Help me understand how a telescope work?

[Andrew*]

Okay, so let's see if I've got this right:

• The telescope projects an image onto the focal plane.
  
• An eyepiece takes some of that image and focuses it into your Mk1 eyeball.
  
• A high magnification eyepiece might only take a small part of the focal plane - this is why such eyepieces seen to have smaller lenses at the scope-ward side? (not sure what this lens would be called). This would be represented by blue.
  
• A low magnification eyepiece might take lots of the image on the focal plane - which is why such eyepieces have bigger lenses on the scope ward side. This would be the yellow lines.
  
• Presumably unused light at the focal plane is rejected (Should I be flocking my eyepieces? )
  
• The light gathered at the focal plane can then be magnified a little - giving a smaller apparent field of view - or a lot, giving a larger field of view. However, the light gathered at the focal plane could remain the same.
  
• Would this explain why higher magnification eyepieces produce a dimmer image - they're only using small part of the image at the focal plane, and then spreading it out lots for the same Apparent FOV - and if you increased the AFOV too, you'd make it even dimmer still?
  

Assuming that I've understood that correctly, this raises a question - what does the focal length of the eyepiece actually mean (I've wondered this since noticing that my 18mm is shorter than my 12mm)?

Yes, you've got that all right. The "scopeward side" lens is called the field lens while the other one is called the eye lens.

I don't understand focal length fully either. Eyepieces series where the longer focal length is physically shorter than a shorter focal length will mostly have barlowing elements that magnify the image further. The further the parallelising elements after the barlowing elements, the higher the magnification, roughly speaking.

Andrew

[Andrew*]

Is focal length the distance from the focal point/plane where the light rays become parallel again?

If so, imagine the converging light rays from the objective foming to the focal plane, then taking the light 3mm away from this. You will have a very small portion of the focal plane and hence a high magnification. Take the light 30mm from the focal plane and the light will have diverged much more and you will take in more of the focal plane. Just a guess though, I really don't know.

[umadog]

Is focal length the distance from the focal point/plane where the light rays become parallel again?

If so, imagine the converging light rays from the objective foming to the focal plane, then taking the light 3mm away from this. You will have a very small portion of the focal plane and hence a high magnification. Take the light 30mm from the focal plane and the light will have diverged much more and you will take in more of the focal plane. Just a guess though, I really don't know.

With the objective only (one lens) the parallel rays converge at the focal point then diverge. However adding a second lens (an eyepiece in this case) changes things. If the eyepiece is located at distance FL_objective + FL_eyepiece then the rays coming out of the back of the eyepiece will be parallel. They need to be parallel because they must enter your eye, be focused by the lens, and the form as an image on the retina.

[ronin]

Yay, glad we got somewhere. Check out the comparison with the refractor, which is actually much simpler!

Andrew that diagram of the refractor is wrong.

The normal light ray selection should include one that goes through the centre of the lens. This ray is undeviated, in the diagram shown any ray at this position would fall either in the wall of the OTA or well off of the aperture where the eyepiece is.

Equally one of the displayed rays will be worse or suffer total reflection and never enter the objective.

In many (most) of the diagrams what I see is the rays are drawn to show an end result and not drawn how they get to that end result, basically "Draw any picture that looks half right, no-one will know." People are drawing any and all incoming rays to go and form an image at the required position for the eyepiece, the truth is they do not.

One other point, mainly on reflectors, is that this seems to be based on the tube being the single and only factor limiting the field of view. It is not, the mirror will have a FoV with or without a tube. Light coming in too much to one side simply gets bounce elsewhere and not to the eyepiece or secondary. A tube may restrict it further but the absence of a tube does not give a 360 degree FoV.

[Andrew*]

Andrew that diagram of the refractor is wrong.

The normal light ray selection should include one that goes through the centre of the lens. This ray is undeviated, in the diagram shown any ray at this position would fall either in the wall of the OTA or well off of the aperture where the eyepiece is.

Equally one of the displayed rays will be worse or suffer total reflection and never enter the objective.

