miguel87

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

Mike

8 minutes ago, Kev M said:

I don't believe this particular reflector can achieve direct focus with a DSLR ( not enough inward travel on the focuser ), you would need to either fit a low profile focuser or move the primary mirror ( not easy to do ).

Alternatively some barlows may work by moving the focal point further out, which I believe you have found works.

 

 

You could find your focus plane by measuring where the barlow front lens sits in your focusser tube when the moon is in focus on your camera. 

Then you can work out if your DSLR camera sensor could reach this spot.

3 minutes ago, GiggaKubicca said:

Firstly I don't want anyone to take this the wrong way because it's not how I intend it as I do really appreciate peoples time and help. However it's starting to get frustrating for me. I seem to have loads of people conflicting with what others have been telling me.

I have tried without a barlow. I have tried with a barlow. I have tried with both a barlow and an eyepiece. The only time I have had focus was with both the barlow on its own and with the barlow and an eyepiece. When I try without a barlow I get zero focus. I can get it all the way till it looks like Jupiter with no detail.

One person says elsewhere that my barlow is cheap and that is why my image quality isn't that good. I have someone else say that the barlow is fine and not to listen to snobs.
One person says try this setting and then someone else says that I shouldn't be using that and use this, neither of which seem to really accomplish much.

I bought the adapter I have now because I was told that it would work with what I am trying to do. However when I tried doing it how I was told I didn't really get anywhere until I used both the barlow and the eyepiece.
I bought an adapter before that because the one before that I bought thinking it was the right thing but it wasn't. Then the second adapter that was recommended didn't work even after having extensive help. In the end I contacted a telescope directly and asked them, which is the adapter that I have now.

I honestly don't know where to go from here because I feel like I have tried everything and I've progresed nowhere and I get told multiple different things regardless of where I ask.

All I want to do is photo the moon and even that seems to be incredibly difficult to achieve, like smashing my head against a damn wall.

I think this is why you are struggling.

The adaptors you have are not familiar to people and sound strange. An adaptor with a 10mm eyepiece in I have never heard of.

Secondly there are.multiple ways of solving this so each person is just recommending their own tried and tested setup.

I dont know the details of what you have been told before but I know a method that will 100% solve the problem. I will try to explain.

You said you reach focus using the barlow attached to the camera, great, there is no mistaking hitting good focus on the moon. The barlow lens reaches further down the focusser tube and can reach the focal plane. When you dont have the barlow in, just the nise piece. The cameras sensor has to reach all the way into.the focal plane which it is not able to do. So to reduce the distance between sensor and focal plane you need to get the nose piece out of the way and the 1.25 inch eyepiece holder out of the way.

The simplest way to do this is a t ring adapter which you already have, and attached to the thread on the t ring adaptor is essentially a copy of the tip of a 2 inch eyepiece. If you google "t ring to 2inch eyepiece adapter" loads come up.

I'm sorry you are being bounced around but all I can give you is what I would do in your situation. I cant imagine how it wouldnt work. Of course there are other problems that a telescope can have but if it is working well for visual use then there shouldn't be an issue.

Good luck

There are adapters on ebay for less than a tenner.

Presuming you have tried the camera with only the nose piece adaptor in place and you still cant teach focus, I would personally get hold of a skywatcher or similar 2 inch to 1.25inch eyepiece holder and unscrew it like I do to use the t ring with. But I'm sure there are other solutions too, theres tons of adaptors available to turn your t ring adapter into a 2 inch eyepiece basically. Better than moving your mirror or cutting the tube IMO.

Ok. Do you not have little finger screws that hold your eyepiece in place when using the scope? The eyepiece holder looks unusual to me.

Try twisting apart the first piece you showed me. I think it is made of two parts. Is your telescope the orion dobsonian?

That first picture looks like it unscrews, there is a silver wear line. Then attach the lower half of it to your t ring.

6 minutes ago, GiggaKubicca said:

 

How do I attach a camera body and just the T-ring to the telescope? I can't see how you'd do it without the body just dropping to the floor when you let go.

Take the piece of the telescope that your 1.25 inch eyepieces go into, it should be attached to the scope with 2 finger screws just like an eyepiece.

Once you have this (1.25 inch eyepiece adapter) unscrew it's two component parts.

The flat part screws directly onto your T-ring on the opposite side to the camera.

I have attached a picture of the part that splits into 2.

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

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

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)?

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?

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?

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!

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.

7 minutes ago, vlaiv said:

Here is how to best understand it:

Further the object is, irrespective of relative sizes of object and aperture angle between two lines that connect object to opposite sides of aperture is shrinking.

Here left edge of triangles is aperture and right vertex is object. Aperture is small relative to distance between objects means that angle at vertex is small. When you have very small angle between two lines - they are effectively parallel to you (and here I mean very very very small angle - like couple light years vs 20cm of aperture small angle - although we don't need to go that far - this holds for moon as well although it is 384000Km away).

Makes sense?

 

Yep that makes sense, I can visualise looking the object from either side of the main mirror. The angle of my gaze would not change for a star.

The same way I can watch the moon out the right hand window in my car and drive 100 miles on a straight road and it will still be there.

