surface brightness and photons

Some questions about a passage in the site:
http://skyandtelescope.com/howto/vis...ticle_78_5.asp
"When an image is magnified by high power, its surface brightness does
indeed grow weaker. But the total number of photons of light entering
the eye remains the same."

What does it mean "remains the same"?? Using higher power would reduce
the exit pupil, so the photons entering the eye should be less, not the
same, right??

Another one:

" (A photon is the fundamental particle of light.
Experiments show that most people can detect as few as 50 to 150
photons per second.) It doesn't really matter that these photons are
spread over a wider area; the retinal image-processing system will cope
with them."

What does it mean "spread over a wider area"?? The exit pupil is
smaller for higher magnification, so only the inner circle of the
retina surface can see them... so what does it mean "spread over a
wider area"??

Ch

Some questions about a passage in the site:
http://skyandtelescope.com/howto/vis...ticle_78_5.asp
"When an image is magnified by high power, its surface brightness does
indeed grow weaker. But the total number of photons of light entering
the eye remains the same."

What does it mean "remains the same"?? Using higher power would reduce
the exit pupil, so the photons entering the eye should be less, not the
same, right??

The exit pupil is smaller at higher power, but this merely means that at
that specific point in the optical train, the total FOV is physically
smaller, not that any light has been subtracted.

Another one:

" (A photon is the fundamental particle of light.
Experiments show that most people can detect as few as 50 to 150
photons per second.) It doesn't really matter that these photons are
spread over a wider area; the retinal image-processing system will cope
with them."

What does it mean "spread over a wider area"?? The exit pupil is
smaller for higher magnification, so only the inner circle of the
retina surface can see them... so what does it mean "spread over a
wider area"??

The exit pupil is smaller, but the image of any individual object (such as a
DSO) as focused on the retina is larger, so the light for that object is
"spread over a wider area."

You may need to draw it out for it to make sense. Unfortunately, ASCII is
not up to it. :-(

Hope this helps.

Chuck Taylor
Do you observe the moon?
Try http://groups.yahoo.com/group/lunar-observing/

Are you interested in understanding optics?
Try http://groups.yahoo.com/group/ATM_Optics_Software/

To reply, remove Delete and change period com to period net

wrote:
Some questions about a passage in the site:
http://skyandtelescope.com/howto/vis...ticle_78_5.asp
"When an image is magnified by high power, its surface brightness
does
indeed grow weaker. But the total number of photons of light entering
the eye remains the same."
snip and
It doesn't really matter that these photons are spread over a wider
area; the retinal image-processing system will cope with them."

Take an object you might be able to see in tonight's night sky - NGC
3242, a PN in Hydra, a.k.a. the "Ghost of Jupiter" nebula. It's about
26 arcseconds in diamater and is located at about 10 hours right
ascension.

Seds entry for NGC 3242
http://www.seds.org/~spider/spider/Misc/n3242.html
"Your Sky" finder chart for NGC 3242
http://www.fourmilab.to/cgi-bin/uncg...&z=1&deepm=8.6
Another NGC 3242 finder chart-
http://www.blackskies.com/pnweek4.htm

You take your 70mm refractor out and bring it up at 20x. The true field
size of the Ghost is about 26 arcseconds. In your apparent field of
view in your eyepiece it looks to be about 520 (20*26) arcseconds or
8.6 arcminutes in diameter.

Then you pop in a high magnification eyepiece; let's say 30x. The
apparent size of Ghost is now about 30*26 arcseconds, or about 780
arcseconds or 13 arcminutes. Because your'r viewing it at a higher
magnification, the image is dimmer. But, it's spread out over a wider
diameter on your fovea.
http://en.wikipedia.org/wiki/Fovea
http://en.wikipedia.org/wiki/Retina

The fovea is about 1.5mm across.
http://webvision.med.utah.edu/facts.html

Athough at higher magnification, the telescope projects the image of
NGC3242 spread-out over a wider circular area on your fovea, the
physics of the light coming from the Ghost of Jupiter hasn't changed.
NGC3242 is still sending out a specific number of photons per second
that are traveling down the objective lens of your refractor. That same
number of photons (which are about 1400 years old) is still spit out
the eyepiece. That same number of photons per second is still hitting
a smaller or larger area of your fovea, even though you increased or
decreased the telescope's magnification.

