Prove that Fo/Fe = magnification
astrongc
is intuitive. But the formula M= fo/fe (where fo
is the focal length of the objective divided by the
focal length of the eyepiece) is not! It is not shown
in the book Telescope Optics how it is derived.
Neither are the many web sites I visited. So I seek
for assistance from those who know more. (so bear
with me again) :)
The height of the object at the focal point is
f*tan (alpha) [Thanks Brian for this formula)]. Now
for a given focal length. The height gets shorter as it
go towards the objective lens from the focal plane.
Now with the formula M=fo/fe. Does it mean that
at 5mm before the focal plane coming from the objective
lens. The eyepiece with 5 mm focal length can image
the same exact height of the image 5 mm prior to
the focal plane?? How is this possible? The lens
of the eyepiece doesn't get the full beam from the
image. Just part of it. So how can you equate focal
length of the telescope divided by focal length of
the eyepiece equals the magnification??
If anyone knows of a web site that can enable you
to move thru interactive java applet the eyepiece
to different position and see the alpha, beta angle, the
image, etc. move corresponding to it. Pls share it. It's
going to be the ultimate tutorial for us newbies. :)
Thanks.
lex
Brian Tung
> The formula Magnification = tan(beta)/tan(alpha)
> is intuitive. But the formula M= fo/fe (where fo
> is the focal length of the objective divided by the
> focal length of the eyepiece) is not! It is not shown
> in the book Telescope Optics how it is derived.
> Neither are the many web sites I visited. So I seek
> for assistance from those who know more. (so bear
> with me again) :)
I couldn't understand any of the detailed questions you asked later
in your post, so I'm afraid I'm going to just have to try to answer
your general question.
A telescope basically works in two steps:
1. The objective converts an object at infinity with a specific
angular size into a real image with a specific linear size.
2. The eyepiece converts a real image into a virtual image (what
you see in the eyepiece) at infinity with a specific angular
size.
The rule in both cases, provided the angles are small, is that the
linear size, divided by the focal length, equals the angular size.
Conversely, the angular size, multiplied by the focal length, equals
the linear size.
So, in the first step, the angular size of the object is, let's say,
alpha. The objective produces a real image of the object with a
linear size equal to fo*alpha, where fo is the focal length of the
objective. OK so far?
In the second step, the linear size is f*alpha, and to get the angular
size of the virtual image, we divide by the focal length--not of the
objective now, but rather of the eyepiece. Let's call that focal
length fe. In that case, the virtual image's angular size is equal
to beta = fo*alpha/fe.
Finally, the magnification is equal to beta divided by alpha, which
as you can see is just fo/fe.
Brian Tung <br...@isi.edu>
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
astrongc
trying to visualize it.
In the latter inquiry. What I mean is this.
If the light beam of angle alpha enters and crosses the
center of the objective lens. The height or distance between
it and the center axis gets larger as it gets nearer the focal
plane with maximum vertical height attained when it reaches
the focal plane. A longer focal length enabling it to get larger
in height from f*tan(theta). Now this means that halfway to
focus. The height of the linear image is divided by 2. Suppose
we have a focal length of 400mm. After the beam passes thru
the objective lens. The height attained at 100mm from
the objective lens is a certain size and it gets taller as it
gets from 200mm to 300mm and finally to 400mm. Now.
at 350mm from the objective lens and 50mm before the
focal plane. The height of the image is a certain height. Does
this height correspond to that of a 50mm eyepiece (as image
linearly)?
Other things. I wonder if magnification in the eyepiece is
connected with eye relief. If one designs a 4mm eyepiece
with eye relief of 25mm. Does the image gets bigger. With
longer eye relief, the angle beta gets shallower so magnification
can become less.
Also can one design an eyepiece with focal length of 20mm
yet uses the same diameter as an eyepiece of 5mm? This means
the rays would be bent sharper hence a 20mm eyepiece can
have magnification equivalent to a 5mm eyepiece?
Of course I know the formula M=fo/fe but I just want to
understand how eyepiece design can be modified that can
altered the magnification from a fixed focal length (if it is
possible at all).
Thanks for the input. You ought to write a book on optics for the amateur
community as you can describe it so clearly :)
lex
br...@isi.edu (Brian Tung) wrote in message news:<bdrjep$f1q$1...@zot.isi.edu>...
AP
"astrongc" <game_...@yahoo.com> a écrit dans le message news:
10ebddec.03063...@posting.google.com...
Look this site, in french:
http://www2.globetrotter.net/astronomie_au_quebec/bibliotheque/biblio.htm
Brian Tung
> If the light beam of angle alpha enters and crosses the
> center of the objective lens. The height or distance between
> it and the center axis gets larger as it gets nearer the focal
> plane with maximum vertical height attained when it reaches
> the focal plane. A longer focal length enabling it to get larger
> in height from f*tan(theta). Now this means that halfway to
> focus. The height of the linear image is divided by 2. Suppose
> we have a focal length of 400mm. After the beam passes thru
> the objective lens. The height attained at 100mm from
> the objective lens is a certain size and it gets taller as it
> gets from 200mm to 300mm and finally to 400mm. Now.
> at 350mm from the objective lens and 50mm before the
> focal plane. The height of the image is a certain height. Does
> this height correspond to that of a 50mm eyepiece (as image
> linearly)?
