Optics question

On Fri, 18 Jun 2004 02:16:00 +0000 (UTC), Brian Tung wrote:

Craig Franck wrote:
It does not strike the mirror in a symmetrical fashion. It hits it
slanted, or "off axis." (I'm not sure this answer will help you, since
I'm not exactly sure what it is you're asking.)

Is it off-axis because the lens or mirror has a curved surface? That makes
sense. But I don't understand why it favors a tail on one side.

That's right; the lens or mirror has an axis of symmetry, and the off-axis
light rays are tilted with respect to that axis. That means that light
rays don't get refracted in a symmetric way by the lens, so that they don't
come together to a point.

Except for spherical mirrors with the entrance pupil at the radius of
curvature. There're no such thing as off-axis rays in that situation,
which is why the Schmidt camera is so incredibly aberration free (ignoring
the presence and correction of spherical aberration).

To see why it favors a tail, I think you would have to do the math, or see
a ray-trace diagram, or something like that. I'm not sure there's a good,
simple, first-order explanation in words alone.

A rotatable 3D diagram would probably be required for an intuitive grasp.

It is interesting that the airy disk gets smaller with larger aperture.
Is that because the wave length of light gets smaller in comparison to
the overall area of the objective?

Hmm, the wavelength of light does get smaller in comparison to the size
of the objective (you can't really compare a length with an area), but I
hesitate to say that that is the *cause* of the trend. The math does work
out that way, though.

It's sort of like being better able to triangulate a position when you
have a longer baseline. I don't know if I can come up with a hard physical
analogue, however.

It's aperture width, not area, that makes the difference. It is, in fact,
a comparison of linear values.

A very complicated diagram would probably be required to get an intuitive
grasp of why the relationship is there.


--
- Mike

Remove 'spambegone.net' and reverse to send e-mail.

"Mike Ruskai" wrote

Brian Tung wrote:

To see why it favors a tail, I think you would have to do the math, or see
a ray-trace diagram, or something like that. I'm not sure there's a good,
simple, first-order explanation in words alone.

A rotatable 3D diagram would probably be required for an intuitive grasp.

I think I have it now. I was confusing the shape of the mirror or lens with
the angle a ray of light hits the slope of the mirror or lens. A round mirror
would always be round, but the shape of a star depends on where on the
mirror the rays of light hit WRT how the face of the mirror is shaped.

So plotting the coma would have two functions: how far from the center of
the field you are, and how far off-focus you are. A star in the center would
have no coma; one on the edge would have maximum coma; in focus least
coma; out of focus the coma would increase until the image disappeared.

From that it follows that coma favors one side of the image because that
corresponds to the points on the mirror or lens that are farthest from the
object.

Would that also mean that an egg shaped mirror would produce egg shaped
stars even in the center of the field?

--
Craig Franck

Cortland, NY

"Mike Ruskai" wrote

Brian Tung wrote:

To see why it favors a tail, I think you would have to do the math, or see
a ray-trace diagram, or something like that. I'm not sure there's a good,
simple, first-order explanation in words alone.

A rotatable 3D diagram would probably be required for an intuitive grasp.

I think I have it now. I was confusing the shape of the mirror or lens with
the angle a ray of light hits the slope of the mirror or lens. A round mirror
would always be round, but the shape of a star depends on where on the
mirror the rays of light hit WRT how the face of the mirror is shaped.

So plotting the coma would have two functions: how far from the center of
the field you are, and how far off-focus you are. A star in the center would
have no coma; one on the edge would have maximum coma; in focus least
coma; out of focus the coma would increase until the image disappeared.

From that it follows that coma favors one side of the image because that
corresponds to the points on the mirror or lens that are farthest from the
object.

Would that also mean that an egg shaped mirror would produce egg shaped
stars even in the center of the field?

--
Craig Franck

Cortland, NY

On Fri, 18 Jun 2004 02:16:00 +0000 (UTC), Brian Tung wrote:

Craig Franck wrote:
It does not strike the mirror in a symmetrical fashion. It hits it
slanted, or "off axis." (I'm not sure this answer will help you, since
I'm not exactly sure what it is you're asking.)

Is it off-axis because the lens or mirror has a curved surface? That makes
sense. But I don't understand why it favors a tail on one side.

That's right; the lens or mirror has an axis of symmetry, and the off-axis
light rays are tilted with respect to that axis. That means that light
rays don't get refracted in a symmetric way by the lens, so that they don't
come together to a point.

Except for spherical mirrors with the entrance pupil at the radius of
curvature. There're no such thing as off-axis rays in that situation,
which is why the Schmidt camera is so incredibly aberration free (ignoring
the presence and correction of spherical aberration).

To see why it favors a tail, I think you would have to do the math, or see
a ray-trace diagram, or something like that. I'm not sure there's a good,
simple, first-order explanation in words alone.

A rotatable 3D diagram would probably be required for an intuitive grasp.

It is interesting that the airy disk gets smaller with larger aperture.
Is that because the wave length of light gets smaller in comparison to
the overall area of the objective?