All of this happens inside the diameter of the exit-pupil (or
exit-pencil of light) that comes out of the eyepiece of the scope. As
you note, the size of the exit-pupil or exit pencil of light got
smaller when you exchanged the 20x eyepiece for the 30x eyepiece, but
the image of the Ghost is well within the sides of either exit pencil
of light.

The quoted text from the Sky & Telescope "how to observe" article means
that the rods and cones in the human fovae has enough dynamic range to
respond to an identical number and rate of photons spread out over a
larger area or a concentrated on a smaller area.

- Peace Canopus56

canopus56 wrote:


The quoted text from the Sky & Telescope "how to observe" article
means
that the rods and cones in the human fovae has enough dynamic range
to
respond to an identical number and rate of photons spread out over a
larger area or a concentrated on a smaller area.

- Peace Canopus56

Thanks for incredible explanations.

Last. The size of the Fovea is 1.5mm... so if the exit pupil is sized
at
0.5mm... only the inner 0.5mm of the Fovea can detect the light right??

If not... What's the significance of smaller or bigger diameter exit
pupil in fovea... like how can smaller diameter exit pupil still
illuminate the entire fovea/retina diameter range with no difference
compared to bigger exit pupil size.

ch.

The surface brightness does indeed NOT grow weaker unless some
catastrophic event takes place with the object itself. However, the
apparent brightness of the surface of the object does indeed grow
weaker the image is magnified.

wrote:
What's the significance of smaller or bigger diameter exit
pupil in fovea... like how can smaller diameter exit pupil still
illuminate the entire fovea/retina diameter range with no difference
compared to bigger exit pupil size.

Okay, so, you want to trace the light coming 1400 light-years away from
the Ghost Nebula the last 35mm. This is where the light travels from
the virtual image created by the telescope at the relief distance of
the eyepiece for another 35mm until it hits your retina.

For anybody counting, 1400 light years is
13,244,739,738,500,714,000,000 millmeters, or in our discussion
example, 13,244,739,738,500,714,000,0_35_ millimeters to chemstud's
retina. http://www.easysurf.cc/cnver15.htm#ly2k

Think of your retina as a digital camera chip, let's say a Meade Deep
Sky Imager, and your brain as the imaging software.

Acute vision is concentrated within the rod-rich fovea, but night
vision also occurs in the mixed rod and cone area of the retina, which
is much larger.
http://hyperphysics.phy-astr.gsu.edu...on/retina.html
http://webvision.med.utah.edu/imageswv/Ostergr.jpeg

The eye has a transverse (pupil to back of the eye) diameter of about
25mm and is about 32mm across laterally (left to right side).
http://webvision.med.utah.edu/anatomy.html

The retina covers about 72% of the backside of the interior globe of
the eye. The active area of the retina is about 1094 sq/mm. The 1.5mm
fovea sits within this active area of the retina. It is rich in
bright-light sensitive cones. The other detectors, the rods, are
dim-light sensitive. The peak concentration of the rods is in a band
around the fovea about 4-5mm in radius from the center of the fovea, or
a diameter of about 8mm to 10mm.
http://webvision.med.utah.edu/facts.html

So, let's estimate the active area of the retina at night as being at
most 16mm or 18mm in diameter across. ( 1094 sq/mm = (18.6mm)^2 * pi
). But the real work is being done in the inner 10mm diameter circle
containing the zone of peak cone concentration.

The distribution of bright-light cone-sensing and low-light
night-vision rods means that some of your telescopic viewing is done
using the other 16-18mm diameter of the active part of the retina
outside the 1.5mm fovea. This area outside the fovea is the source of
"averted" vision.

What does this mean for the pencil of parallel light coming out of the
exit pupil, which decreases in size and brightness as magnification
increases?

What we call "vision" is really a mental construction. Your eyeball is
rapidly scanning small portions of the view in front of you and your
brain uses memory to construct a mental picture of the physical world.
Think of brain as the software in a Meade DSI that assembles a final
picture from a series of shorter exposures.