Ah, no. The eyepiece is not placed *at* the focal plane. It would
be placed 50 mm (or whatever the focal length of the eyepiece is)
behind the focal plane. The real image is formed at the focal plane,
and in order to view that real image, the eyepiece has to be 50 mm
behind that real image, just as it would have to be 50 mm behind an
actual object if you were to use the eyepiece as a magnifying glass.
Thus, the optical length of a refractor is fo+fe. (The physical
length may vary because of dew shields, diagonals that fold the
optical path, and so on.)
As you've apparently noticed, for objects that are off the axis, some
of the light rays focused by the objective may not enter the eyepiece,
because they are too "high." Only some of the light rays get in.
The scale of the image is not affected by this, but the image does
get darker as the object goes further off the axis (alpha gets
larger), and fewer and fewer of the light rays enter the eyepiece.
We call this vignetting. There is usually some vignetting at the
edge of the field of view, but for visual purposes, this ordinarily
doesn't present a problem. Only when vignetting is severe does it
really become noticeable--typically, for example, I don't notice 50
percent vignetting, because my attention is generally focused on the
center of the field of view.
> Other things. I wonder if magnification in the eyepiece is
> connected with eye relief. If one designs a 4mm eyepiece
> with eye relief of 25mm. Does the image gets bigger. With
> longer eye relief, the angle beta gets shallower so magnification
> can become less.
Well, not really. The eye relief is principally a function of two
parameters: the diameter of the eye lens of the eyepiece (the end
you look through, as opposed to the field lens, which is the end
closer to the objective), and the apparent field of view. If you
place your eye the proper distance away from the eye lens--this is
the eye relief--the apparent field of view is obviously bounded by
how large an angle the eye lens appears to your eye. If the diameter
of the eye lens is de, and the eye relief is er, then
tan(AFOV/2) <= (de/2)/er
> Also can one design an eyepiece with focal length of 20mm
> yet uses the same diameter as an eyepiece of 5mm? This means
> the rays would be bent sharper hence a 20mm eyepiece can
> have magnification equivalent to a 5mm eyepiece?
Yes. The Tele Vue Radians, for example, have a reasonably constant
apparent field of view (about 60 degrees) with a reasonably constant
eye relief (about 20 mm), so the eye lens must be just about the
same in all of them, from the 3 mm Radian to the 18 mm.
> Of course I know the formula M=fo/fe but I just want to
> understand how eyepiece design can be modified that can
> altered the magnification from a fixed focal length (if it is
> possible at all).
No, for practical purposes, the magnification is determined by the
focal lengths of the eyepiece and objective. Fortunately, we have
optical wizards among us who can give us practically any focal length
eyepiece we want, with wide fields of view and excellent correction.
Unfortunately, you may have to pay a bit for them.
> Thanks for the input. You ought to write a book on optics for the amateur
> community as you can describe it so clearly :)
I would love to do that, but (a) I don't feel I'm really qualified to
do so, and (b) I don't know that there is really a market for Yet
Another Book on Telescope Optics.
Chuck Olson
"astrongc" <game_...@yahoo.com> wrote in message
news:10ebddec.03063...@posting.google.com...
> The formula Magnification = tan(beta)/tan(alpha)
> is intuitive. But the formula M= fo/fe (where fo
> is the focal length of the objective divided by the
> focal length of the eyepiece) is not! It is not shown
> in the book Telescope Optics how it is derived.
> Neither are the many web sites I visited. So I seek
> for assistance from those who know more. (so bear
> with me again) :)
Maybe I can paint a mechanical picture that is a little more intuitive than the optics or
mathematics. If we observe that the side rays through a lens only serve to define the distance
to the focus plane, we can simplify the picture to use only the central rays that are straight
and unbent. Imagine the central ray through the objective lens coming straight from the top of a
far away planet to the bottom of the planet's image. Then the eyepiece sends that point at the
bottom of the image through the eyepiece center straight into the eye. Each lens center can be
visualized as a pivot point for these straight central rays as other points on the planet are
imaged. Now since the central rays are always straight, visualize these rays as sticks pivoting
at the objective and eyepiece lens centers but with a connecting hinge at the image plane. The
longer the stick that pivots at the center of the objective lens is, the greater will be the
movement of the hinge for any entry angle. The shorter the stick that pivots at the center of
the eyepiece, the greater will be the exit angle. The ratio of entry angle to exit angle is the
magnification, and the distances from the lens pivot points to the hinge are the lens focal
lengths. With this analogy, it should be obvious that the magnification is close to fo/fe, and
with Brian's mathematics, we see it is exactly equal to fo/fe.
Any better?
