High data rate space transmissions through visible lightcommunication.

On Jun 29, 2:10*am, "Chris" wrote:
Been there done that.

The data rate depends on noise and the signalling rate. With optical
signaling the signalling rate can be gigahertz but in the presence of noise
the data rate goes down due to the need of retransmission due to missing
bytes.

To reduce noise you increase the signal to noise ratio. You can receive
information with low data rate when the signal is below noise level. You
just use a very narrow bandwidth and a low signalling rate.

To increase the signal to noise ratio you have to concentrate the energy
into a beam intercepting the receiver.

And the receiver needs a similar beam antenna intercepting the transmitter.

On Jun 28, 8:31*pm, Robert Clark wrote:
On Jun 28, 1:41 pm, Richard Clark wrote:
On Sun, 28 Jun 2009 17:14:14 +1000, "Peter Webb"
wrote:
Furthermore, the radiation from a reflected area is isotropic - goes in all
directions - and hence very little is directed towards the earth.

Actually, it is lambertian in its distribution, and it would have a
major lobe that was directed in rather typical fashion (at the same,
but negative angle to the norm to the surface). *However, as is the
intent of your response, very much less will find its way to the
intended target.
Richard Clark, KB7QHC

*This describes the reflection from the Iridium antennas as specular
where most of the reflected light is concentrated in a single
direction:
SeeSat-L Apr-98: Method for predicting flare.http://satobs.org/seesat/Apr-1998/0175.html
*About specular reflection:
Specular reflection.http://en.wikipedia.org/wiki/Specular_reflection
*We could get even higher concentration of the image by using
parabolic mirror reflectors.


We might want to test this out in Earth orbit first before sending
to the Moon. There are low cost CubeSats only 10 cm on a side that can
be launched to orbit at low cost:

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

Say we made the reflective surface on the CubeSat be a square 10 cm
wide. This is .01 m^2 in area, so a factor of 6.25x10^(-3) smaller
than the area of the Iridium antennas. This would result in the
apparent magnitude being dimmed to -2.5.
This is still quite bright and would be brighter than the star
Sirius, the brightest star in the sky. Hmmm. I'm wondering about that.
The sunlight at the Earth's distance amounts to about 1,000 watts/m^2.
So even if this 10 cm wide reflector in space reflected all the light
that fell on it to the Earth's surface it would still be only 10
watts. So this is saying a 10 watt light source at a distance of 780
km would be as bright as a -2.5 magnitude star.
Is this right even if you put behind it say a parabolic reflector as
with a flashlight so all the light was directed forward?


Bob Clark

On Jun 29, 7:55*pm, Robert Clark wrote:
...
* We might want to test this out in Earth orbit first before sending
to the Moon. There are low cost CubeSats only 10 cm on a side that can
be launched to orbit at low cost:

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

*Say we made the reflective surface on the CubeSat be a square 10 cm
wide. This is .01 m^2 in area, so a factor of 6.25x10^(-3) smaller
than the area of the Iridium antennas. This would result in the
apparent magnitude being dimmed to -2.5.
*This is still quite bright and would be brighter than the star
Sirius, the brightest star in the sky. Hmmm. I'm wondering about that.
The sunlight at the Earth's distance amounts to about 1,000 watts/m^2.
So even if this 10 cm wide reflector in space reflected all the light
that fell on it to the Earth's surface it would still be only 10
watts. So this is saying a 10 watt light source at a distance of 780
km would be as bright as a -2.5 magnitude star.
*Is this right even if you put behind it say a parabolic reflector as
with a flashlight *so all the light was directed forward?


This should be a standard question in telescope optics if anyone
knows the answer. You have a mirror of a certain size. How large is
the reflected image according to distance to the imaging screen (the
Earth in this case)? For a flat mirror? For a parabolic mirror? For a
bright source, how bright can the image be at the imaging screen?



Bob Clark

On Jun 29, 4:55*pm, Robert Clark wrote:
On Jun 28, 8:31*pm, Robert Clark wrote:



On Jun 28, 1:41 pm, Richard Clark wrote:
On Sun, 28 Jun 2009 17:14:14 +1000, "Peter Webb"
wrote:
Furthermore, the radiation from a reflected area is isotropic - goes in all
directions - and hence very little is directed towards the earth.