The relaxed human eye has a focal length of about 17mm. It's focal
length can vary as your eye tries to focus on nearby or distant
objects.
http://hypertextbook.com/facts/2002/...retskaya.shtml

When you look in the eyepiece, you match this 17mm focal length of your
eye with the virtual image of the celestial object placed at the eye
relief distance behind the eyepiece. So the middle of your eye's simple
positive lens, the cornea, sits about 17mm behind the eye relief
distance of the eyepiece.

At one focal length or 17mm behind the middle of your cornea, the
telescopic virtual image of the celestial object sitting at the eye
relief distance of the eyepiece is refocused by your cornea _without
any magnification_ inside the back of your eye and on top of the 10mm
diameter of the "working" part of the retina at night.

The maximum size of the virtual image that can be transfered by the
cornea is limited by the eye's iris or pupil. The relaxed eye pupil
size is 6mm or 7mm in most people. So the maximum size of the circular
disk projected on your retina is 6 or 7mm. If the pupil is fully
dilated to 6mm, we are using about 36% of the working area of the
retina at night. ( (6^2) / 10^2 ).

At the lowest useable magnification of a telescope, the diameter of the
virtual image produced at the eye relief distance, what you call the
exit pupil size, is 6mm or 7mm. This is the maximum circular area of
light that the iris will allow to pass into the eye and that is
deposited on this 10mm diameter working area of the retina. The maximum
number of rods that can recieve a signal are triggered by this 6mm
diameter circle of light. The eyeball scans this image and the mind
forms a picture in your mind. Now you see the Ghost of Jupiter nebula.

Because the maximum number of rods in your retina are firing, it looks
it's brightest.

As magnification increases, the exit pupil diameter decreases and the
relative brightness of the light in the exit pupil pencil of light
descreases. The result is that less rods and cones are triggered on
the retina and fovea. The eyeball scans this lower level of light, less
neurons are triggered, but the "software" of the mind can compensate
and still build up a picture.

You see the Ghost of Jupiter nebula, but because fewer low-light
sensing rods and bright-light sensing cones are firing, it looks
dimmer.

As this dimming process continues, the cones no longer have enough
bright light to trigger and eye sees with "averted vision" in the rods.


Eventually, at some point (usually where the exit pencil falls below
about 0.3mm in size), the number of photons in the dim exit pupil beam
is too small, only a few photons reach the retina, but too few rods are
triggered in the retina. The mind does not get enough of a signal to
form a mental picture.

The diameter of the exit pupil, or the diameter of the virtual image
inside the exit pupil diameter, isn't controlling; it's the number and
rate of photons that is important.

The eyeball is still scanning the exit pupil virtual image projected
onto the retina. Even though it may be only 0.3mm in diameter and is
smaller than even the 1.5mm diameter of the fova, the eyes runs this
small virtual image across all its detectors in the 10mm working area
of the night-time retina.

And there are alot of those detectors in the eye - 110,000,000 to
125,000,000 rods and 6,400,000 cones. This compares to a 3.3 Megapixel
digital camera chip, that produces generates an image with 2048 x 1536
pixels.
http://webvision.med.utah.edu/facts.html (rods and cones count)
http://www.apogeephoto.com/sept2001/...ms092001.shtml (chip)

But in the 0.3mm exit pupil beam, there simply are not enough energtic
photons in its dim light to trigger a chemical reaction in the
low-light seeing rods of the retina.

You can confirm that the diameter of the extended object or exit pupil
is not controlling with a simple test. Look through a telescope field
of view using moderate magnification at a single bright star - a point
source - in an otherwise blank field of view. A star is a point object,
not an extended object. It has no effective diameter (other than the
apparent diameter of the Airy disk). But you can still see it because
the number and rate of photons coming from it are sufficient to trigger
receptors in your eye.