Chuck
Unknown
Beautifully described.
james
On Tue, 1 Jul 2003 09:15:05 +0000 (UTC), br...@isi.edu (Brian Tung)
wrote:
astrongc
> Maybe I can paint a mechanical picture that is a little more intuitive than the optics or
> mathematics. If we observe that the side rays through a lens only serve to define the distance
> to the focus plane, we can simplify the picture to use only the central rays that are straight
> and unbent. Imagine the central ray through the objective lens coming straight from the top of a
> far away planet to the bottom of the planet's image. Then the eyepiece sends that point at the
> bottom of the image through the eyepiece center straight into the eye. Each lens center can be
> visualized as a pivot point for these straight central rays as other points on the planet are
> imaged. Now since the central rays are always straight, visualize these rays as sticks pivoting
> at the objective and eyepiece lens centers but with a connecting hinge at the image plane. The
> longer the stick that pivots at the center of the objective lens is, the greater will be the
> movement of the hinge for any entry angle. The shorter the stick that pivots at the center of
> the eyepiece, the greater will be the exit angle. The ratio of entry angle to exit angle is the
> magnification, and the distances from the lens pivot points to the hinge are the lens focal
> lengths. With this analogy, it should be obvious that the magnification is close to fo/fe, and
> with Brian's mathematics, we see it is exactly equal to fo/fe.
>
> Any better?
>
> Chuck
To understand your analogy. I have to first understand the wire control of
an airplane or F16. Then apply it to optics. :) Thanks anyway. I'll return
to it after reviewing aviation.
lex
astrongc
I have researched about magnifier, virtual image and how the
eyepiece is exactly like that. However. Some inquries that need
a bit of clarification.
Suppose the real image at the focal plane formed by the objective
lens is 1/2 inch (just an example). And you view this with a 20mm
eyepiece. You would get a certain virtual image size which is larger.
Now supposed you have an actual physical object like a microprocessor that
is exactly 1/2 inch in height and you view this thru an eyepiece (like
a magnifying glass). What would form a larger virtual image. The
objective real image (height 1/2 inch) at the focal plane or the
processor (height 1/2 inch) when both of them are viewed directly
with an eyepiece (A while back, I got my Televue plossl to check if
it also serves as magnifying glass and when I insert my finger to the
field stop inside the barrel of the eyepiece, I can see some hair in
my fingers). Now I'm asking if the objective real image or the
processor is larger because when you view an object thru a
magnifying glass, the beam of the heighest part is exactly horizonal
to the eyepiece (or magnifying glass) while that of the objective, it's
kind of slant so the virtual image formed by this should be larger,
right? If they are both the same, then some part still escapes my
understanding and hope someone clarify it.
Also I wonder what is the formula that would equate the
relationship between the real image and the virtual image like
how big exactly is the virtual image compared to the real image.
A certain Chris Peterson wrote in the other thread: "Now, you are
going to view the image of the Airy disk with an EP, which is really
just a simple magnifier with power proportional to 1/focal length."
What does he meant by that? I mean. If the focal length of the
scope is 400mm. Then the power of the magnifier is proportional
to 1/focal length or 1/400?? It just makes the object much much
smaller!! Wonder where is the typo.
Also, looking at the illustration of the image formation of a telescope.
I can see that it is possible to put the eyepiece before the focal
plane rather than after it. So if you have a 20mm eyepiece, you can
put this behind the focal plane at distance 20mm before the focal
plane. The implication of this is that you can see the image erect up
and down and non-reverse left to right so it's perfect for terrestrial
application. How come no such eyepiece is designed?? Or if it's not
possible to make, why?
Gee. It sure is fun to know how telescope optics work in addition
to viewing the sky thru them. I wonder how many percentage of
the people in this newsgroup actually understand in detail the
image formation principles of a telescope, like what is beta and alpha, etc.
20%? 50%? 70%? It sure would be interesting to find out.
lex
br...@isi.edu (Brian Tung) wrote in message news:<bdsemg$gcb$1...@zot.isi.edu>...
Brian Tung
> Suppose the real image at the focal plane formed by the objective
> lens is 1/2 inch (just an example). And you view this with a 20mm
> eyepiece. You would get a certain virtual image size which is larger.
The virtual image is at infinity, typically, so it has no well-defined
linear size. Instead, it has an angular size, which is equal to about
2*arctan(1/4 inch divided by 20 mm) = 2*arctan(0.32) = 35 degrees
(You ought to try to choose consistent units.)
> Now supposed you have an actual physical object like a microprocessor that
> is exactly 1/2 inch in height and you view this thru an eyepiece (like
> a magnifying glass). What would form a larger virtual image. The
> objective real image (height 1/2 inch) at the focal plane or the
> processor (height 1/2 inch) when both of them are viewed directly
> with an eyepiece
They are the same size.
> Now I'm asking if the objective real image or the
> processor is larger because when you view an object thru a
> magnifying glass, the beam of the heighest part is exactly horizonal
> to the eyepiece (or magnifying glass) while that of the objective, it's
> kind of slant so the virtual image formed by this should be larger,
> right? If they are both the same, then some part still escapes my
> understanding and hope someone clarify it.
You've been misled by diagrams that show parallel beams of light from
an object being focused to a point. In actuality, each part of the
object or real image emits an entire fan of rays. In the case of the
object, light from each part of the object enters every part of the
lens. Yes, that means the light rays cross. There's nothing at all
paradoxical about that--it's much like how four people seated at a
square table can carry on two conversations, even if they "cross" one
another.
In the case of the real image, the fan may be incomplete. Trace two
lines, one from the top of the objective, and one from the bottom,
through the top of the real image. These two lines delimit the fan
of rays that come from the real image. The extent to which this fan
enters the eyepiece determines the illumination of that part of the
image. If you do the same two lines, but passing through the center
of the real image, you get the fan of light rays coming from there.