Actually, it is lambertian in its distribution, and it would have a
major lobe that was directed in rather typical fashion (at the same,
but negative angle to the norm to the surface). *However, as is the
intent of your response, very much less will find its way to the
intended target.
Richard Clark, KB7QHC

*This describes the reflection from the Iridium antennas as specular
where most of the reflected light is concentrated in a single
direction:
SeeSat-L Apr-98: Method for predicting flare.http://satobs.org/seesat/Apr-1998/0175.html
*About specular reflection:
Specular reflection.http://en.wikipedia.org/wiki/Specular_reflection
*We could get even higher concentration of the image by using
parabolic mirror reflectors.

* We might want to test this out in Earth orbit first before sending
to the Moon. There are low cost CubeSats only 10 cm on a side that can
be launched to orbit at low cost:

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

*Say we made the reflective surface on the CubeSat be a square 10 cm
wide. This is .01 m^2 in area, so a factor of 6.25x10^(-3) smaller
than the area of the Iridium antennas. This would result in the
apparent magnitude being dimmed to -2.5.
*This is still quite bright and would be brighter than the star
Sirius, the brightest star in the sky. Hmmm. I'm wondering about that.
The sunlight at the Earth's distance amounts to about 1,000 watts/m^2.
So even if this 10 cm wide reflector in space reflected all the light
that fell on it to the Earth's surface it would still be only 10
watts. So this is saying a 10 watt light source at a distance of 780
km would be as bright as a -2.5 magnitude star.
*Is this right even if you put behind it say a parabolic reflector as
with a flashlight *so all the light was directed forward?

* * *Bob Clark

Cheap satellites for the greater good of science and humanity are not
allowed.

For less than 10% of just one Apollo mission, we could had a large
scale platform of science and astronomy instruments parked within the
Earth-moon L1 (Selene L1), as of before the first Apollo test flight
to/from our moon.

~ BG

On Jul 2, 2:04*pm, Robert Clark wrote:
On Jun 29, 7:55*pm, Robert Clark wrote:
...
* We might want to test this out in Earth orbit first before sending
to the Moon. There are low cost CubeSats only 10 cm on a side that can
be launched to orbit at low cost:

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

*Say we made the reflective surface on the CubeSat be a square 10 cm
wide. This is .01 m^2 in area, so a factor of 6.25x10^(-3) smaller
than the area of the Iridium antennas. This would result in the
apparent magnitude being dimmed to -2.5.
*This is still quite bright and would be brighter than the star
Sirius, the brightest star in the sky. Hmmm. I'm wondering about that.
The sunlight at the Earth's distance amounts to about 1,000 watts/m^2.
So even if this 10 cm wide reflector in space reflected all the light
that fell on it to the Earth's surface it would still be only 10
watts. So this is saying a 10 watt light source at a distance of 780
km would be as bright as a -2.5 magnitude star.
*Is this right even if you put behind it say a parabolic reflector as
with a flashlight *so all the light was directed forward?

*This should be a standard question in telescope optics if anyone
knows the answer. You have a mirror of a certain size. How large is
the reflected image according to distance to the imaging screen (the
Earth in this case)? For a flat mirror? For a parabolic mirror? For a
bright source, how bright can the image be at the imaging screen?


I found this page with the equation for concave mirrors:

The Mirror Equation - Concave Mirrors.
http://www.glenbrook.k12.il.us/gbssc...ln/u13l3f.html

The mirror equation gives the relationship between the focal length,
the object distance, and the image distance:

1/f = 1/do + 1/di

The object and image distances determine the magnification, which is
the ratio of the image size to object size:

M = hi/ho = - di/do, where the minus sign indicates the image would be
inverted.

The object distance is the distance to the Sun at about 150,000,000
km. The image distance would be the distance from the Moon to the
Earth at 380,000 km. Because the solar distance is so much larger than
the lunar distance, the Earth would be quite close to the focus.
The size of the image is obtained from the equation hi/ho = - di/do,
hi/1,400,000 = - 380,000/150,000,000, using 1,400,000 km for the size
of the Sun. So hi = -3,547 km, with an inverted image.
What I'm still puzzling about, assuming the idea that a 4 m mirror at
lunar distance would have an apparent magnitude of +3 is correct, is
would this mean all the observers in that 3,547 km wide area would
view it as being of +3 magnitude?