In many (most) of the diagrams what I see is the rays are drawn to show an end result and not drawn how they get to that end result, basically "Draw any picture that looks half right, no-one will know." People are drawing any and all incoming rays to go and form an image at the required position for the eyepiece, the truth is they do not.

One other point, mainly on reflectors, is that this seems to be based on the tube being the single and only factor limiting the field of view. It is not, the mirror will have a FoV with or without a tube. Light coming in too much to one side simply gets bounce elsewhere and not to the eyepiece or secondary. A tube may restrict it further but the absence of a tube does not give a 360 degree FoV.

Thank you for putting me right. I don't quite understand why a ray entering through the centre of the lens would hit the wall. It would arrive at the focal plane like all the others.

You are making statements but not explaining how they can be understood. Could you please draw an equivalent and correct version of my diagram to help us understand?

[umadog]

Andrew, I think one of Ronin's points is that the minimal ray diagram required to characterise a lens requires three rays. It looks like this: http://www.physicsclassroom.com/class/refrn/u14l5da3.gif

There's the central, undeviated ray; the ray that leaves the object parallel with the optical axis; and the ray that leaves the lens parallel with the optical axis. Together, these three rays demonstrate that you have an image-forming condition and show you where the image is. You have an image-forming condition when all rays that leave one point on the object (the upward arrow in the link above) converge onto one point in the image plane. When the object and the image are on opposite side of the lens, we say that we have formed a "real image." This is one that you could record by placing, say, a CCD chip at this location. If the object and the image are on the same side of the lens, we call the image a "virtual image." A virtual image cannot be recorded by placing a CCD camera at that location. When you use a lens as a "magnifying glass", you're looking at a virtual image. A virtual image is formed when the object is a distance of less than one focal length from the lens. No image is formed when the object is at exactly one focal length.

[Andrew*]

Oh, right, that makes a lot of sense, thank you very much, and also serves to explain non-infinity sources. Basically the light entering the scope from the same source across the aperture are in fact not parallel, but are close to parallel for infinity sources.... Here's a revised diagram.

[Macavity]

And another one - Including a slightly "clunky" calculator.

http://hyperphysics....opt/teles2.html

The "simple" telescope is perhaps not that simple? <G> Probably why spotty 'erberts were "drilled" with endless ray diagram problems for single (convex, concave) lenses with various object (image) distances, then mirrors - If you were really "lucky" things with two (more) lenses... the telescope, even / maybe? Probably to convince aspiring scientists of the merit of equations... calculators... computers, even!

I suppose the only formula I have come to rely on (with luck, mentally!) is to take a "chip" diagonal (or field stop diameter) divide by the focal length of the scope, multiply by 57.3 (to convert from Radians to Degrees) ==> Telescopic Field of view! If you use (Teleview) Radians, you could use the e.p. focal length, I guess...

Interesting thought re. the "calculator" and non-infinite things... Many of us (myopics) cannot see sharp images at infinity. But we can refocus scopes - In my case presumably to give an image at ~10 -15cm?!? Those with 20:20 vision might have tried focussing at other (non-relaxed) distance. A bigger image? Not something that regularly makes the "news" tho'?

[NeilFawcett]

And another one - Including a slightly "clunky" calculator.

http://hyperphysics....opt/teles2.html

The "simple" telescope is perhaps not that simple? <G> Probably why spotty 'erberts were "drilled" with endless ray diagram problems for single (convex, concave) lenses with various object (image) distances, then mirrors - If you were really "lucky" things with two (more) lenses... the telescope, even / maybe? Probably to convince aspiring scientists of the merit of equations... calculators... computers, even!

I suppose the only formula I have come to rely on (with luck, mentally!) is to take a "chip" diagonal (or field stop diameter) divide by the focal length of the scope, multiply by 57.3 (to convert from Radians to Degrees) ==> Telescopic Field of view! If you use (Teleview) Radians, you could use the e.p. focal length, I guess...

Interesting thought re. the "calculator" and non-infinite things... Many of us (myopics) cannot see sharp images at infinity. But we can refocus scopes - In my case presumably to give an image at ~10 -15cm?!? Those with 20:20 vision might have tried focussing at other (non-relaxed) distance. A bigger image? Not something that regularly makes the "news" tho'?