Hmm, the wavelength of light does get smaller in comparison to the size
of the objective (you can't really compare a length with an area), but I
hesitate to say that that is the *cause* of the trend. The math does work
out that way, though.

It's sort of like being better able to triangulate a position when you
have a longer baseline. I don't know if I can come up with a hard physical
analogue, however.

It's aperture width, not area, that makes the difference. It is, in fact,
a comparison of linear values.

A very complicated diagram would probably be required to get an intuitive
grasp of why the relationship is there.


--
- Mike

Remove 'spambegone.net' and reverse to send e-mail.

Craig Franck wrote:
It does not strike the mirror in a symmetrical fashion. It hits it
slanted, or "off axis." (I'm not sure this answer will help you, since
I'm not exactly sure what it is you're asking.)

Is it off-axis because the lens or mirror has a curved surface? That makes
sense. But I don't understand why it favors a tail on one side.

That's right; the lens or mirror has an axis of symmetry, and the off-axis
light rays are tilted with respect to that axis. That means that light
rays don't get refracted in a symmetric way by the lens, so that they don't
come together to a point.

To see why it favors a tail, I think you would have to do the math, or see
a ray-trace diagram, or something like that. I'm not sure there's a good,
simple, first-order explanation in words alone.

Yes. You only see the secondary mirror if the eye piece is out of
focus, so it seems the diffraction effect should be in that off-focus
focal plane.

That's not the way that diffraction works. You see diffraction effects
because the wave front has a hole in it. The wave front does come to a
focus at the focal point, but because some parts of the wave front are
missing, the focus is disturbed. This disturbance can be seen in the
eyepiece as diffraction effects.

It isn't, precisely speaking, an edge effect in the sense that it happens
*only* at the edge. And the edge of the telescope tube *does* diffract
the light. If it didn't, there would be no such thing as the Airy disc.
Light would focus down to an infinitesimal point, rather than the Airy
disc.

So with a lens or mirror it is the combination of waves of light being
gathered over the entire surface and combining that causes the airy disc.

Yes, that's right.

It is interesting that the airy disk gets smaller with larger aperture.
Is that because the wave length of light gets smaller in comparison to
the overall area of the objective?

Hmm, the wavelength of light does get smaller in comparison to the size
of the objective (you can't really compare a length with an area), but I
hesitate to say that that is the *cause* of the trend. The math does work
out that way, though.

It's sort of like being better able to triangulate a position when you
have a longer baseline. I don't know if I can come up with a hard physical
analogue, however.

Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt

"Brian Tung" wrote

Craig Franck wrote:
In 4.2.2 they discuss coma and attribute it to "the intersection of
rays not being symmetrical." Shouldn't "off axis light" come into the
telescope in a symmetrical fashion when confronting an evenly distributed
light source?

It does not strike the mirror in a symmetrical fashion. It hits it
slanted, or "off axis." (I'm not sure this answer will help you, since
I'm not exactly sure what it is you're asking.)

Is it off-axis because the lens or mirror has a curved surface? That makes
sense. But I don't understand why it favors a tail on one side.

WRT diffraction spikes, if the secondary mirror and struts are not in the
plane of focus, why would the diffraction effect occur where the mirror
itself is not visible?

I'm not quite sure what you mean by "where the mirror itself is not
visible." Which mirror, the secondary mirror? Do you mean, why do you
see a diffraction effect even though you don't see the secondary mirror
itself in the eyepiece?

Yes. You only see the secondary mirror if the eye piece is out of
focus, so it seems the diffraction effect should be in that off-focus
focal plane.

And if it's an "edge effect," way doesn't the edge of
the telescope tube diffract the light as well in a way that is visible?

It isn't, precisely speaking, an edge effect in the sense that it happens
*only* at the edge. And the edge of the telescope tube *does* diffract
the light. If it didn't, there would be no such thing as the Airy disc.
Light would focus down to an infinitesimal point, rather than the Airy
disc.

So with a lens or mirror it is the combination of waves of light being
gathered over the entire surface and combining that causes the airy disc.

It is interesting that the airy disk gets smaller with larger aperture. Is that
because the wave length of light gets smaller in comparison to the overall
area of the objective?

--
Craig Franck

Cortland, NY

Craig Franck wrote:
In 4.2.2 they discuss coma and attribute it to "the intersection of
rays not being symmetrical." Shouldn't "off axis light" come into the
telescope in a symmetrical fashion when confronting an evenly distributed
light source?

It does not strike the mirror in a symmetrical fashion. It hits it
slanted, or "off axis." (I'm not sure this answer will help you, since
I'm not exactly sure what it is you're asking.)

If one were to rotate the lens or mirror, would the coma rotate as well?

No, generally not, unless there is something extra wrong with the lens
or mirror.

WRT diffraction spikes, if the secondary mirror and struts are not in the
plane of focus, why would the diffraction effect occur where the mirror
itself is not visible?