Bob Clark

"Peter Webb" wrote in message
...

"Robert Clark" wrote in message
...
On Jul 2, 2:04 pm, Robert Clark wrote:
On Jun 29, 7:55 pm, Robert Clark wrote:
...
We might want to test this out in Earth orbit first before sending
to the Moon. There are low cost CubeSats only 10 cm on a side that can
be launched to orbit at low cost:

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

Say we made the reflective surface on the CubeSat be a square 10 cm
wide. This is .01 m^2 in area, so a factor of 6.25x10^(-3) smaller
than the area of the Iridium antennas. This would result in the
apparent magnitude being dimmed to -2.5.
This is still quite bright and would be brighter than the star
Sirius, the brightest star in the sky. Hmmm. I'm wondering about that.
The sunlight at the Earth's distance amounts to about 1,000 watts/m^2.
So even if this 10 cm wide reflector in space reflected all the light
that fell on it to the Earth's surface it would still be only 10
watts. So this is saying a 10 watt light source at a distance of 780
km would be as bright as a -2.5 magnitude star.
Is this right even if you put behind it say a parabolic reflector as
with a flashlight so all the light was directed forward?

This should be a standard question in telescope optics if anyone
knows the answer. You have a mirror of a certain size. How large is
the reflected image according to distance to the imaging screen (the
Earth in this case)? For a flat mirror? For a parabolic mirror? For a
bright source, how bright can the image be at the imaging screen?


I found this page with the equation for concave mirrors:

The Mirror Equation - Concave Mirrors.
http://www.glenbrook.k12.il.us/gbssc...ln/u13l3f.html

The mirror equation gives the relationship between the focal length,
the object distance, and the image distance:

1/f = 1/do + 1/di

The object and image distances determine the magnification, which is
the ratio of the image size to object size:

M = hi/ho = - di/do, where the minus sign indicates the image would be
inverted.

The object distance is the distance to the Sun at about 150,000,000
km. The image distance would be the distance from the Moon to the
Earth at 380,000 km. Because the solar distance is so much larger than
the lunar distance, the Earth would be quite close to the focus.
The size of the image is obtained from the equation hi/ho = - di/do,
hi/1,400,000 = - 380,000/150,000,000, using 1,400,000 km for the size
of the Sun. So hi = -3,547 km, with an inverted image.

Having now re-read your response, I see you are talking about lunar
distances.

You wil note that the formula does not involve the aperture of the mirror,
and the formula just says the apparent angles for the source and the image
are the same. As I said in my other post, it is effectively a pin-hole
camera, it doesn't need a lens, and its silly to talk about a mirror with a
focal length of 300,000 kms - it would be flat to less than subatomic
distances. All the mathematics for an incoherent non-point source will work
the same for a flat mirror as a nominally parabolic one at these distances.



What I'm still puzzling about, assuming the idea that a 4 m mirror at
lunar distance would have an apparent magnitude of +3 is correct, is
would this mean all the observers in that 3,547 km wide area would
view it as being of +3 magnitude?

I won't do the recalculation of whether +3 is correct. Lets assume it is.

What almost every person who looks up at the mirror will see is one of two
things:

1. A relection of a empty space, or at least not the sun.

2. A reflection of the sun illuminating the mirror, and covering it
completely, because the mirror is a tiny pin hole compared to the angular
size of the sun.

That is all or nothing.

A tiny tiny percentage will have for a fraction of a second a reflection of
the exact edge of the sun, like the tiny pinhole was pointed at the exact
edge of the Sun.

Whilst your fully illuminated portion is 3,500 kms wide, a 2 metre wide
mirror 380,000 kms away subtends and angle of 2/380*10^6 metres = 2/(4*10^8)
= 5 * 10^-9 radians. The width of the sun is 10^-2 radians, so the width of
the zone with brightness between zero and full is:

3,500 kms * 5 * 10^-9 / 10^-2
= 3.5 * 10^6 * 5 * 10^-7
= about 20 cms.

(In practice it will be somewhat larger due to single slit diffraction, but
still very small).