The image on that site shows my confusion with the eyepiece?

Why, if you move the eyepieces further away, does the FOV get larger (& objects smaller). If you follow the lines drawn, and imagine the lens further away, it would suggest the opposite behaviour!?

[umadog]

Why, if you move the eyepieces further away, does the FOV get larger (& objects smaller). If you follow the lines drawn, and imagine the lens further away, it would suggest the opposite behaviour!?

I don't really understand your question. Try playing with this applet and see if it makes sense: http://www.mtholyoke.edu/~mpeterso/classes/phys301/geomopti/twolenses.html

It allows you change the lens fl, add an object, add an infinity light source, etc. It plots the locations of virtual and real images too. No image is formed when the focal points of two lenses intersect. Although it should be possible for a third lens to take the light and form an image, this applet seems to just lose the image under these circumstances. I guess that's a bug.

[Rockrae78]

Just to add a spanner in the works, there is an interesting article in September's Sky @ Night magazine entitled 'How does my telescope work?'

Instead of focussing on the way a scope shepherds light (using mirrors and lenses), it looks at how a telescope must be a kind of electrical equipment. Very interesting read after being baffled by most the science discussed in this thread!

[NeilFawcett]

Just to add a spanner in the works, there is an interesting article in September's Sky @ Night magazine entitled 'How does my telescope work?'

Instead of focussing on the way a scope shepherds light (using mirrors and lenses), it looks at how a telescope must be a kind of electrical equipment. Very interesting read after being baffled by most the science discussed in this thread!

Interesting! I've just signed up for 12 months of that magazine for an offer at £29... Hopefully I'll get that issue then

[NeilFawcett]

I don't really understand your question. Try playing with this applet and see if it makes sense: http://www.mtholyoke.../twolenses.html

It allows you change the lens fl, add an object, add an infinity light source, etc. It plots the locations of virtual and real images too. No image is formed when the focal points of two lenses intersect. Although it should be possible for a third lens to take the light and form an image, this applet seems to just lose the image under these circumstances. I guess that's a bug.

OK... Had a good play with that, but I still can't see why for example a 12mm eye piece will give twice the magnification as a 24mm eyepiece

[dph1nm]

but I still can't see why for example a 12mm eye piece will give twice the magnification as a 24mm eyepiece

Remember that it is not just the distance from the focal plane which changes with these two eyepieces. The 12mm eyepiece will bend the light twice as sharply as the 24mm one.

NigelM

[NeilFawcett]

Remember that it is not just the distance from the focal plane which changes with these two eyepieces. The 12mm eyepiece will bend the light twice as sharply as the 24mm one.

NigelM

Yes, but I still can't picture why, when we're looking at the moon, why the eyepiece at 12mm makes the moon appear twice as big as our 24mm one?

I've tried drawing myself diagrams but can't fathom it. I drew diagrams originally in this thread to work out the FOV of a telescope and that explains it perfectly now. So what I'm after now is the next (last) logical step, how an eye piece zooms in on a subset of that FOV.

[TwoPi]

I might regret posting this(!), but is it not simply a case that the smaller the focal length of the eyepiece, the closer you are to the image plane and so the bigger the image appears. So halving the distance between the eyepiece lens and the focal point on your earlier diagram means the image appears twice as big to the eye.

<sits and waits to be shot down in flames >

[TwoPi]

(meant to put a smiley after my last post but hit the Post button too soon - feel free to correct me!)

[NeilFawcett]

I might regret posting this(!), but is it not simply a case that the smaller the focal length of the eyepiece, the closer you are to the image plane and so the bigger the image appears. So halving the distance between the eyepiece lens and the focal point on your earlier diagram means the image appears twice as big to the eye.

<sits and waits to be shot down in flames >

Do I need to think of it in the way, at the telescopes image plane, there's in effect the image sitting their of what's been captured by the telescope FOV? What I don't understand though is how a lens at 12mm zooms in twice as much as the 24mm. And what happens to all the rest of the light from that image (FOV)? I just cannot picture the workings in my mind

[TwoPi]

Yes, there's an image at the image plane - this is what the eyepiece focuses on and what a camera takes a picture of when you attach the body of an SLR (no lens) directly to your telescope. The camera sensor (or film plate) is at the focal point of the telescope, with the telescope acting like a telephoto lens. If there was no image there, you wouldn't see/capture anything!

