect - blue rays are from one star - they converge in one point on focal plain, red are from another star and green are from third star.

All three sets of parallel rays exit - and have same "diameter" - or same circle between them - blue rays pass imaginary circle of certain diameter, green do as well and red do as well. They all cross at same place - and this is where all three imaginary circles come to same place - to form a single circle all parallel rays come thru - this is the exit pupil and distance of it is eye relief.

[miguel87]

Thanks Vlaiv, I certainly have a better understanding than before. Appreciate you taking the time to explain.

Mike

[andrew s]

2 hours ago, miguel87 said:

Thanks Vlaiv, I certainly have a better understanding than before. Appreciate you taking the time to explain.

Mike

Mike I am glad @vlaiv managed to explain things better than I could. I thought the following might be helpful.

a) show vlaiv's diagram above and in b) I have blanked out all but the bit shown in you Blue diagarm posted above (but missing the eye).

In c) I have shown how the eyepiece forms an image of the objective at the exit pupil (note the rays are an inverted mirror image but at a steeper angle due to the angular magnification) with d) showing the same thing for a double headed arrow.

Regards Andrew

Edited May 9, 2020 by andrew s

[miguel87]

4 hours ago, andrew s said:

Mike I am glad @vlaiv managed to explain things better than I could. I thought the following might be helpful.

a) show vlaiv's diagram above and in b) I have blanked out all but the bit shown in you Blue diagarm posted above (but missing the eye).

In c) I have shown how the eyepiece forms an image of the objective at the exit pupil (note the rays are an inverted mirror image but at a steeper angle due to the angular magnification) with d) showing the same thing for a double headed arrow.

Regards Andrew

Thanks, I do have a better understanding but not perfect. I am content for the moment 😂 In fact, as I write this and look at diagram c) I do have a bit of a eureka moment of understanding. Its cuts to the core of my misunderstanding. I will try to explain.

Here is what I was writing before:

I look at the different parallel Ray's coming out of the exit pupil (let's imagine a 5mm exit pupil) and think, if i shrink that pupil, cut it off at the top and bottom so that it is now 2.5mm. Then the parallel photons coming along the optical axis and passing through the exit will reduce by 50%. The parallel photons coming from the edge of field see the exit at an angle (it therefore appears narrower, like a doorway seen from a sharp angle) and therefore a smaller amount of photons will pass through....

Then I realised I was wrong because the angled photons also saw the original 5mm exit from an angle. The relative losses from any particular angle will still be 50%, they just may not have started at the same brightness.

So I was wrong all along, sorry guys.

If your pupil shrinks and cuts off the edge of the exit pupil, the whole image will dim by an equal amount.

✌

[andrew s]

3 minutes ago, miguel87 said:

If your pupil shrinks and cuts off the edge of the exit pupil, the whole image will dim by an equal amount.

✌

Bingo. Regards Andrew 

[Don Pensack]

On 05/05/2020 at 09:33, bingevader said:

And this is still true if the magnification remains the same but the diameter of the EP increases (so no reduction in magnification to brighten the image)?

I will quite happily stand corrected, I've been living with a misconception for some years!

There was a vogue for larger and large EPs at one point and I remember the same conversation then.

Maybe I didn't remember it and that's the problem!

I think you meant if the field size increased, not the diameter of the eyepiece (which has little relevance).

The exit pupil (brightness) and magnification go hand in hand.  A larger apparent field spreads the light farther into your peripheral vision, but does not brighten the image.

If it did, we'd all want to use 150° eyepieces.

The purpose for larger eyepieces is to get wider true fields, because a 32mm 50° eyepiece will have the same brightness as a 32mm 100° eyepiece, but the latter has 4x the field area.

Remember, every point on the focal plane of both eyepieces is illuminated by the entire primary (* see below)

 

* In practice, we do not choose secondary sizes that illuminate the edges of the field to 100%, we choose secondaries to have about a 30% light drop off at the edge (we don't see it, though a camera can,

so photographic secondary sizes are larger).  So if a 32mm 50° eyepiece has a 30% light loss at the edge, a 32mm 100° eyepiece would have significantly more light loss at the edge.

That's why we choose the size of our secondary mirrors to illuminate the field stop of our lowest power, largest field, eyepieces to 70° at the edge.  At some point, as the magnification goes up and the field stop of the higher power eyepieces get smaller, the illumination at the edge reaches 100% because the effects of secondary edge of field light loss gradually fall to zero.

[Don Pensack]

22 hours ago, miguel87 said:

If an eyepiece 'outputs' parallel lines, I dont understand how the exit pupil comes about, hovering above the eyepiece?

The exit pupil is an image of the primary mirror.  It is not the image of the sky.  It is the distance we hold our eye from the eyepiece to see

the entire field of the eyepiece.  To show an eyepiece is not focusing the light, back away from the eyepiece.  The image stays in focus, but you are progressively seeing less and less field.

Exit pupil and field go hand in hand, but focus is in the eye.

[miguel87]

20 minutes ago, Don Pensack said:

To show an eyepiece is not focusing the light, back away from the eyepiece.  The image stays in focus, but you are progressively seeing less and less field.

You could say the eyepiece is refocusing the image. From the 'real' image on the telescope focal plane to the unfocused light travelling out of the exit pupil.

Thanks 👍 it's always nice to raise one's level of understanding.

[miguel87]

34 minutes ago, Don Pensack said:

The exit pupil is an image of the primary mirror.  It is not the image of the sky.  It is the distance we hold our eye from the eyepiece to see

the entire field of the eyepiece.  To show an eyepiece is not focusing the light, back away from the eyepiece.  The image stays in focus, but you are progressively seeing less and less field.

Exit pupil and field go hand in hand, but focus is in the eye.

I think for novices like me, when people say the exit pupil is the image of the primary it doesn't help. It may even make things work, because I imagine a reflection of the night sky in a mirror. I understand now why this isnt the case, but...

I point a mirror at the sky, what do I think the image of the primary mirror is? Probably the night sky.

[vlaiv]

1 hour ago, Don Pensack said:

The exit pupil is an image of the primary mirror.  It is not the image of the sky. 

Same here. Although I know what you mean - when you say image - I instantly think in terms of focused light (probably due to too much time spent contemplating imaging rather than observing).