# Thread: secondary light projection shape

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## secondary light projection shape

This question will show what a novice I am. I believe the light from the primary to the secondary is a cone and many diagrams of newts show this. They also show the light projected from the secondary to the eyepiece as a cone. I would have thought the light projected from the secondary would be a cylinder because the secondary is a flat mirror. Can someone tell me what shape is it?

Thanks,
henrylr

2. A cone converging behind a flat mirror will remain a cone upon reflection.

You don't have to take it on faith, you can measure the individual angles of incidence and reflection in each of these two example light rays, and see that the reflected rays are indeed still converging.

Last edited by DaveC426913; 2017-Nov-04 at 02:35 AM.

3. Thus:

newt-scope.png

(click me)
Last edited by DaveC426913; 2017-Nov-04 at 04:13 AM.

4. Originally Posted by henrylr
This question will show what a novice I am. I believe the light from the primary to the secondary is a cone and many diagrams of newts show this. They also show the light projected from the secondary to the eyepiece as a cone. I would have thought the light projected from the secondary would be a cylinder because the secondary is a flat mirror. Can someone tell me what shape is it?

Thanks,
henrylr
To change converging incident rays to parallel reflected rays, you would need some sort of a convex surface.

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Will A + B always = 90%?

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Is there some equation, property of light bouncing of a mirror, or something else, that determines the incident angle of a cone of light leaving a flat mirror at a 45 deg. angle to a primary mirror? What would result from a secondary at 60 degrees to a primary?

7. Originally Posted by henrylr
Will A + B always = 90%?
Yes.

8. Originally Posted by henrylr
Is there some equation, property of light bouncing of a mirror, or something else, that determines the incident angle of a cone of light leaving a flat mirror at a 45 deg. angle to a primary mirror? What would result from a secondary at 60 degrees to a primary?
The emerging beam will be leaning 30 degrees away from the usual perpendicular to the tube, but otherwise unchanged.

9. Originally Posted by Hornblower
Originally Posted by henrylr
Will A + B always = 90%?
Yes.
Wait, why would A+B always equal 90?

That would only be the case if we carefully chose two ideal light rays that are equidistant from the centre/edge AND the scope is perfectly collimated. I mean, those light rays are arbitrarily chosen.

Not sure if I'm being pedantic here.
Last edited by DaveC426913; 2017-Nov-05 at 03:35 PM.

10. Originally Posted by henrylr
Is there some equation, property of light bouncing of a mirror, or something else, that determines the incident angle of a cone of light leaving a flat mirror at a 45 deg. angle to a primary mirror? What would result from a secondary at 60 degrees to a primary?
The thing that is always true for reflected light is that for a given ray, the angle of incidence will be equal to the angle of reflection. You can use this property (as DaveC426913 did) to determine the path of light for a mirror at any arbitrary angle. If you do this for a ray on each side of your converging cone, you'll be able to see the path of the entire beam after reflection.

11. I was taking those rays as defined by a cone centered on the optical axis. Of course others from intermediate positions will give a different sum.

12. Originally Posted by Hornblower
I was taking those rays as defined by a cone centered on the optical axis.
Right. That would do it.

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I did a search for "properties of light and got many hits that were, university or scientific, sites related to physics. One thing in common was the statement, "Reflected light obeys the law of reflection, that the angle of reflection equals the angle of incidence". While I don't understand wave or particle theory, the search was fun and enlightening (no pun intended).

14. Originally Posted by henrylr
I did a search for "properties of light and got many hits that were, university or scientific, sites related to physics. One thing in common was the statement, "Reflected light obeys the law of reflection, that the angle of reflection equals the angle of incidence". While I don't understand wave or particle theory, the search was fun and enlightening (no pun intended).
My bold. Good for you. We all have to start somewhere, and you are getting off to a good start in learning some of the geometric properties of optics and understanding why a Newtonian reflector works the way it does.

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## more optics questions

I'm still trying to understand more of the optical properties of a reflector. Numerous physics documents on paraboloids state the focal length of a paraboloid is a point where all the reflected rays converge. Based on that it seems, in the case of a newtonian, the focal plane would be some distance away from the focal point, and would vary, based on the eyepiece and distance to the object being viewed. Is that true?

Thanks,
henrylr
Last edited by henrylr; 2017-Nov-07 at 08:59 PM.

16. Originally Posted by henrylr
I'm still trying to understand more of the optical properties of a reflector. Numerous physics documents on paraboloids state the focal length of a paraboloid is a point where all the reflected rays converge. Based on that it seems, in the case of a newtonian, the focal plane would be some distance away from the focal point, and would vary, based on the eyepiece and distance to the object being viewed. Is that true?

Thanks,
henrylr
The surface we call the focal plane includes the axial focal point shown in the diagram, along with the focal points from objects that are not on the optical axis. It is not separated from them. It is the location of a sharply focused image of an extended object such as the Moon.

17. Addendum: The eyepiece is incorrectly positioned in that sketch. If the telescope is sharply focused on a star, the field stop of the eyepiece will be even with the primary mirror's focal point, which we call the prime focus. Another star, a small angle off the axis, will make a focal point laterally displaced from the one shown in the sketch, but still in the plane of the field stop. This plane is the focal plane of the system.

The position of the focal point shown here is that for incoming parallel rays, which we get from stars. If we aim the scope at a nearby object, the focus will be pushed back a short distance, and we will have to pull the eyepiece out in its drawtube to achieve a sharp focus.

If you are having trouble grasping all of this, believe me, I understand. As a 10-year-old beginner I could use a telescope effectively, but I was all fouled up in trying to make sense of these ray tracing exercises. My father got me straightened out by making suitable sketches during a face to face dialogue. Doing that in a forum like this is more awkward.