For telescopes and extended objects, this experience of descreasing
exit pupil diameter and decreasing brightness is captured in Knisley's
"useful magnification" table, which expresses the size of the exit
pupil on a uniform scale of inches of aperature per applied power.
Knisely's table can be found by using usenet search engines on this
usenet group for the "Knisely useful magnification ranges visual
observing astronomical telescopes" and his Dec. 30, 2002 post in the
thread "Maximum magnification" or his May 14, 2004 post in the thread
"Eyepiece advice, again." In an abbreviated form, the table is:

=================
Knisely's Useful Magnifications

LOW POWER (3.7 to 9.9x per inch of aperture)(6.9mm to 2.6mm exit
pupil)
MEDIUM POWER (10x to 17.9x per inch of aperture)(2.5mm to 1.4mm exit
pupil)
HIGH POWER (18x to 29.9x per inch of aperture)(1.4mm to 0.8mm exit
pupil)
VERY HIGH POWER (30x to 41.9x per inch of aperture)(0.8mm to 0.6mm exit
pupil)
EXTREME POWER (42x to 75x per inch)(0.6mm to 0.3mm exit pupil)
EMPTY MAGNIFICATION (100x per inch and above)

From D. Knisely, 5/14/2004 sci.astro.amateur Usenet post, Thread
"Eyepiece advice, again".
=================

Another interesting effect occurs as the exit pupil reaches the 0.3mm
diameter. The beam becomes so small, that, IMHO, it illuminates
optical defects in your eye's lens, the cornea.

You can test your eyes' corneas for roughness defects at your kitchen
table using a method suggested in Suiter's 1994 classic, _Star Testing
Astronomical Telescopes_ (Willman-Bell), (on p. 240). I have tried this
test, it works and it is a entertaining astronomical diversion for a
rainy night. Suiter described the test as follows -

============
"Your eye also suffers from medium scale roughness. Take aluminum foil
and perforate it with a pin. Hold the foil about 8 to 15 cm in front
of your eye and look through the pinhole at a frosted incandescent
light bulb. Try to focus your eye on the lamp, not the pinhole, and
cover the other eye. If you have punch the right size hole in the
foil, you should see a mottled disk that roughly approximates the
out-of-focus patterns seen in this book. . . . . The appearance may be
cleared up slighty by placing a colored [telescope lens] filter between
the lamp and the pinhole.
As you blink, horizontal lines appear briefly on the defocused disk. .
.. . [Y]ou may also see some dim radial spikes outside the disk. These
spikes may be caused by diffraction . . . or streaks in the roughness.

The roughness is visible as coarseness in the expanded field. This
coarseness does not vary from blink to blink. . . . .

The human eye is not even close to diffraction-limited. An eye with a
3-mm iris opening . . . can theoretically resolve lines separated by
0.6 arcminutes, but a person who resolves lines only 1 arcminute apart
is deemed to have excellent vision."
============

Making a good small pinhole in foil takes some practice. You may have
to try several tries before making a suitable small hole. The idea is
to just punch the tip of the pin through the other side of the foil in
order to make the smallest hole possible.

When you look through an amateur telescope at a bright object like the
Moon using the highest possible maximum magnification, the exit pupil
size becomes so small that you will see an image with a similar
"mottled disk" appearance as seen in Suiter's aluminum foil and light
bulb test. IMHO, it is caused by the exit pupil becoming so small that
it is acting like the pinhole in the foil. The dot of light reveals
roughness defects in your cornea.

In summary, the reduced size of the exit pupil diameter does not
control whether some kind of image image can be seen. The point is
that there has to be enough photons in the exit pupil diameter to
trigger a chemical reaction in the cones of your retina.

- Enjoy Canopus56

P.S. - When replying, please remember to delete most of the quote of
this post text, so other users do not have to scroll through a repost
of the long response.

canopus56 wrote:

Thanks for the indepth explanations. Bush must name the next
Hubble Telescope after you

At the lowest useable magnification of a telescope, the diameter of
the
virtual image produced at the eye relief distance, what you call the
exit pupil size, is 6mm or 7mm. This is the maximum circular area of
light that the iris will allow to pass into the eye and that is
deposited on this 10mm diameter working area of the retina. The
maximum
number of rods that can recieve a signal are triggered by this 6mm
diameter circle of light. The eyeball scans this image and the mind
forms a picture in your mind. Now you see the Ghost of Jupiter
nebula.