The apparent angular width of the circular pinhole which is the mirror 2
metres across at the distance of the moon - not even resolvable with the
most powerful telescopes on earth - is insignificant compared to the
apparent angular width of the sun, which is as big as the moon and easily
seen with the naked eye.

Hence the chances of this pinhole partially overlapping the edge of the sun
are effectively zero.

On Jul 6, 1:58 pm, Robert Clark wrote:
On Jul 2, 2:04 pm, wrote:
On Jun 29, 7:55 pm, wrote:
...
We might want to test this out in Earth orbit first before sending
to theMoon. There are low cost CubeSats only 10 cm on a side that can
be launched to orbit at low cost:

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

Say we made the reflective surface on the CubeSat be a square 10 cm
wide. This is .01 m^2 in area, so a factor of 6.25x10^(-3) smaller
than the area of the Iridium antennas. This would result in the
apparent magnitude being dimmed to -2.5.
This is still quite bright and would be brighter than the star
Sirius, the brightest star in the sky. Hmmm. I'm wondering about that.
The sunlight at the Earth's distance amounts to about 1,000 watts/m^2.
So even if this 10 cm wide reflector in space reflected all the light
that fell on it to the Earth's surface it would still be only 10
watts. So this is saying a 10 watt light source at a distance of 780
km would be as bright as a -2.5 magnitude star.
Is this right even if you put behind it say a parabolic reflector as
with a flashlight so all the light was directed forward?

This should be a standard question in telescope optics if anyone
knows the answer. You have a mirror of a certain size. How large is
the reflected image according to distance to the imaging screen (the
Earth in this case)? For a flat mirror? For a parabolic mirror? For a
bright source, how bright can the image be at the imaging screen?

I found this page with the equation for concave mirrors:

The Mirror Equation - Concave Mirrors.http://www.glenbrook.k12.il.us/gbssc...ln/u13l3f.html

The mirror equation gives the relationship between the focal length,
the object distance, and the image distance:

1/f = 1/do + 1/di

The object and image distances determine the magnification, which is
the ratio of the image size to object size:

M = hi/ho = - di/do, where the minus sign indicates the image would be
inverted.

The object distance is the distance to the Sun at about 150,000,000
km. The image distance would be the distance from the Moon to the
Earth at 380,000 km. Because the solar distance is so much larger than
the lunar distance, the Earth would be quite close to the focus.
The size of the image is obtained from the equation hi/ho = - di/do,
hi/1,400,000 = - 380,000/150,000,000, using 1,400,000 km for the size
of the Sun. So hi = -3,547 km, with an inverted image.
What I'm still puzzling about, assuming the idea that a 4 m mirror at
lunar distance would have an apparent magnitude of +3 is correct, is
would this mean all the observers in that 3,547 km wide area would
view it as being of +3 magnitude?


Let's see how big the mirror would have to be at Mars to be visible
with a telescope a (serious) amateur might have. The distance between
Earth and Mars varies from around 60 million km to 400 million km. So
we have to take into account the diminution with distance compared to
the lunar distance of 380,000 km. However, the distance of Mars from
the Sun is also larger than the Earth's so the reflected image would
be dimmer because of that as well.
The furthest distance of Mars from the Sun is about 250 million km.
This is about 1.6 times as far as the Earth distance so the mirror
reflection would be dimmed by a factor of 1.6^2 = 2.56 to begin with.
At the closest Earth-Mars distance of 60 million km, this is 158 times
further than the Earth-Moon distance so the reflection seen at the
Earth would be dimmer by an additional factor of 158^2, about 25,000
times dimmer. At the furthest Earth-Mars distance of 400 million km,
this is 1053 times further than the Moon, so it would be dimmed by a
factor of 1,100,000 times.
So the total diminution at Mars' closest approach would be (1/2.56)x
(1/25,000) = 1/64,000. And at furthest distance, it would be (1/2.56)x
(1/1,100,000) = 1/2,816,000. The closest distance dimming of 1/64,000
amounts to a change of apparent magnitude of 12, so to +15 for the
same 4 m sized mirror. The furthest distance dimming of 1/2,816,000
amounts to a change of apparent magnitude of 16 so to +19.
This article gives the apparent magnitude an 8-inch telescope might
detect as +14 under good seeing conditions:

