Multiple pinhole cameras for extrasolar planet detection.

Dear mmeron:

wrote in message
...
....
Yes. Nothing of significance changed here. And,
additional complexity has been introduced since
everyting reflects to some extent at glancing
angles and glancing angles is what you're going to
get inside the tube. And I see little reason for this.
Keep in mind that a telescop is effectively a pinhole,
with large light gathering capacity added. I do not
see the point of cripling it by reducing the light
gathering capacity.

May or may not be related to the topic under discussion...
http://www.sciencedaily.com/releases...1002154230.htm

David A. Smith

In article , "N:dlzc D:aol T:com \(dlzc\)" writes:
Dear mmeron:

wrote in message
...
...
Yes. Nothing of significance changed here. And,
additional complexity has been introduced since
everyting reflects to some extent at glancing
angles and glancing angles is what you're going to
get inside the tube. And I see little reason for this.
Keep in mind that a telescop is effectively a pinhole,
with large light gathering capacity added. I do not
see the point of cripling it by reducing the light
gathering capacity.

May or may not be related to the topic under discussion...
http://www.sciencedaily.com/releases...1002154230.htm

It is related to techniques, but not to principles. Techniques deal
with sqeezing the most out of the information available. Principles
set the limits on what may be available.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"

wrote:
In article .com, "Robert Clark" writes:
wrote:
....

Yes. I was informed via email that a single small pinhole would be
limited in resolution by the Rayleigh criterion for the diffraction
limit:

Angular resolution.
http://en.wikipedia.org/wiki/Angular...on#Explanation

The formula given is in terms of the sine of the angle between the two
bodies. But for large distances this is about the same as the ratio of
the distance between the bodies to the distance to the system. So for a
distance of 10 light-years to the system, 100 trillion km, a 10 million
km distance from the planet to the star and and an optical wavelength
of 500 nm, the formula results in a diameter of about 6 meters to
resolve the two bodies.

Yes, about this. Note that this is not a "small pinhole" anymore.
Note also that the Rayleigh criterion, as presented in rudimentary
sources, is a great oversimplification as it implicitly assumes two
objects of comparable brightness. Thus ...

Then optical scopes already available would be
able to resolve them if that were the only problem.
However, the main problem in detection actually is removing the glare
produced by the star that washes out the light from the planet. The
star could be perhaps a billion times brighter than the planet.

Yep. And that's where the problem with a simple minded use of
Raleyigh's criterion enters. The criterion takes two objects as
resolved once the center of the diffraction pattern of one falls out
of the Airy disk (the zero order peak) of the other. For objects of
comparable brightness that's fine. When there is enormoud brightness
disparity (as in the case above, one has to consider that the higher
diffraction orders of the brighter source may still be much brighter
than the zeroeth order of the dimmer one. That's the glare you get,
even if all else is perfect.

The main objective in future planet detection scopes is actually to mask
out the light from the star to allow the planet to be detected, such as
by nulling interferometers or by occulting disks.



I may I need to correct my diagram. Fig. 1a in the report "Optimum
Pinhole Camera Design",
http://www.huecandela.com/hue-x/pin-...%20Wellman.pdf , shows
the geometrical blur is produced by the light rays hitting both edges
of the aperture and being continued on in straight-lines. Then the
diffraction blur is produced on both sides of the geometrical blur disk
due to the light bending on each edge of the aperture.
In the diagram below, the length C is still the distance along the
imaging screen to the image of the star. But the geometrical blur disk
is outside this distance. So the condition for the light from the star
including the diffracted light to hit the sides of the tube is
1.22λf/D C-D or D + 1.22λf/D C. The length C is still C =
f*(a/s) by similar triangles so the inequality becomes D + 1.22λf/D
f*(a/s) .
As you said though this only includes the central Airy disk:

Airy disc.
http://en.wikipedia.org/wiki/Airy_pattern

For a star a billion times brighter than the planet we will need to
block also several of the diffraction rings beyond the central disk.
The "Airy disc" page gives the formula for the intensity for the rings
for a circular aperture.
It turns out it decreases quite rapidly with distance from the center.
However, the intensity decreases even faster with a rectangular
apertu

Diffraction & Resolution.
http://vision.berkeley.edu/roordalab...Resolution.pdf

See the formulas on p. 43. The formulas are easier to calculate as
well than for the circular aperature. The intensity I is the square of
the amplitude E formula given for a rectangular aperture. If we take
the rectangle to be equal sided for simplicity we see the maximum
intensity of the diffraction ring will decrease by a factor of k^4 if
the distance from the center is increased by a factor of k.
So if the diffraction rings blocked were 33 times further out than
the central disk, the intensity would be even smaller by an additional
factor of a million. The length f of the tubes would have to be
increased by this factor as well however.
It probably is also the case that there are apertures for which the
intensity decreases at an even faster rate than for rectangular
apertures.
Note that it may be the need for the long distances for f may be
obviated if it is indeed the case the light from the separate apertures
can be combined interferometrically. In that case the size of the
diffraction rings would correspond to the size of single large aperture
and the length f would be proportional smaller.
It is certainly the case that the light can be combined if the tubes
were not present since this is how multiple mirror telescopes work. The
question is would the blocking of the light by the tubes prevent the
superposition needed to form the image as from a single large aperture
scope.
It is true that the tubes would block some of the light from the
planet as well. But this is true for the occulting disk method for
explanet detection also:

Question: Why does the occulting screen need to be so far away?
http://www-int.stsci.edu/~jordan/umb..._far_away.html


Bob Clark


S P
*\__a__*
\\ |
\ \ |s
\ \|
\ |\
\ | \
\| \
|__D___\
|\ | \
| \ | \
| \ | \
| \ | \
| \ | \
| \| \
f| | \
| |\ \
| | \ Image of S \
| | \ | \
| | \ v \
|______|_|__\+_________________\
C ^ Bg = D
|
|
Bd/2 = 1.22λf/D

In article .com, "Robert Clark" writes:
wrote:
In article .com, "Robe=
rt Clark" writes:
wrote:
.=2E.

Yes. I was informed via email that a single small pinhole would be
limited in resolution by the Rayleigh criterion for the diffraction
limit:

Angular resolution.
http://en.wikipedia.org/wiki/Angular...on#Explanation

The formula given is in terms of the sine of the angle between the two
bodies. But for large distances this is about the same as the ratio of
the distance between the bodies to the distance to the system. So for a
distance of 10 light-years to the system, 100 trillion km, a 10 million
km distance from the planet to the star and and an optical wavelength
of 500 nm, the formula results in a diameter of about 6 meters to
resolve the two bodies.

Yes, about this. Note that this is not a "small pinhole" anymore.
Note also that the Rayleigh criterion, as presented in rudimentary
sources, is a great oversimplification as it implicitly assumes two
objects of comparable brightness. Thus ...

Then optical scopes already available would be
able to resolve them if that were the only problem.
However, the main problem in detection actually is removing the glare
produced by the star that washes out the light from the planet. The
star could be perhaps a billion times brighter than the planet.

Yep. And that's where the problem with a simple minded use of
Raleyigh's criterion enters. The criterion takes two objects as
resolved once the center of the diffraction pattern of one falls out
of the Airy disk (the zero order peak) of the other. For objects of
comparable brightness that's fine. When there is enormoud brightness
disparity (as in the case above, one has to consider that the higher
diffraction orders of the brighter source may still be much brighter
than the zeroeth order of the dimmer one. That's the glare you get,
even if all else is perfect.

The main objective in future planet detection scopes is actually to mask
out the light from the star to allow the planet to be detected, such as
by nulling interferometers or by occulting disks.



I may I need to correct my diagram. Fig. 1a in the report "Optimum
Pinhole Camera Design",
http://www.huecandela.com/hue-x/pin-...%20Wellman.pdf , shows
the geometrical blur is produced by the light rays hitting both edges
of the aperture and being continued on in straight-lines. Then the
diffraction blur is produced on both sides of the geometrical blur disk
due to the light bending on each edge of the aperture.
In the diagram below, the length C is still the distance along the
imaging screen to the image of the star. But the geometrical blur disk
is outside this distance. So the condition for the light from the star
including the diffracted light to hit the sides of the tube is
1=2E22=CE=BBf/D C-D or D + 1.22=CE=BBf/D C. The length C is still C =
=3D
f*(a/s) by similar triangles so the inequality becomes D + 1.22=CE=BBf/D
f*(a/s) .
As you said though this only includes the central Airy disk:

Airy disc.
http://en.wikipedia.org/wiki/Airy_pattern

For a star a billion times brighter than the planet we will need to
block also several of the diffraction rings beyond the central disk.
The "Airy disc" page gives the formula for the intensity for the rings
for a circular aperture.
It turns out it decreases quite rapidly with distance from the center.

Unfortunately, not that rapidly. The intensity of a ring is
proportional to 1/n^3 where n is the ring number. Takes lits of rings
to drop by a factor of a billion.

However, the intensity decreases even faster with a rectangular
apertu

Somewhat faster but still algebraic drop.