But if the exit pupil is 1mm... it should still register on
the same 10mm working area of the retina similar to that done
by an exit pupil of 6-7mm, right?? because the exit pupil only
determines the brightness of the point that will reach a
certain rod/cone or pixel detector of the retina. Agreed?

ch

wrote:
canopus56 wrote:
But if the exit pupil is 1mm... it should still register on
the same 10mm working area of the retina similar to that done
by an exit pupil of 6-7mm, right?? because the exit pupil only
determines the brightness of the point that will reach a
certain rod/cone or pixel detector of the retina. Agreed?

I feel we are close to the same place.

As magnification increases, the exit pupil beam gets relatively smaller
_and_ dimmer (that is relative to the high magnification exit pupil
beam). So the dimmer high magnification 1mm diameter exit pupil only
makes a 1mm diameter circle of light on the 10mm diameter working area
of retina. The eye's lens does not magnify the 1mm diameter exit pupil
into a 10mm circle.

At low magnification, the exit pupil beam gets relatively bigger _and_
brighter (that is relative to the low magnification exit pupil beam).
The brighter 7mm diameter exit pupil at low magnification makes a 7mm
diameter circle on the 10mm diameter working area of retina. The light
in the 7mm circle is also brighter. The eye's lens does not magnify the
7mm diameter exit pupil into a 10mm circle.

But whether any rods fire in response to the 1mm circle or the 7mm
circle of light depends, in physics speak, on the luminous flux
contained within the boundaries of the exit pupil's beam. The flux is
more important than the diameter of the beam itself. If the flux is not
high enough; a rods will not fire and send a message to the brain.
http://en.wikipedia.org/wiki/Flux
http://en.wikipedia.org/wiki/Lumen_%28unit%29

Agreed.

Anyone else have any comments? Any wrong ideas in this description?

- Canopus56

canopus56 wrote:
wrote:
canopus56 wrote:
But if the exit pupil is 1mm... it should still register on
the same 10mm working area of the retina similar to that done
by an exit pupil of 6-7mm, right?? because the exit pupil only
determines the brightness of the point that will reach a
certain rod/cone or pixel detector of the retina. Agreed?

I feel we are close to the same place.

As magnification increases, the exit pupil beam gets relatively
smaller
_and_ dimmer (that is relative to the high magnification exit pupil
beam). So the dimmer high magnification 1mm diameter exit pupil only
makes a 1mm diameter circle of light on the 10mm diameter working
area
of retina. The eye's lens does not magnify the 1mm diameter exit
pupil
into a 10mm circle.

At low magnification, the exit pupil beam gets relatively bigger
_and_
brighter (that is relative to the low magnification exit pupil beam).
The brighter 7mm diameter exit pupil at low magnification makes a 7mm
diameter circle on the 10mm diameter working area of retina. The
light
in the 7mm circle is also brighter. The eye's lens does not magnify
the
7mm diameter exit pupil into a 10mm circle.

But whether any rods fire in response to the 1mm circle or the 7mm
circle of light depends, in physics speak, on the luminous flux
contained within the boundaries of the exit pupil's beam. The flux is
more important than the diameter of the beam itself. If the flux is
not
high enough; a rods will not fire and send a message to the brain.
http://en.wikipedia.org/wiki/Flux
http://en.wikipedia.org/wiki/Lumen_%28unit%29

Agreed.

Anyone else have any comments? Any wrong ideas in this description?

- Canopus56

Wait. But the eye is like an objective lens. When parallel beam
hit it, it converges into a point. Likewise, the parallel beam
from the exit pupil converges into a point at the retina. So what
do you mean the 7mm exit pupil will make a 7mm circle at the
10mm working area of the retina?? I think the 7mm exit pupil
should converge into a point in the retina. Not??