I'm not quite sure what you mean by "where the mirror itself is not
visible." Which mirror, the secondary mirror? Do you mean, why do you
see a diffraction effect even though you don't see the secondary mirror
itself in the eyepiece?

And if it's an "edge effect," way doesn't the edge of
the telescope tube diffract the light as well in a way that is visible?

It isn't, precisely speaking, an edge effect in the sense that it happens
*only* at the edge. And the edge of the telescope tube *does* diffract
the light. If it didn't, there would be no such thing as the Airy disc.
Light would focus down to an infinitesimal point, rather than the Airy
disc.

Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt

On 2004-06-17, Craig Franck wrote:

In 4.2.2 they discuss coma and attribute it to "the intersection of
rays not being symmetrical." Shouldn't "off axis light" come into the
telescope in a symmetrical fashion when confronting an evenly distributed
light source? If one were to rotate the lens or mirror, would the coma
rotate as well?

Look at a cross section of a parabolic mirror. With light rays parallel
to the axis of the parabola the rays striking the left hand side of the
mirror make the same angle with the surface of the glass as the rays
striking the right hand side of the mirror and the rays all meet at the
center. With rays coming in off axis the rays striking the left hand
side of the mirror no longer make the same angles as rays striking the
right hand side of the mirror and the rays don't all meet at the center.
This is what causes coma. With parabolic mirrors the image of an off
axis star spreads away from the center. The diagrams in the book
show exactly what happens. Since lenses and mirrors are figures of
rotation, rotating the lens or mirror on it's axis doesn't change
anything.

On 2004-06-17, Craig Franck wrote:

In 4.2.2 they discuss coma and attribute it to "the intersection of
rays not being symmetrical." Shouldn't "off axis light" come into the
telescope in a symmetrical fashion when confronting an evenly distributed
light source? If one were to rotate the lens or mirror, would the coma
rotate as well?

Look at a cross section of a parabolic mirror. With light rays parallel
to the axis of the parabola the rays striking the left hand side of the
mirror make the same angle with the surface of the glass as the rays
striking the right hand side of the mirror and the rays all meet at the
center. With rays coming in off axis the rays striking the left hand
side of the mirror no longer make the same angles as rays striking the
right hand side of the mirror and the rays don't all meet at the center.
This is what causes coma. With parabolic mirrors the image of an off
axis star spreads away from the center. The diagrams in the book
show exactly what happens. Since lenses and mirrors are figures of
rotation, rotating the lens or mirror on it's axis doesn't change
anything.

"Craig Franck" wrote in message .. .

In 4.2.2 they discuss coma and attribute it to "the intersection of
rays not being symmetrical." Shouldn't "off axis light" come into the
telescope in a symmetrical fashion when confronting an evenly distributed
light source? If one were to rotate the lens or mirror, would the coma
rotate as well?

Coma is caused by the geometry of optical surfaces, resulting in the
bundles of parallel (normally) off-axis light being focused
differently by different zones of one or more optical surfaces. In
general, only the very center of the surface focuses such bundle of
rays into an on-axis point (talking geometrical optics); every next
concentric zone on the optical surface focuses it into an off-axis
centered circle. Both circle diameter and its off-axis shift increase
with the zone hight, reaching the maximum for the outer edge of the
optical surface.

For a concave mirror, the circle radius is given by hr^2/16F^2, and
the off-axis shift of its center is twice as much, hr^2/8F^2, with "h"
being the off-axis distance of a point in the focal plane, "r" the
aperture radius normalized to 1, and F the F#. The entire length of
comatic blur is 3h/16F^2. For example,
a 200mm f/5 parabola would have comatic blur length 1mm off-axis of
0.0075mm; central area of the mirror would focus at the axial point;
concentric zone at half the mirror radius (so r=0.5) would focus into
a circle of 0.000625mm radius, with its center shifted off-axis by
0.00125mm. And the edge zone (r=1) would focus into a circle of
0.025mm radius, shifted off-axis by 0.05mm.

WRT diffraction spikes, if the secondary mirror and struts are not in the
plane of focus, why would the diffraction effect occur where the mirror
itself is not visible? And if it's an "edge effect," way doesn't the edge of
the telescope tube diffract the light as well in a way that is visible?

It is not an "edge effect", and not a result of "ray bending" etc.
Those are popular misconceptions. What causes diffraction can be
illustrated by a converging wavefront, whose every point emits waves -
so called wavelets - in all directions. If you replace those wavelets
by "raylets", you see that all raylets coming to the focus point have
identical path length, regardless of from what point on the wave front
they arrive (this is because the focal point is a centar of the
wavefront sphere). Since the path length is identical, all these
raylets meet in phase, resulting in maximum wave interference and
highest light intensity. For points slightly off-axis in the focal
plane, the raylets don't have identical path lengths, and don't meet
in phase. The interference and intensity weaken, dropping to zero at
the Airy radius (first minima), then partially recover through the
first bright ring, hit the second minima, and so on, producing ever
fainter rings.

Any obstruction placed in the light path will alter net interference
of the raylets in the focal plane by blocking out portion of the
wavefront.

Vlad