apparent magnitude through scope

Hello. How does one calculate the apparent magnitude of star as viewed
through a telescope? For example, if a star has a known naked-eye apparent
magnitude of 3, what would that star's apparent magnitude be, as viewed
through a 5" scope?

I am looking for the theoretical answer. I understand that there are issues
like central obstruction (in a cassegrain design), light absorption in the
mirrors and lenses, and so forth. On that note, what would be a realistic
correction to apply to the theoretical answer? For example, perhaps there
is a range of light transmission that applies to 95% of scopes, like 1% to
5% light loss (?)

Thanks for sharing your knowledge!
Peter

"Peter Michelson" wrote in message
news:_DORd.10710$Ps.10226@okepread06...
Hello. How does one calculate the apparent magnitude of star as viewed
through a telescope? For example, if a star has a known naked-eye
apparent
magnitude of 3, what would that star's apparent magnitude be, as viewed
through a 5" scope?

I am looking for the theoretical answer. I understand that there are
issues
like central obstruction (in a cassegrain design), light absorption in the
mirrors and lenses, and so forth. On that note, what would be a realistic
correction to apply to the theoretical answer? For example, perhaps there
is a range of light transmission that applies to 95% of scopes, like 1% to
5% light loss (?)

Thanks for sharing your knowledge!
Peter


I don't know your answer but...

....how does one compare the magnitude of a point source like a star, versus
that of a large planet, easily seen as an extended object in a small 'scope,
such as Saturn ?

--
M Stewart
Milton Keynes, UK
http://www.megalith.freeserve.co.uk/oddimage.htm

Wasn't it Peter Michelson who wrote:
Hello. How does one calculate the apparent magnitude of star as viewed
through a telescope? For example, if a star has a known naked-eye apparent
magnitude of 3, what would that star's apparent magnitude be, as viewed
through a 5" scope?

I am looking for the theoretical answer. I understand that there are issues
like central obstruction (in a cassegrain design), light absorption in the
mirrors and lenses, and so forth. On that note, what would be a realistic
correction to apply to the theoretical answer? For example, perhaps there
is a range of light transmission that applies to 95% of scopes, like 1% to
5% light loss (?)

Technically, a star with apparent magnitude 4.5 always has apparent
magnitude 4.5 whatever telescope you're looking through.

However, I guess I know what you mean. The amount of light captured from
a star is proportional to the area of the aperture. So you have to
compare the area of your telescope's lens or mirror to the area of your
(dark adapted) pupil, and this gives the brightness ratio. To convert a
brightness ratio to a difference in magnitude, take the logarithm to
base 2.514.

A typical pupil diameter is 0.25", so a 5" scope has 400 times the area,
so the brightness ratio is 400, and stars appear to be 6.5 magnitudes
brighter. A mag 3 star would appear to be mag -3.5.

If your calculator doesn't do logs to base 2.512, then calculate it as
log(brightness_ratio)/log(2.512)

--
Mike Williams
Gentleman of Leisure

"Mike Williams" wrote in message
...
The amount of light captured from
a star is proportional to the area of the aperture. So you have to
compare the area of your telescope's lens or mirror to the area of your
(dark adapted) pupil, and this gives the brightness ratio. To convert a
brightness ratio to a difference in magnitude, take the logarithm to
base 2.514.

A typical pupil diameter is 0.25", so a 5" scope has 400 times the area,
so the brightness ratio is 400, and stars appear to be 6.5 magnitudes
brighter. A mag 3 star would appear to be mag -3.5.

If your calculator doesn't do logs to base 2.512, then calculate it as
log(brightness_ratio)/log(2.512)


Thank you - that's clear and makes sense to me. However, it occurs to me
that in order to fairly compare naked eye viewing with typical scope
viewing, one would have to take into account the use of two pupils. But I
am not sure if it is simply an additive effect (i.e., 200 times the area of
two pupils instead of 400 times the area of one pupil). I supposed one
would have to take into account the synthesizing properties of the early
visual system. Any thoughts on this?