A practical guide to buying telescopes.
By Jeff Kanipe
Special to SPACE.com
posted: 04:05 pm ET
19 June 2000
http://www.space.com/scienceastronom...escope_II.html

So at Mars closest approach you would need a larger mirror to amount
to a change in apparent magnitude of +1, so one of 2.51 times greater
collecting area than the 4 meter one, so to 6.4 meters across.
However, for the furthest distance magnitude of +19 you would need a
larger reflecting surface than the 4 meter one to amount to a change
in apparent magnitude of +5 in order to be visible by the 8-inch
scope. So it would need to be larger by a factor of 100 in collecting
area, so to 40 meters across.
Certainly a large reflective surface. But the Echo 2 satellite used a
microwave reflective balloon surface of 41 meter diameter in the
1960's so it is technically feasible:

Echo satellite.
http://en.wikipedia.org/wiki/Echo_satellites

The image size as before would be calculated from the image and
object distances, and the Sun's diameter. It would range from about
1/4 the Sun's size to about 1-1/2 times the Sun's size. This again
raises the question: would all observers within the large observing
region really see the reflecting surface as having those apparent
magnitudes calculated?


Bob Clark

"Robert Clark" wrote in message
...
On Jul 6, 1:58 pm, Robert Clark wrote:
On Jul 2, 2:04 pm, wrote:
On Jun 29, 7:55 pm, wrote:
...
We might want to test this out in Earth orbit first before sending
to theMoon. There are low cost CubeSats only 10 cm on a side that can
be launched to orbit at low cost:

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

Say we made the reflective surface on the CubeSat be a square 10 cm
wide. This is .01 m^2 in area, so a factor of 6.25x10^(-3) smaller
than the area of the Iridium antennas. This would result in the
apparent magnitude being dimmed to -2.5.
This is still quite bright and would be brighter than the star
Sirius, the brightest star in the sky. Hmmm. I'm wondering about
that.
The sunlight at the Earth's distance amounts to about 1,000
watts/m^2.
So even if this 10 cm wide reflector in space reflected all the light
that fell on it to the Earth's surface it would still be only 10
watts. So this is saying a 10 watt light source at a distance of 780
km would be as bright as a -2.5 magnitude star.
Is this right even if you put behind it say a parabolic reflector as
with a flashlight so all the light was directed forward?

This should be a standard question in telescope optics if anyone
knows the answer. You have a mirror of a certain size. How large is
the reflected image according to distance to the imaging screen (the
Earth in this case)? For a flat mirror? For a parabolic mirror? For a
bright source, how bright can the image be at the imaging screen?

I found this page with the equation for concave mirrors:

The Mirror Equation - Concave
Mirrors.http://www.glenbrook.k12.il.us/gbssc...ln/u13l3f.html

The mirror equation gives the relationship between the focal length,
the object distance, and the image distance:

1/f = 1/do + 1/di

The object and image distances determine the magnification, which is
the ratio of the image size to object size:

M = hi/ho = - di/do, where the minus sign indicates the image would be
inverted.

The object distance is the distance to the Sun at about 150,000,000
km. The image distance would be the distance from the Moon to the
Earth at 380,000 km. Because the solar distance is so much larger than
the lunar distance, the Earth would be quite close to the focus.
The size of the image is obtained from the equation hi/ho = - di/do,
hi/1,400,000 = - 380,000/150,000,000, using 1,400,000 km for the size
of the Sun. So hi = -3,547 km, with an inverted image.
What I'm still puzzling about, assuming the idea that a 4 m mirror at
lunar distance would have an apparent magnitude of +3 is correct, is
would this mean all the observers in that 3,547 km wide area would
view it as being of +3 magnitude?