Note that the two fans may intersect.
Each fan of light is diverging as it passes through the real image.
The eyepiece collimates each fan into a parallel, collimated beam, so
that the virtual image appears at infinity, with an angular size
determined by the focal length of the eyepiece and the linear size of
the real image, as explained in the previous post.
> Also I wonder what is the formula that would equate the
> relationship between the real image and the virtual image like
> how big exactly is the virtual image compared to the real image.
As I mentioned in my earlier post, the virtual image is at infinity,
and has an angular size roughly equal to the linear size of the real
image, divided by the focal length of the eyepiece.
> A certain Chris Peterson wrote in the other thread: "Now, you are
> going to view the image of the Airy disk with an EP, which is really
> just a simple magnifier with power proportional to 1/focal length."
He means that the eyepiece is just like a magnifying glass, and so
is subject to the usual rule: halve the focal length, and you double
the magnification. The magnification is not dimensionless (you're
relating the linear size of an object or real image to the angular
size of the virtual image) so he can't say it's equal to 1 over the
focal length. That's why he (rightly) says that it's proportional.
> What does he meant by that? I mean. If the focal length of the
> scope is 400mm. Then the power of the magnifier is proportional
> to 1/focal length or 1/400?? It just makes the object much much
> smaller!! Wonder where is the typo.
Note that he said "the eyepiece [EP]." You've applied his rule to the
objective, which is not how he meant it. Remember the two-stage
virtual image formation:
object at infinity --> real image in scope --> virtual image at infinity
In the first case, the greater the focal length, the larger the real
image, because the linear size of the real image is the angular size
of the celestial object *times* the focal length of the objective.
In the second case, the greater the focal length, the smaller the
virtual image, because the angular width of the virtual image is equal
to the linear size of the real image *divided* by the focal length of
the eyepiece. That's why the two focal lengths are on opposite sides
of the fraction in the magnification formula.
> Also, looking at the illustration of the image formation of a telescope.
> I can see that it is possible to put the eyepiece before the focal
> plane rather than after it. So if you have a 20mm eyepiece, you can
> put this behind the focal plane at distance 20mm before the focal
> plane. The implication of this is that you can see the image erect up
> and down and non-reverse left to right so it's perfect for terrestrial
> application. How come no such eyepiece is designed?? Or if it's not
> possible to make, why?
You can only put it before the focal plane if it's a negative eyepiece,
as in a Galilean refractor. These aren't made very often, apparently
because they are subject to severe distortion, and a narrow field of
view.
> Gee. It sure is fun to know how telescope optics work in addition
> to viewing the sky thru them. I wonder how many percentage of
> the people in this newsgroup actually understand in detail the
> image formation principles of a telescope, like what is beta and alpha, etc.
> 20%? 50%? 70%? It sure would be interesting to find out.
I don't know, but in my experience, not a lot of people understand it
very well. (I'm not sure I understand many of the finer aspects of it
myself!) People tend to memorize rules of thumb, which is all well and
good, but those rules of thumb often hide the reasoning that leads to
them.
astrongc
> You've been misled by diagrams that show parallel beams of light from
> an object being focused to a point. In actuality, each part of the
> object or real image emits an entire fan of rays. In the case of the
> object, light from each part of the object enters every part of the
> lens. Yes, that means the light rays cross. There's nothing at all
> paradoxical about that--it's much like how four people seated at a
> square table can carry on two conversations, even if they "cross" one
> another.
>
Hi Brian,
I've been thinking and visualizing what you said for quite some time.
And I think you may have missed something. Note that the image formed
at the focal plane by the objective lens doesn't reflect light. That
is. It's not an actual physical object so when you examine it with
a magnifier (or an eyepiece). It doesn't comply with the behavior of
real object. Going to the same example of examining a 0.5 inch
microprocessor vs 0.5 inch focal plane image. The 0.5" processor (real
object)
produces fan like light beam in all parts of the object (thru
reflection). And the angle of the light beam determines the position
of each part of the object
that's why our eyes don't see the world as just compose of a blur of
all mixed up lights. Now. In the top most position of the 0.5" real
object
processor. It would emit a fan like beam and the magnifier can
determine
it's position by the angle emitted. So it would produce a
corresponding
virtual image. Now about the 0.5" image at the focal plane. The
telescope image
itself doesn't emit fan-like beam of light. It comes from the
objective
lens itself. As the two parallel rays from top and bottom of objective
lens reach the focal plane from below the axis. It diverges into the
eyepiece from the focal plane at a more slant angle. So the parallel
beam produced by the eyepiece would be higher. This would determine
the
top most position of the virtual image of the object which would be a
bit
taller than that of a real object with same dimension viewed under the
eyepiece (or magnifier) because one reflects light and the other
doesn't
(but instead comes from the objective lens at more slant angle).
Right? What's wrong if I still got it wrong though. Thanks a lot for
the assistance.
Say. Did Galileo really figured out all those formula by himself.
To understand those things. One has got to really understand virtual
image as it pertains to an almost holographic real image at the focal
plane. What
inspire Galileo or other pioneers who think of that. If it's not
Galileo.