Diffraction & Resolution.
http://vision.berkeley.edu/roordalab...s/Diffraction=
&Resolution.pdf

See the formulas on p. 43. The formulas are easier to calculate as
well than for the circular aperature. The intensity I is the square of
the amplitude E formula given for a rectangular aperture. If we take
the rectangle to be equal sided for simplicity we see the maximum
intensity of the diffraction ring will decrease by a factor of k^4 if
the distance from the center is increased by a factor of k.
So if the diffraction rings blocked were 33 times further out than
the central disk, the intensity would be even smaller by an additional
factor of a million. The length f of the tubes would have to be
increased by this factor as well however.
It probably is also the case that there are apertures for which the
intensity decreases at an even faster rate than for rectangular
apertures.
Note that it may be the need for the long distances for f may be
obviated if it is indeed the case the light from the separate apertures
can be combined interferometrically. In that case the size of the
diffraction rings would correspond to the size of single large aperture
and the length f would be proportional smaller.
It is certainly the case that the light can be combined if the tubes
were not present since this is how multiple mirror telescopes work. The
question is would the blocking of the light by the tubes prevent the
superposition needed to form the image as from a single large aperture
scope.
It is true that the tubes would block some of the light from the
planet as well. But this is true for the occulting disk method for
explanet detection also:

Question: Why does the occulting screen need to be so far away?
http://www-int.stsci.edu/~jordan/umb..._far_away.html


Bob Clark


S P
*\__a__*
\\ |
\ \ |s
\ \|
\ |\
\ | \
\| \
|__D___\
|\ | \
| \ | \
| \ | \
| \ | \
| \ | \
| \| \
f| | \
| |\ \
| | \ Image of S \
| | \ | \
| | \ v \
|______|_|__\+_________________\
C ^ Bg =3D D
|
|
Bd/2 =3D 1.22=CE=BBf/D


Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"

You will need to use a fixed-width font such as Courier in your news
reader to see the diagrams properly.
Alternatively, you can see the diagram he

sci.astro Multiple pinhole cameras for extrasolar planet detection.
http://groups.google.com/group/sci.a...e=source&hl=en

Unfortunately, Google puts some extraneous characters in in
interpreting the equations. But the diagram is readable enough.


Bob Clark

Robert Clark wrote:
wrote:
In article .com, "Robert Clark" writes:
wrote:
...

Yes. I was informed via email that a single small pinhole would be
limited in resolution by the Rayleigh criterion for the diffraction
limit:

Angular resolution.
http://en.wikipedia.org/wiki/Angular...on#Explanation

The formula given is in terms of the sine of the angle between the two
bodies. But for large distances this is about the same as the ratio of
the distance between the bodies to the distance to the system. So for a
distance of 10 light-years to the system, 100 trillion km, a 10 million
km distance from the planet to the star and and an optical wavelength
of 500 nm, the formula results in a diameter of about 6 meters to
resolve the two bodies.

Yes, about this. Note that this is not a "small pinhole" anymore.
Note also that the Rayleigh criterion, as presented in rudimentary
sources, is a great oversimplification as it implicitly assumes two
objects of comparable brightness. Thus ...

Then optical scopes already available would be
able to resolve them if that were the only problem.
However, the main problem in detection actually is removing the glare
produced by the star that washes out the light from the planet. The
star could be perhaps a billion times brighter than the planet.

Yep. And that's where the problem with a simple minded use of
Raleyigh's criterion enters. The criterion takes two objects as
resolved once the center of the diffraction pattern of one falls out
of the Airy disk (the zero order peak) of the other. For objects of
comparable brightness that's fine. When there is enormoud brightness
disparity (as in the case above, one has to consider that the higher
diffraction orders of the brighter source may still be much brighter
than the zeroeth order of the dimmer one. That's the glare you get,
even if all else is perfect.

The main objective in future planet detection scopes is actually to mask
out the light from the star to allow the planet to be detected, such as
by nulling interferometers or by occulting disks.



I may I need to correct my diagram. Fig. 1a in the report "Optimum
Pinhole Camera Design",
http://www.huecandela.com/hue-x/pin-...%20Wellman.pdf , shows
the geometrical blur is produced by the light rays hitting both edges
of the aperture and being continued on in straight-lines. Then the
diffraction blur is produced on both sides of the geometrical blur disk
due to the light bending on each edge of the aperture.
In the diagram below, the length C is still the distance along the
imaging screen to the image of the star. But the geometrical blur disk
is outside this distance. So the condition for the light from the star
including the diffracted light to hit the sides of the tube is
1.22λf/D C-D or D + 1.22λf/D C. The length C is still C =
f*(a/s) by similar triangles so the inequality becomes D + 1.22λf/D
f*(a/s) .
As you said though this only includes the central Airy disk:

Airy disc.
http://en.wikipedia.org/wiki/Airy_pattern

For a star a billion times brighter than the planet we will need to
block also several of the diffraction rings beyond the central disk.
The "Airy disc" page gives the formula for the intensity for the rings
for a circular aperture.
It turns out it decreases quite rapidly with distance from the center.
However, the intensity decreases even faster with a rectangular
apertu