Let's see how big the mirror would have to be at Mars to be visible
with a telescope a (serious) amateur might have. The distance between
Earth and Mars varies from around 60 million km to 400 million km. So
we have to take into account the diminution with distance compared to
the lunar distance of 380,000 km. However, the distance of Mars from
the Sun is also larger than the Earth's so the reflected image would
be dimmer because of that as well.
The furthest distance of Mars from the Sun is about 250 million km.
This is about 1.6 times as far as the Earth distance so the mirror
reflection would be dimmed by a factor of 1.6^2 = 2.56 to begin with.
At the closest Earth-Mars distance of 60 million km, this is 158 times
further than the Earth-Moon distance so the reflection seen at the
Earth would be dimmer by an additional factor of 158^2, about 25,000
times dimmer. At the furthest Earth-Mars distance of 400 million km,
this is 1053 times further than the Moon, so it would be dimmed by a
factor of 1,100,000 times.
So the total diminution at Mars' closest approach would be (1/2.56)x
(1/25,000) = 1/64,000. And at furthest distance, it would be (1/2.56)x
(1/1,100,000) = 1/2,816,000. The closest distance dimming of 1/64,000
amounts to a change of apparent magnitude of 12, so to +15 for the
same 4 m sized mirror. The furthest distance dimming of 1/2,816,000
amounts to a change of apparent magnitude of 16 so to +19.
This article gives the apparent magnitude an 8-inch telescope might
detect as +14 under good seeing conditions:

A practical guide to buying telescopes.
By Jeff Kanipe
Special to SPACE.com
posted: 04:05 pm ET
19 June 2000
http://www.space.com/scienceastronom...escope_II.html

So at Mars closest approach you would need a larger mirror to amount
to a change in apparent magnitude of +1, so one of 2.51 times greater
collecting area than the 4 meter one, so to 6.4 meters across.
However, for the furthest distance magnitude of +19 you would need a
larger reflecting surface than the 4 meter one to amount to a change
in apparent magnitude of +5 in order to be visible by the 8-inch
scope. So it would need to be larger by a factor of 100 in collecting
area, so to 40 meters across.
Certainly a large reflective surface. But the Echo 2 satellite used a
microwave reflective balloon surface of 41 meter diameter in the
1960's so it is technically feasible:

Echo satellite.
http://en.wikipedia.org/wiki/Echo_satellites

The image size as before would be calculated from the image and
object distances, and the Sun's diameter. It would range from about
1/4 the Sun's size to about 1-1/2 times the Sun's size. This again
raises the question: would all observers within the large observing
region really see the reflecting surface as having those apparent
magnitudes calculated?


Bob Clark

Cool !
Nice calculations, Bob !
One thing that comes to mind : when the mirror is actually on the Mars
surface (or the Lunar surface), wouldn't it have to outshine the normal
reflection that the planet (or the moon) radiates back to us, before we can
see it with a telescope ?
How does that change the mirror sizes ?

Rob

On Jul 15, 2:30*am, "Rob Dekker" wrote:
"Robert Clark" wrote in message

...



On Jul 6, 1:58 pm, Robert Clark wrote:
On Jul 2, 2:04 pm, wrote:
On Jun 29, 7:55 pm, wrote:
...
* We might want to test this out in Earth orbit first before sending
to theMoon. There are low cost CubeSats only 10 cm on a side that can
be launched to orbit at low cost:

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

*Say we made the reflective surface on the CubeSat be a square 10 cm
wide. This is .01 m^2 in area, so a factor of 6.25x10^(-3) smaller
than the area of the Iridium antennas. This would result in the
apparent magnitude being dimmed to -2.5.
*This is still quite bright and would be brighter than the star
Sirius, the brightest star in the sky. Hmmm. I'm wondering about
that.
The sunlight at the Earth's distance amounts to about 1,000
watts/m^2.
So even if this 10 cm wide reflector in space reflected all the light
that fell on it to the Earth's surface it would still be only 10
watts. So this is saying a 10 watt light source at a distance of 780
km would be as bright as a -2.5 magnitude star.
*Is this right even if you put behind it say a parabolic reflector as
with a flashlight *so all the light was directed forward?

*This should be a standard question in telescope optics if anyone
knows the answer. You have a mirror of a certain size. How large is
the reflected image according to distance to the imaging screen (the
Earth in this case)? For a flat mirror? For a parabolic mirror? For a
bright source, how bright can the image be at the imaging screen?