Who are the pioneers who first understand them. Most amateur
astronomers know
Einstein figures out E=mc(square), but they don't know who figures out
that a telescope magnification is equal to focal length of the
objective divided by focal length of the eyepiece. So it's time we all
know :)
lex
Chuck Simmons
What's wrong is that you aren't seeing properly that the objective forms
a real image. That is, the real image may be examined by using a ground
glass screen. If you examine the real image with a magnifier (eyepiece),
the real image will be magnified just the same as if it were a real
object. If you trace many rays from the object through the telescope,
you will see that the rays from the real image are indistinguishable
from rays from a real object at the image plane of the objective.
> Say. Did Galileo really figured out all those formula by himself.
> To understand those things. One has got to really understand virtual
> image as it pertains to an almost holographic real image at the focal
> plane. What
> inspire Galileo or other pioneers who think of that. If it's not
> Galileo.
> Who are the pioneers who first understand them. Most amateur
> astronomers know
> Einstein figures out E=mc(square), but they don't know who figures out
> that a telescope magnification is equal to focal length of the
> objective divided by focal length of the eyepiece. So it's time we all
> know :)
Galileo used a Galilean telescope. This type of telescope uses a
negative eyepiece in front of the focal plane of the objective and forms
a virtual image (there is no real image except on the retina of the
eye). If you ray trace this type of telescope you can see that it has a
very narrow field of view. What is remarkable is that Galileo made such
wonderful discoveries with such a dismal telescope.
Probably Newton knew how to calculate the power of telescopes but it
could have been known earlier.
Chuck
--
... The times have been,
That, when the brains were out,
the man would die. ... Macbeth
Chuck Simmons chr...@webaccess.net
William Hamblen
>I have researched about magnifier, virtual image and how the
>eyepiece is exactly like that. However. Some inquries that need
>a bit of clarification.
Telescopes make a real image that you can see without an eyepiece.
Try it some time by aiming your telescope at the moon. You can see an
upside down image of the moon apparently floating in the air at the
focal plane of your telescope or if that is too difficult caught on a
piece of frosted plastic or wax paper held at the focal plane. The
size of the image depends on the focal length of the telescope, call
it F, and the angular size of the object. If you haven't developed
presbyopia you can get as close as 10 inches or 250 mm to the real
image. At that distance the apparent size of the image will be F/250
times the apparent size of the object in the sky. Your eyepiece is
really a small magnifying glass. The magnification of a magnifying
glass usually is given as 250/f because a magnifier of focal length f
millimeters lets you get optically that close to the object. Multiply
one magnification times the other and you get F/f as the total
magnification.
Microscope eyepieces work the same way. Microscopes have more-or-less
standardized dimensions so microscope eyepieces are rated by
magnification instead of focal length. A 10X microscope eyepiece has
a focal length of 25 mm. Microscope objectives likewise are rated by
magnification and the total magnification is the product of the
eyepiece magnification times the objective magnification.
William Hamblen
<chr...@webaccess.net> wrote:
>Probably Newton knew how to calculate the power of telescopes but it
>could have been known earlier.
Kepler had it figured out.
PrisNo6
> If anyone knows of a web site that can enable you
> to move thru interactive java applet the eyepiece
> to different position and see the alpha, beta angle, the
> image, etc. move corresponding to it. Pls share it. It's
> going to be the ultimate tutorial for us newbies. :)
After looking at some of the two ray tracing online Java applet
programs, I can probably walk you through how two basic magnification
equations are derived, using two webpages:
http://www.mtholyoke.edu/~mpeterso/classes/phys301/geomopti/lenses.html
http://www.phy.ntnu.edu.tw/java/Lens/lens_e.html
The first Java applet ray-trace program (lenses.html) are part of an
online resources page for Prof. Mark A. Peterson's Physic's 301 class
at Mount Holyoke College.
The second applet (lens_e.html) was written by Dr. Hsiang-Wu Huang, a
physics professor, at the National Taiwan Normal University.
I have no association with either of the professors or colleges, and
mention them here to give proper credit for these helpful online
resources.
My prior post discusses derivation of the equation for the angular
magnification of extended objects yielding the magnification equation:
M = f_obj_mm / f_eyepiece (1.1)
In this post, we'll concentrate on using the objective/exit pupil
equation for magnification:
M = Diameter_obj / Diameter_exit_pupil (1.2)
using ray tracing of a beam source and the telescope simulator at:
http://www.mtholyoke.edu/~mpeterso/classes/phys301/geomopti/twolenses.html
Some of the controls in Dr. Peterson's webpage are initially difficult
to understand, so I'll lead you through which buttons to push. It may
be helpful to print this post first, to be able to refer the text
while you use the telescope simulator.
I. Objective/exit pupil magnification equation using ray tracing of a
beam source
One equation for figuring magnification of a telescope is:
M = Diameter_obj / Diameter_exit_pupil
a) Load Dr. Peterson's telescope simulator at:
http://www.mtholyoke.edu/~mpeterso/classes/phys301/geomopti/twolenses.html
The telescope simulator Java appelt is in top 1/4 of the page. In the
paragraph above the simulator is the sentence containing a hyperlink:
"One of the simplest and most useful lens combinations is the
_astronomical telescope (below)_." Click the hyperlink "astronomical
telescope" to reset the applet to display a simulated telescope.