Diffraction & Resolution.
http://vision.berkeley.edu/roordalab...Resolution.pdf

See the formulas on p. 43. The formulas are easier to calculate as
well than for the circular aperature. The intensity I is the square of
the amplitude E formula given for a rectangular aperture. If we take
the rectangle to be equal sided for simplicity we see the maximum
intensity of the diffraction ring will decrease by a factor of k^4 if
the distance from the center is increased by a factor of k.
So if the diffraction rings blocked were 33 times further out than
the central disk, the intensity would be even smaller by an additional
factor of a million. The length f of the tubes would have to be
increased by this factor as well however.
It probably is also the case that there are apertures for which the
intensity decreases at an even faster rate than for rectangular
apertures.
Note that it may be the need for the long distances for f may be
obviated if it is indeed the case the light from the separate apertures
can be combined interferometrically. In that case the size of the
diffraction rings would correspond to the size of single large aperture
and the length f would be proportional smaller.
It is certainly the case that the light can be combined if the tubes
were not present since this is how multiple mirror telescopes work. The
question is would the blocking of the light by the tubes prevent the
superposition needed to form the image as from a single large aperture
scope.
It is true that the tubes would block some of the light from the
planet as well. But this is true for the occulting disk method for
explanet detection also:

Question: Why does the occulting screen need to be so far away?
http://www-int.stsci.edu/~jordan/umb..._far_away.html


Bob Clark


S P
*\__a__*
\\ |
\ \ |s
\ \|
\ |\
\ | \
\| \
|__D___\
|\ | \
| \ | \
| \ | \
| \ | \
| \ | \
| \| \
f| | \
| |\ \
| | \ Image of S \
| | \ | \
| | \ v \
|______|_|__\+_________________\
C ^ Bg = D
|
|
Bd/2 = 1.22λf/D

George Dishman wrote:
wrote in message
...
In article . com, "Robert
Clark" writes:
I was looking up references to pinhole cameras when I came upon this
page that gives the size of the image compared to the size of the
source:

Finding the Size of the Sun and Moon.
http://cse.ssl.berkeley.edu/AtHomeAs...tivity_03.html

The page explains that the ratio of the size of the source to the
distance of the source to the pinhole is the same as the ratio of the
size of the image to the distance between the pinhole and the imaging
screen.
Then couldn't this be used to resolve two far away point sources of
light that are at a very small angular distance from each other?

Diffraction limits.

http://en.wikipedia.org/wiki/New_Worlds_Imager

Cash's proposal was presented at the October
meeting this year:

http://www.niac.usra.edu/library/mee...ual/oct06.html

The presentation:

http://www.niac.usra.edu/files/libra...6/1200Cash.pdf

George

The latest version is an occulter rather than a pinhole and could fly
as soon as 2013 to be used with the James Webb Space Telescope.
Webster Cash gives a nice description of it in the video on this TV
news site:

CU Researcher Invents Star Shade To Study Planets.
Shaun Boyd
Reporting
"(CBS4) BOULDER, Colo. A researcher at the University of Colorado in
Boulder is getting international attention for his invention that helps
scientists find never-before-seen planets and map their make-up."
....
"If we can find planets similar to Earth and actually chemically
analyze what's going on in their atmospheres, we may be able to for the
first time take a look at alien life," Dr. Cash said.
"The hope is to build and launch the star shade within the next
decade."
http://cbs4denver.com/topstories/loc...186222050.html

Another link on it:

Starshade brings fresh hope in search for alien life
By Roger Highfield, Science Editor
Last Updated: 12:55am BST 06/07/2006
http://www.telegraph.co.uk/news/main...6/wspace06.xml


Bob Clark

Robert Clark wrote:
I was looking up references to pinhole cameras when I came upon this
page that gives the size of the image compared to the size of the
source:

Finding the Size of the Sun and Moon.
http://cse.ssl.berkeley.edu/AtHomeAs...tivity_03.html

The page explains that the ratio of the size of the source to the
distance of the source to the pinhole is the same as the ratio of the
size of the image to the distance between the pinhole and the imaging
screen.
Then couldn't this be used to resolve two far away point sources of
light that are at a very small angular distance from each other?


There is some research at using very many time pinhole cameras to
form large space scopes. They get around the diffraction limits of the
small apertures by using the pinholes together to make a single large
aperture.
The advantage apparently is the savings in weight:

Photon sieve telescope: imaging with 10 million pinholes.
Geoff Andersen
"Telescope designers are developing a diffractive optic with millions
of holes for future large space-based telescopes."
http://newsroom.spie.org/x4189.xml?highlight=x541

October 22, 2006
Telescope idea could transform spy work.
http://www.gazette.com/display.php?id=1325576&secid=1


Bob Clark

Robert Clark wrote:
There is some research at using very many time pinhole cameras to
form large space scopes. They get around the diffraction limits of the
small apertures by using the pinholes together to make a single large
aperture.

"very many time pinhole cameras"?

That should be "very many tiny pinhole cameras".


Bob Clark