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Originally Posted by Hornblower
Addendum: The eyepiece is incorrectly positioned in that sketch. If the telescope is sharply focused on a star, the field stop of the eyepiece will be even with the primary mirror's focal point, which we call the prime focus. Another star, a small angle off the axis, will make a focal point laterally displaced from the one shown in the sketch, but still in the plane of the field stop. This plane is the focal plane of the system.

.
My search for info on the physics of light and optics, and your's and other's replies, have helped me understand quit a bit more. One thing that had confused me were the diagrams of newts, many without eyepieces, showing the focal point from the secondary converging in the focuser tube, some well below the top of the tube...hence my sketch with the focal point away from the field stop. You cleared that up for me and the attached diagram, from this site https://www.handprint.com/ASTRO/ae2.html, was also very helpful.

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A question about secondary alignment. I've read that centering the secondary in a sight tube will result in the secondary being offset towards the primary. I've also read that physically centering the secondary in the focuser tube will not create an offset. I would have thought both techniques would give the same result, since crosshairs in a sight tube would be coaxial with the focuser tube axis. Why are the results different?

Thanks,
henrylr

20. Originally Posted by henrylr
A question about secondary alignment. I've read that centering the secondary in a sight tube will result in the secondary being offset towards the primary. I've also read that physically centering the secondary in the focuser tube will not create an offset. I would have thought both techniques would give the same result, since crosshairs in a sight tube would be coaxial with the focuser tube axis. Why are the results different?

Thanks,
henrylr
I am not sure what you mean by "centering the secondary in a sight tube" and "physically centering the secondary in the focuser tube". How do crosshairs figure in positioning the secondary?

21. Addendum: After scratching my head a bit, I know how I would do it with crosshairs in the focusing tube. I would calculate where on the face of the secondary the optical axis of the primary would touch it, and put a dot there. Centering the dot on the crosshairs with the longitudinal movement, after having decentered the secondary laterally in the main tube by the right amount, will do the job.

I did the same thing geometrically in my big Dob mechanically. I measured the lateral offset to get that component right. Then I ran a straight steel rod through accurately centered holes in disks I had placed in the ends of the focusing tube. I could do this because I have a lathe. I placed the tip of the rod at the centerline of the main tube and then moved the secondary longitudinally until it just touched the rod. After doing all of this I noticed that the circumference of the secondary appeared to be exactly centered in the focusing tube. This is due to the perspective effect involving the reflected cone of light. Had I recognized this ahead of time I could have saved myself the trouble of the work at the lathe. Actually I am glad I did this, because sighting through these holes gives me a way of touching up the collimation in the field after hauling the scope in my truck to remote observing sites.

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Originally Posted by Hornblower
I am not sure what you mean by "centering the secondary in a sight tube" and "physically centering the secondary in the focuser tube". How do crosshairs figure in positioning the secondary?
I was confused by a few posts on another site. Also, I should not have included crosshairs in my post. However I do have a question about the first step in my collimation process listed below. I've read that offset calculations are not required, since offsets happen automatically during the process of collimation. But I do use the calculation for my 1st step. Is it unnecessary?

1st: Offset the secondary away from the OTA along the focuser axis, (offset = minor dia. / (4 x focal ratio).

2nd: Place a card between the primary and secondary, blocking primary reflection, and a sheet of light colored paper against the tube wall behind the secondary. Then, via the center bolt, rotate and, move the secondary along the OTA until is appears as a circle concentric with the focuser tube inner edge.

3rd: Remove card between secondary and primary. Use the three secondary tilt screws to get the edge of the primary concentric with the focuser (primary clips equally visible).

4th: Use three primary tilt screws to get reflections, of center dot and crosshair, stacked up.

Do these steps make sense?

23. Originally Posted by henrylr
I was confused by a few posts on another site. Also, I should not have included crosshairs in my post. However I do have a question about the first step in my collimation process listed below. I've read that offset calculations are not required, since offsets happen automatically during the process of collimation. But I do use the calculation for my 1st step. Is it unnecessary?

1st: Offset the secondary away from the OTA along the focuser axis, (offset = minor dia. / (4 x focal ratio).

2nd: Place a card between the primary and secondary, blocking primary reflection, and a sheet of light colored paper against the tube wall behind the secondary. Then, via the center bolt, rotate and, move the secondary along the OTA until is appears as a circle concentric with the focuser tube inner edge.

3rd: Remove card between secondary and primary. Use the three secondary tilt screws to get the edge of the primary concentric with the focuser (primary clips equally visible).

4th: Use three primary tilt screws to get reflections, of center dot and crosshair, stacked up.

Do these steps make sense?
I would not count on accurate offset happening automatically. You could have a considerable error the lateral offset and still get the appearance of having the circumference of the secondary centered in the focuser tube. I would definitely start with your step 1 and set the lateral offset. Your simple formula is close enough for most purposes, but the more complex one in the linked reference will more accurately center the fully illuminated field. This really is not super critical for ordinary observing.

Your method is adequate for a long focus scope of f/8 or longer. For a short one like my f/4.5 Dob, where accurate collimation is critical for eliminating coma from the center of the field, I wish to be sure of having the center spot on the primary precisely on the optical axis of the focuser tube. That is where accurately centered crosshairs and an accurately centered small hole at the eye end come into play. When that is done and the primary is tilted until the reflection of the end of the focuser tube is centered on the center spot, you will have good collimation even if the position of the secondary is slightly off. All a slight position error does there is slightly decenter the field of full illumination, which as I mentioned is not a serious concern for most observing. I don't have crosshairs, but my disks at each end of the focuser tube do very well.

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