Another. I think both the 2 mm and 7mm exit pupil should register
on the entire 10mm working area of the retina. Here's why I think.
Imagine looking at a 15mm plossl, then adding a barlow, the
exit pupil will decrease by half. But you are still seeing the
entire image of the lens. So the parallel beam from the exit
pupil converges into a point in the retina and the bigger
exit pupil just gives a brighter image since there is strong
point of converging light at the retina.

What do you think?

ch.

wrote:
Wait. But the eye is like an objective lens. When parallel beam
hit it, it converges into a point. Likewise, the parallel beam
from the exit pupil converges into a point at the retina. So what
do you mean the 7mm exit pupil will make a 7mm circle at the
10mm working area of the retina?? I think the 7mm exit pupil
should converge into a point in the retina. Not??

The image of extended objects are not formed with a single bundle of
parallel rays coming down the telescope tube.

Extended objects, by definition, have an arcsecond size. For an
extended object like the Moon, some of the bundles of parallel rays
come from center of the object, as you visualized in your post -

http://members.csolutions.net/fisher...Telescope1.gif

- and some come from one side of the object at a slight angle of
divergence to the optical axis -

http://members.csolutions.net/fisher...Telescope3.gif

- and some come from the other side, also with a slight angle of
divergence -

http://members.csolutions.net/fisher...Telescope4.gif

At the intersection where these three bundles of parallel light meet on
the observer side of the eyepiece -

http://members.csolutions.net/fisher...Telescope5.gif

- is where the virtual image of the extended object forms -

http://members.csolutions.net/fisher...Telescope7.gif

The distance between this virtual image and the eyepiece is the
eyepiece's eye relief distance.

I recommend that you take a few minutes to play with a telescope ray
tracing Javascript applet, put on the web by Professor Mark Peterson of
Mount Holyoke College -

http://www.mtholyoke.edu/~mpeterso/c...twolenses.html

Using this ray tracing simulater, you can put three bundles of parallel
light through the telescope and angle two of them with respect optical
axis. Put one on the optical axis, a second parallel to the top of the
lens and a third parallel to the bottom of the lens. Do this by -

1) Selecting the "astrnomical telescope" link to put a telescope in the
simulator.

2) Use the "Beam" button, to add two beams.

3) Once selected, there are drag "dots" on the beams that allow you to
angle them.

4) After you are practiced at using the simulater add a virtual eye and
retina using the "add an 'eye' at the far right" link below the
simulator window. (I found easier to create the desired simulation by
manually adding another lens at the far right using the "Lens" button
and a field stop using the "Aperature" button.)

As you note, the parallel bundle of rays that comes into the telescope
directly parallel to and centered down the middle of the objective's
optical axis do end up focusing to a point on the retina. The bundle
of parallel rays that fill up the entire objective lens and that are
parallel to the optical axis become what we have been calling the "exit
pupil" beam.

But the exit pupil bundle of parallel light rays are only one of the
many bundles of rays of parallel light that create the image on the
retina.

Many other bundles are at what is called a slight "angle of divergence"
from the optical axis of the scope. They travel through a disk area on
the optical axis of the objective, called paraxial disk. These focus
elsewhere on the plane of the retina. Together, many of these bundles
create the entire image - which is in the shape of a disk, not a point.


You can see this effect by focusing your telescope on the bright full
Moon and filling the eyepiece with Moon's disk. You can step back about
6-10 inches from the eyepiece and still see a dim image of the Moon.
IMHO (and I may wrong about this) the dim image of the Moon at a
distance form the eyepiece is the exit pupil bundle of light travelling
parallel out of the eyepiece. This image is dim because it is not all
the light that makes up the extended image in a telescope. Now move
your eye up to its normal eye relief viewing position. The image will
be several magnitudes brigther. This is because the other bundles of
light that are at an angle of divergence from the optical axis are now
entering your eyepupil and are reaching your retina.

- Peace - Canopus56

P.S. - If it is any consolation, I did the same brain fart when I first
starting to study how light travels through a refractor. This is
because most diagrams in beginning astronomy texts, for some
inexplicable reason, do not show all three bundles of light. They just
show one bundle traveling parallel to the central axis. Probably they
do this to make a clearer illustration of Galileo's M = D_o /
D_exit_pupil = D_fl / EP_fl relationship.