The beam source (the parallel rays coming from a distant extended
object or a point source star with no parallax) is on the left; the
objective is the left lens; the eyepiece is the right lens.
Note the beam source is a stream of parallel light rays that fill the
entire diameter of the objective's aperature (Diameter_obj).
The applet's default schematic is missing something: the "third" lens
in all telescopes - the human eye.
b) Let's add a simulated human eye to the diagram.
Click on the "Lens" control. _To the right_ of the eyepiece lens,
click anywhere in the diagram. The lens of the human eye will be
added.
c) Now let's add the iris of the human eye and simulate the Ramsden
disk of the eye. The relaxed human pupil is about 7mm or .3 inches in
persons under 40 years of age.
Click on the "Aperature" control. _Between the eyepiece and the
human eye lens_, click anywhere in the diagram. Grab one of white
dots at the end of the aperature and reduce the size of the aperature
until the "opening" value is "0.3".
You can see that the light from the beam source on the left-hand side
travels through the objective, travel through the focal point of the
objective lens, and then expand out as they approach the eyepiece
lens. The eyepiece lens converts the rays into a set of parallel rays
that then enter the human eye through the iris (the Ramsden disk).
The diameter of this stream of parallel light is the "exit pupil"
diameter (Diameter_exit_pupil) of a telescope or binoculars. In the
simulator this exit pupil diameter is now approximately equal to the
diameter of the human exit pupil (Diameter_human_exit_pupil), or the
Ramsden disk. The simulater now shows an eyepiece lens with a nearly
optimized exit pupil stream, or:
Diameter_exit_pupil = Diameter_human_exit_pupil (2.0)
The magnification of the system is given by dividing the size of the
beam source coming into the objective by the size of the exit pupil
exiting the eyepiece lens, or:
M = Diameter_obj / Diameter_exit_pupil (1.2)
d) Now let's put a higher powered eyepiece lens in our simulated
telescope, refocus and let's see what happens to the exit_pupil stream
of light.
Click on the eyepiece lens. Two white dots appear at the focal length
of the eyepiece lens - one on either side of the lens. Note the focal
length of this piece is 1" or 25.4 mm.
Let's insert a high powered lens in our simulated telescope with a
focal length of 7.6 mm or .3 inches.
Click and grab one of the white dot focal length controls and drag it
closer to the eyepiece lens. Continue to drag it to the eyepiece lens
until the value of the "FL" focal length read-out is "0.3".
Now refocus the telescope by moving the eyepiece lens. Click and grab
anywhere inside the eyepiece lens. Drag the eyepiece lens and the
left-most focal point indicator until the light rays in the exit pupil
stream are again parallel. This point is near where the left most
focal point indicator will fall over the focal point of the objective.
Note that the exit pupil diameter is now much small than the Ramsden
disk of the iris.
f) Visualize the maximum useful magnification of a telescope.
The figure now illustrates the limitation of the highest useful
magnification of a telescope. At progressively higher magnifications
are used, the exit pupil stream of parallel light becomes smaller. It
is perceived in the human eye as dimmer, since the stream of exit
pupil light no longer uses all of the available iris. At some point,
the virtual image reaching the eye is so small and dim, that the
brightness of the virtual image falls below the ability of the human
eye to see it - at around 6.5 mag.
David Knisley periodically posts a summary of useful eyepiece
magnifications. I have summarized his table here. Note that he
describes useful magnifications, partially in terms of the size of the
exit pupil produced by the lens. It may be helpful to compare
Peterson's telescope simulator with Knisely's table to better
understand how exit pupil size relates to useful maximum
magnification:
===========
LOW POWER (3.7 to 9.9x [of magnification] per inch of aperture)(6.9mm
to 2.6mm exit pupil): Useful for finding objects and for observing
ones of large angular size like open clusters.
MEDIUM POWER (10x to 17.9x [of magnification] per inch of
aperture)(2.5mm to 1.4mm exit pupil): Useful for observing details in
many deep-sky objects.
HIGH POWER (18x to 29.9x [of magnification] per inch of
aperture)(1.4mm to 0.8mm exit pupil): A very useful power range for
observing the planets.
VERY HIGH POWER (30x to 41.9x [of magnification] per inch of
aperture)(0.8mm to 0.6mm exit pupil): Useful for high resolution study
of planetary detail, and resolving double stars near or just above the
resolution limit of the instrument.
EXTREME POWER (42x to 75x [of magnification] per inch of
aperture)(0.6mm to 0.3mm exit pupil). Mainly used for resolution of
double stars at the resolution limit of the instrument.
EMPTY MAGNIFICATION (100x per inch and above) [less than 0.3mm of exit
pupil.] Basically useless powers.
=================
g) Visualize the minimum useful magnification of a telescope.
Now let's put a low magnfication eyepiece in the telescope. Set the
focal length in the simulator to fl=1.2 inches or an eyepiece with a
focal length of about 30.5 mm (as explained above). Refocus the
eyepiece (as explained above) so the rays of light in the exit pupil
stream are parallel.
The figure now illustrates the limitation of the lowest useful
magnification of a telescope. The exit pupil stream of light now
covers more than the area of the human iris. Light is wasted. The
image will again become dimmer because all available light is not
used.