*I found this page with the equation for concave mirrors:

The Mirror Equation - Concave
Mirrors.http://www.glenbrook.k12.il.us/gbssc...ln/u13l3f.html

*The mirror equation gives the relationship between the focal length,
the object distance, and the image distance:

1/f = 1/do + 1/di

*The object and image distances determine the magnification, which is
the ratio of the image size to object size:

M = hi/ho = - di/do, where the minus sign indicates the image would be
inverted.

*The object distance is the distance to the Sun at about 150,000,000
km. The image distance would be the distance from the Moon to the
Earth at 380,000 km. Because the solar distance is so much larger than
the lunar distance, the Earth would be quite close to the focus.
The size of the image is obtained from the equation hi/ho = - di/do,
hi/1,400,000 = - 380,000/150,000,000, using 1,400,000 km for the size
of the Sun. So hi = -3,547 km, with an inverted image.
*What I'm still puzzling about, assuming the idea that a 4 m mirror at
lunar distance would have an apparent magnitude of +3 is correct, is
would this mean all the observers in that 3,547 km wide area would
view it as being of +3 magnitude?

Let's see how big the mirror would have to be at Mars to be visible
with a telescope a (serious) amateur might have. The distance between
Earth and Mars varies from around 60 million km to 400 million km. So
we have to take into account the diminution with distance compared to
the lunar distance of 380,000 km. However, the distance of Mars from
the Sun is also larger than the Earth's so the reflected image would
be dimmer because of that as well.
The furthest distance of Mars from the Sun is about 250 million km.
This is about 1.6 times as far as the Earth distance so the mirror
reflection would be dimmed by a factor of 1.6^2 = 2.56 to begin with.
At the closest Earth-Mars distance of 60 million km, this is 158 times
further than the Earth-Moon distance so the reflection seen at the
Earth would be dimmer by an additional factor of 158^2, about 25,000
times dimmer. At the furthest Earth-Mars distance of 400 million km,
this is 1053 times further than the Moon, so it would be dimmed by a
factor of 1,100,000 times.
*So the total diminution at Mars' closest approach would be (1/2.56)x
(1/25,000) = 1/64,000. And at furthest distance, it would be (1/2.56)x
(1/1,100,000) = 1/2,816,000. The closest distance dimming of 1/64,000
amounts to a change of apparent magnitude of 12, so to +15 for the
same 4 m sized mirror. The furthest distance dimming of 1/2,816,000
amounts to a change of apparent magnitude of 16 so to +19.
This article gives the apparent magnitude an 8-inch telescope might
detect as +14 *under good seeing conditions:

A practical guide to buying telescopes.
By Jeff Kanipe
Special to SPACE.com
posted: 04:05 pm ET
19 June 2000
http://www.space.com/scienceastronom...escope_II.html

So at Mars closest approach you would need a larger mirror to amount
to a change in apparent magnitude of +1, so one of 2.51 times greater
collecting area than the 4 meter one, so to 6.4 meters across.
However, for the furthest distance magnitude of +19 you would need a
larger reflecting surface than the 4 meter one to amount to a change
in apparent magnitude of +5 in order to be visible by the 8-inch
scope. So it would need to be larger by a factor of 100 in collecting
area, so to 40 meters across.
Certainly a large reflective surface. But the Echo 2 satellite used a
microwave reflective balloon surface of 41 meter diameter in the
1960's so it is technically feasible:

Echo satellite.
http://en.wikipedia.org/wiki/Echo_satellites

The image size as before would be calculated from the image and
object distances, and the Sun's diameter. It would range from about
1/4 the Sun's size to about 1-1/2 times the Sun's size. This again
raises the question: would all observers within the large observing
region really see the reflecting surface as having those apparent
magnitudes calculated?

* * Bob Clark

Cool !
Nice calculations, Bob !
One thing that comes to mind : when the mirror is actually on the Mars
surface (or the Lunar surface), wouldn't it have to outshine the normal
reflection that the planet (or the moon) radiates back to us, before we can
see it with a telescope ?
How does that change the mirror sizes ?

Rob

The mirror(s) or whatever array of corner cubes could easily be narrow
bandpass coated. The Apollo missions entirely forgot about such,
along with several dozen other essentials and/or better alternatives.

A sufficient laser cannon and its TRACE or modified Hubble like
observatory situated within the Earth-moon L1 (aka Selene L1) should
more than do the trick.

~ BG