Thanks again,
Peter

If your calculator doesn't do logs to base 2.512, then calculate it as
log(brightness_ratio)/log(2.512)

m(telescope) = m(naked eye) - 2.5 * log[brightness ratio] (note that the 2.5
here is NOT an approximation for the 2.512..... quoted above)

If binocular vision is exactly additive to perceived brightness (which I
suspect that it isn't quite)

Area(eye) = 2 * pi * r(pupil)^2
Area(telescope) = pi * r(telescope)^2

Brightness ratio = r(telescope)^2/(2 * r(pupil)^2)

m(telescope) = m(naked eye) - 5 * log [r(telescope)] + 2.5 log(2) + 5 *
log[r(pupil)]
m(telescope) = m(naked eye) + 5 * log[r(pupil)] - 5 * log[r(telescope)] +
0.7526

If a typical pupil diameter is 0.25", then
m(telescope) = m(naked eye) - 5 * log(telescope aperture in inches) - 2.2577

For an 8" scope, and mag 3 star, it now appears m = -3.77

If you can normally see to mag 6 with the naked eye, then you might expect
to see down to mag 12.77

Nicely laid out. Thank you.

I've done a little research and come up with the following:

1) *Detection* of a dim visual stimulus is enhanced by binocular vision by a
factor of 1.4 (the square root of 2). This has implications for the
limiting visual magnitude but NOT perceived brightness

2) *Perceived brightness* using binocular vision is approximately an average
of the brightness perceived by each eye when contour information is present.
However, in the absence of contour information (if you just put a patch over
one eye), the perceived brightness does not change.

The implications seem clear:

1) if you use both eyes to view the night sky instead of one, you will
detect fainter stars, but none of the stars will appear brighter.

2) if you use both eyes to view the night sky and put a weak neutral filter
(like a sunglass lens) over one eye, stars will appear dimmer than if you
just closed that eye.

3) a further implication, perhaps unrelated to our discussion, is that using
a binocular eyepiece on a telescope actually reduces both detection and
perceived brightness.

So the important distinction here is between detection sensitivity and
perceived brightness.

For the purposes of our calculations, if we decide that limiting magnitude
is the criterion, we should assume that binocular vision affords 1.4 times
the detection capability of monocular vision. Thus, Area(eye) = 1.4 * pi *
r(pupil)^2.

However, if we want to make predictions about perceived brightness, I think
we should assume that binocular vision is equivalent to monocular vision.
As a simple test, close one eye: does everything appear half as bright? Of
course not. Equally bright - yes. In this case, Area(eye) = pi *
r(pupil)^2.

Does this analysis make sense or am I missing something?

Thanks,
Peter

References:
http://arapaho.nsuok.edu/~salmonto/V.../Lecture10.pdf
http://www.cns.bu.edu/Profiles/Grossberg/GroKel99.pdf

"Richard Bullock" wrote in message
...
If your calculator doesn't do logs to base 2.512, then calculate it as
log(brightness_ratio)/log(2.512)

m(telescope) = m(naked eye) - 2.5 * log[brightness ratio] (note that the
2.5
here is NOT an approximation for the 2.512..... quoted above)

If binocular vision is exactly additive to perceived brightness (which I
suspect that it isn't quite)

Area(eye) = 2 * pi * r(pupil)^2
Area(telescope) = pi * r(telescope)^2

Brightness ratio = r(telescope)^2/(2 * r(pupil)^2)

m(telescope) = m(naked eye) - 5 * log [r(telescope)] + 2.5 log(2) + 5 *
log[r(pupil)]
m(telescope) = m(naked eye) + 5 * log[r(pupil)] - 5 * log[r(telescope)] +
0.7526

If a typical pupil diameter is 0.25", then
m(telescope) = m(naked eye) - 5 * log(telescope aperture in inches) -
2.2577

For an 8" scope, and mag 3 star, it now appears m = -3.77

If you can normally see to mag 6 with the naked eye, then you might expect
to see down to mag 12.77