II. Angular magnification equation using an extended object with
angular measurement
My prior post discusses derivation of the equation for angular
magnification of extended objects:
M = f_obj_mm / f_eyepiece (1.1)
I won't go throught the entire proof again, but Prof. wang's single
lens ray trace program is helpful in understand how this equation
works.
a) Load Dr. Wang's thin lens simulator at:
http://www.mtholyoke.edu/~mpeterso/classes/phys301/geomopti/lenses.html
The simulator page loads a floating Java applet that is helpful in
understanding the angular magnification equation:
M = f_obj_mm / f_eyepiece (1.1)
as discussed in my prior post.
The default applet shows an objective lens of the telescope with an
extended object (that has an angular size to the left of the lens.)
The program traces three light rays. It is only the center light ray
that passes through the mid-point of the lens that we are concerned
with.
At the far right of the diagram is a white vertical line. This is a
focal plane line. Click and drag it over the virtual image of object
on the right-hand side of the lens at the focal point of the objective
lens.
b) Visualize angular magnification.
Again, it is only the center light ray that passes through the
mid-point of the lens that we are concerned with. Ignore the other
two other ray trace lines.
Normally the angular magnification and the extended object now
illustrated in the applet's figure are used to discuss objects that
are a known, finite distance in front of the objective lens.
For our purposes, the actual height of the extended object and its
distance from the objective lens does not matter. Only the angular
size of the object is relevant. It does not matter that the object is
a baseball sitting on a fence post thirty feet away (with an angular
size of 30 arcminutes or a globular cluster 1500 parsecs away (that
also has an angular size of 30 arcminutes). Both the globular cluster
and the baseball have the same angular size.
In the applet, the extended object is represented by the blue vertical
line with an arrow on top. This is to the left of the objective lens.
The extended object might be visualized as a globular cluster,
centered on the centerline. 1/2 the height of the globular cluster is
shown by the blue-line with the arrow on top.
Click and grab the top of the extended object to the left of the
objective lens. Move it up and down in the "y" plane while holding it
steady in the "x" plane. (Holding the object steady in the "X" plane
is intended to simulate an object at an infinite distance.)
Note that a virtual image of object is traced in the focal plane on
the right-hand size of the objective.
Also note that the angle of the central ray going from the top of the
extended object and crossing the mid-point of the lens at the
centerline, is always equal to the angle at which the ray leaves the
objective lens. This is angle theta_obj discussed in my prior post.
Rays leaving the extended object and passing through the lens form a
new virtual image at the focal point of the objective with a linear
height of:
h_objective_mm = f_obj_mm x tan(theta_obj)
At this point, please see my prior post regarding proof that
magnification (M) can be computed by:
M = f_obj_mm / f_eyepiece (1.1)
Dr. Wang's applet may be of use in visualizing how the proof works.
I hope the above helps you to better visualize how telescopic
magnfication works and how that magnification can be related to both
the linear height of the image produced by the objective lens (using
one magnification equation) _and_ the diameter of the exit pupil
stream (using the other magnification equation).
For a simulator that shows what it looks like through the eyepiece for
various objective focal lengths, eyepiece focal lengths and
magnifications, see the online web application at:
Regards - Kurt
astrongc
reflect light because it is not a hologram. It produces its
own light transmitting fan like beam in all directions coming
from every part of the image, right?
If this is what makes it focus to the magnifier like real object.
Then what's the purpose of the diverging rays after passing
thru the focal plane. If the diversing rays are not what
produces the image in the magnifier (or eyepiece) but light
from the image at the focal plane itself with slightly different
angle, then isn't it that the diversing beam from the objective
lens can only muddy the image at the magnifier.
Let us use a hologram as an example. Suppose you can
project the image of a car in front of the street. The car
would reflect its own light acting like real object. However,
the beam from the hologram projector just creates the
car image at the street and then the beam just cuts itself
in empty space. Going back to the telescope. Suppose
the converging rays from the objective lens just end up
right there at the focal plane without any diverging rays
expanding again from the focal plane towards the
eyepiece. Can you still see the image at the focal plane?
I figure the image would produce its own light or wavefront
different from the original one from the objective lens, right?
(just like the hologram of the car producing its own light
and reflections unrelated to the original holographic projecting
beam)
Speaking of hologram. If there is not smoke or other
medium to produce the screen upon which to project
the hologram. Is there a principle wherein it is just
possible to project a hologram in empty space without
any screen material?? Like self-referencing wavefront or
those using so called scalar wave or higher order
electromagnetics?? Sorry for a bit off topic but it's the
ultimate application of knowledge of optics and I'd just
like to have an idea where the future would bring us.
lex
William Hamblen <william...@earthlink.net> wrote in message news:<lor6gv48t01b42pgm...@4ax.com>...
astrongc
It would keep me busy for days to come analyzing and visualizing them :)
lex
fish...@csolutions.net (PrisNo6) wrote in message news:<9865fa0b.0307...@posting.google.com>...
Chuck Simmons
>
> Hmm... even though the image at the focal plane doesn't
> reflect light because it is not a hologram. It produces its
> own light transmitting fan like beam in all directions coming
> from every part of the image, right?
>
> If this is what makes it focus to the magnifier like real object.