The 12mm EP 'zooms in' twice as much as a 24mm EP because the it's closer to the focal point and so the exit pupil is smaller at that point (look back at your diagram). The magnification you experience is (near as makes no difference) = Diameter of objective / Diameter of exit pupil. So if you halve the exit pupil, you double the magnification (as the diameter of the objective or primary mirror is fixed in a telescope).

In terms of what happens to the rest of the FOV, that is projected outside the field stop of the eyepiece (or in the case of imaging, to the side of the camera's sensor), so it doesn't get transmitted to the eye / recorded in the photo.

I don't claim to be an optics expert so happy to be corrected on the above, but that's the way I think of it anyway. Hope it helps!

[NeilFawcett]

Yes, there's an image at the image plane - this is what the eyepiece focuses on and what a camera takes a picture of when you attach the body of an SLR (no lens) directly to your telescope. The camera sensor (or film plate) is at the focal point of the telescope, with the telescope acting like a telephoto lens. If there was no image there, you wouldn't see/capture anything!

The 12mm EP 'zooms in' twice as much as a 24mm EP because the it's closer to the focal point and so the exit pupil is smaller at that point (look back at your diagram). The magnification you experience is (near as makes no difference) = Diameter of objective / Diameter of exit pupil. So if you halve the exit pupil, you double the magnification (as the diameter of the objective or primary mirror is fixed in a telescope).

In terms of what happens to the rest of the FOV, that is projected outside the field stop of the eyepiece (or in the case of imaging, to the side of the camera's sensor), so it doesn't get transmitted to the eye / recorded in the photo.

I don't claim to be an optics expert so happy to be corrected on the above, but that's the way I think of it anyway. Hope it helps!

OK, that makes sense. What I'd now need - and can't seem to find anywhere on the internet - is a simple diagram of for example a 12mm and 24mm eyepiece, showing why the 12mm magnifies twice as much... In my noddy attempts I simply see all the FOV image being bent by the lens intot eh eye, but it gives the impression the further away the eyepiece, the less of the image enters the eye (ie: magnification increases with distance).

[TwoPi]

Have you seen this? It explains rather nicely what magnification is and its relation to the angles between objects (e.g. the angle between the two sides of the moon or two stars). There's a diagram that shows what angles we're talking about and if you drew the diagram again with the eyepiece lens half the distance from the focal point you will see how the ratio of angles changes, hence why the image appears bigger.

At the end of the day it all boils down to the fact that "different focal length eyepieces really are just moving you closer to, or further from, the image plane of the objective. When you get closer, the image is bigger (higher magnification) and when you get further, the image is smaller (lower magnification). This is a very effective (and very accurate) way of thinking about how the scope works."

You halve the distance between yourself and an object (in our case an image of an object), it appears twice as big!

[NeilFawcett]

Yeh, I've seen that. What'd I'd need to see (& I've tried drawing it myself) is what it would look like with the lens further back, and seeing why an object then looks smaller?

[TwoPi]

Well if you move the eyepiece lens to the right, theta_e (no idea how to get a greek letter on here!) gets smaller, whereas distance h (and angle theta_o) stay the same. In other words we perceive distance h as shorter. I.e. the object looks smaller.

This is like looking at a ruler with 1cm markings on right in front of your face versus holding it at arms length. Obviously the distance between the 1cm markings hasn't changed, it's the angle between the two markings that has changed. This is what our perception of size is. Smaller angle = smaller object.

[umadog]

The trick is to examine ray diagram. From that you can see how the angles have changed to get a larger image. It's true that in the demo I linked to they're squashed and it's hard to see them. I'll look for a better image...

EDIT: Ok! This is a good start: http://brunelleschi.imss.fi.it/esplora/cannocchiale/dswmedia/esplora/eesplora2.html

[TwoPi]

Like it, good explanation and diagram.