3 minutes ago, vlaiv said:

Want to go bit by bit and see where you get stuck?

From a point (for our purposes this can be a star) that is very distant (like really distant) incoming "rays" of light are parallel.

Do you understand why is this?

I try to understand this with the relative sizes of the star and the telescope aperture.

And if you had a aperture wider than the light source then the lines would NOT be parallel?

23 minutes ago, vlaiv said:

Here it is - this image explains it all.

If you have a star / point at some angle Alpha to optical axis, following will happen:

- all rays from that point will be parallel before they reach aperture - same angle

- after objective they will start to converge and finally converge at focal point - all light from original star falls into single point on focal plane - this is why star is in focus on camera sensor (provided it is focused well) and it also means that field stop won't remove any light - it only limits angle that can be seen as bigger angle means point on focal plane is further away from center.

- then rays start to diverge (just happily go on their own way and since they came to a point they continue now to spread)

- eyepiece catches those rays and makes them parallel again. Few things to note - angle is now different - that is magnification. All parallel rays occupy certain "circle" - that was aperture earlier and now it is exit pupil. Ratio of angles and ratio of sizes of these pupils is magnification.

- Eye is same thing as telescope - it is device that again focuses parallel rays.

Thus field stop can't act as aperture stop because all rays from "aperture" have been squeezed into single point on focal plane.

I wish I understood all of that 😁

I think my problem is perhaps drawing a parallel between the telescopes primary image created in an empty focusser tube and the exit pupil of an eyepiece.

What your saying is the exit pupil is not a 2d image but a 'window's that the images passes through?

I hope I am getting there?!

I gather that most people disagree with me here and also that I am likely wrong about it all 😂

But I'm just stating how I understand it from my own experience and knowledge.

Apologies if I am frustrating anybody with all my comments.

And thanks to everyone for a very engaging conversation (even if it is out of my depth)

On a closely related but slightly different note:

We are discussing the issue of brightness being lost if an exit pupil is too big.

How about the field stop? If the field stop is smaller than the 'real image' created by the telescope arent we already losing aperture? Regardless of what the eyepiece then does after the field stop.

17 minutes ago, Ags said:

This is incorrect. The exit pupil is the beam of light for every part of the image formed by the eyepiece. If you have an exit pupil of 5 mm, that means that each star in the field is represented by a beam of parallel light 5 mm in diameter emerging from the eye lens of the eyepiece. All of those bundles of light converge to a point above the eye lens (forming a disc 5 mm wide), and the height of that point of convergence is the eyepiece's eye relief. So when your pupil meets the disk at the eye relief height, it occludes all parts of the image equally (assuming the eye pupil is smaller than the exit pupil).

You say parallel lines converge, but that is impossible.

Also if every star is represented by a 5mm wide beam, they cant all occupy the same space or they would not be parallel.

There is information within the exit pupil that would be lost if it was a uniformly bright, plain white disc of light.

Also if you use an eyepiece that has a too big exit pupil, you will notice that you cannot quite see all of the image at the edges. If all parts of the exit pupil, contained all parts of the image then even viewing the central 1mm portion would show the whole image.

8 minutes ago, Louis D said:

I reach focus using a GSO coma corrector attached via a T-thread to M48 thread adapter.  It also flattens the field and corrects coma at the same time.

That's great but it's not exactly 'complete beginner' information.

1 minute ago, Stu said:

Doesn’t contrast improve because the sky background is dimmed by increasing magnification, effectively the same as an extended object?

Yes exactly. But a more compact point source like star will not be dimmed (if the exit pupil airy disc has the brightness of the star spread across it's entire area, then reducing its size would dim the star too) the star is not represented in all areas of the exit pupil.

Images of stars behave differently. A star is essentially a point source of light so no amount of magnification will produce an extended image. (On the assumption of perfect optics and seeing!) Therefore, the brightness of an image of a star depends on the diameter of the object glass or mirror, but not on the magnification. Thus, a high magnification eyepiece will not dim stars but can dim extended objects such as light pollution, and can be useful in increasing the contrast between stars and the background sky.

that is copied from Orwell astronomy society. If brightness was uniform then increased mag would not improve sky contrast

The exit pupil is not far from prime focus, and as you see with a camera or lens, as you move away from prime focus the sharp points spread out and become halos with details of the primary. But even when a long way out of focus, the light across the whole image is not uniform. You can easily get a sense bright areas and dark areas because every light source across the image will turn into it's own little halo, but stay in location. So even out of focus, the image is not uniformly bright.

If the exit pupil at the lens of the human eye was like a perfectly uniform white piece of paper then there is nothing to bring into focus. There has to be information there for the eye to interpret, just like the out of focus view we see in an eyepiece before we move the lens to sit right at the focus point.

Just because the light has been collimated and parallel does not mean it is uniformly bright.

And it is at infinity because the target is at infinity! Not a property of the exit pupil but dependent on the target. If you were looking at a person and not a star it would not be at infinity.

There is an image at an exit pupil! You can see it for yourself floating in the air upside down with a pair of binoculars on a sunny day.