> Then what's the purpose of the diverging rays after passing
> thru the focal plane. If the diversing rays are not what
> produces the image in the magnifier (or eyepiece) but light
> from the image at the focal plane itself with slightly different
> angle, then isn't it that the diversing beam from the objective
> lens can only muddy the image at the magnifier.
>
> Let us use a hologram as an example.
Let us not. Holograms are unrelated to this and add enormous
complication.
> Suppose you can
> project the image of a car in front of the street. The car
> would reflect its own light acting like real object. However,
> the beam from the hologram projector just creates the
> car image at the street and then the beam just cuts itself
> in empty space.
You see? The hologram is a complicated critter and does not belong in a
geometrical optics discussion.
> Going back to the telescope. Suppose
> the converging rays from the objective lens just end up
> right there at the focal plane without any diverging rays
> expanding again from the focal plane towards the
> eyepiece. Can you still see the image at the focal plane?
> I figure the image would produce its own light or wavefront
> different from the original one from the objective lens, right?
> (just like the hologram of the car producing its own light
> and reflections unrelated to the original holographic projecting
> beam)
If you pick a point on the real distant object and trace many rays from
the object through the objective, the rays all converge at the
corresponding point on the real image and then diverge just as if they
were coming from a point on a real object located at the position of the
real image. Thus the real image produced by the objective is exactly
like a real object as far as geometrical optics is concerned.
> Speaking of hologram. If there is not smoke or other
> medium to produce the screen upon which to project
> the hologram. Is there a principle wherein it is just
> possible to project a hologram in empty space without
> any screen material?? Like self-referencing wavefront or
> those using so called scalar wave or higher order
> electromagnetics?? Sorry for a bit off topic but it's the
> ultimate application of knowledge of optics and I'd just
> like to have an idea where the future would bring us.
What? What screen? No screen is needed for a hologram. What is "higher
order electromagnetics?" You are grossly complicating a simple
geometrical optics problem by dragging in wave optics and interference.
Because the wavelength of light is very short, the telescope can be
understood very well using geometrical optics alone.
astrongc
> In the case of the real image, the fan may be incomplete. Trace two
> lines, one from the top of the objective, and one from the bottom,
> through the top of the real image. These two lines delimit the fan
> of rays that come from the real image. The extent to which this fan
> enters the eyepiece determines the illumination of that part of the
> image. If you do the same two lines, but passing through the center
> of the real image, you get the fan of light rays coming from there.
> Note that the two fans may intersect.
http://www.physics.cornell.edu/courses/p101-102/p102/14/java/convex/
Whew. It tooks me several days after consulting many web sites (and finding
the above url) that I finally understood what Brian meant above. So the fan
of light from the object in the focal plane is incomplete compare to the real
object. This means the focal plane image would be dimmer than that of a real
object assuming both are of the same size (and have the same virtual size).
Right? (Now if I'm wrong again it means I didn't understand it thoroughly
again).
Suppose I got it now. Is it possible to put something in the focal plane (an
exotic material or technology) so that the limited fan of light can be made
into full fan of light illuminating the entire eyepiece lens. This means using
a 4mm eyepiece at high magnification would produce a much brighter image with
exil pupil of 7mm or more instead of just 0.5mm?? (of course I know the rule
the exil pupil is smaller as the magnification increases but if a relay system
can be put in the focal plane then that rule can be broken and the next
generation of high tech eyepiece with more brightness than ordinary eyepiece is
possible (perhaps part of Markus promised Project Magic Super Eyepiece?)).
lex
Chuck Olson
> br...@isi.edu (Brian Tung) wrote in message news:<bdu1hd$ie1$1...@zot.isi.edu>...
>
> > In the case of the real image, the fan may be incomplete. Trace two
> > lines, one from the top of the objective, and one from the bottom,
> > through the top of the real image. These two lines delimit the fan
> > of rays that come from the real image. The extent to which this fan
> > enters the eyepiece determines the illumination of that part of the
> > image. If you do the same two lines, but passing through the center
> > of the real image, you get the fan of light rays coming from there.
> > Note that the two fans may intersect.
>
> http://www.physics.cornell.edu/courses/p101-102/p102/14/java/convex/
>
> Whew. It tooks me several days after consulting many web sites (and finding
> the above url) that I finally understood what Brian meant above. So the fan
> of light from the object in the focal plane is incomplete compare to the real
> object. This means the focal plane image would be dimmer than that of a real
> object assuming both are of the same size (and have the same virtual size).
> Right? (Now if I'm wrong again it means I didn't understand it thoroughly
> again).
>
> Suppose I got it now. Is it possible to put something in the focal plane (an
> exotic material or technology) so that the limited fan of light can be made
Sure. It's called an image converter, and is quite expensive.
Chuck
astrongc
> >
> > Suppose I got it now. Is it possible to put something in the focal plane (an
> > exotic material or technology) so that the limited fan of light can be made
> > into full fan of light illuminating the entire eyepiece lens?
>
> Sure. It's called an image converter, and is quite expensive.
>
> Chuck
Really? Is the image converter purely lens material or does it needs
a battery. If I'd put this in the focal plane of my eyepiece, then
my 5mm can increase its exil pupil to 3 times its size with much
greater brightness. How much is it and where is it available? Has
anyone here uses one?
lex
Chuck Olson
I'm going to leave those questions as an exercise for you,. the student, to answer using search
engines.
Chuck