Someone please explain this to me (maximum magnification)

Hi,

For quite a while, I've accepted the informal rule for
maximum useful magnification of a telescope -- roughly
50 times the aperture in inches.

After thinking calmly about it and trying to explain it
to someone else (someone even less experienced than I am),
I find myself struggling to understand the physical/
mathematical/geometric reasons why this is so.

The more I think about it, the less it makes sense.
Limiting visual magnitude I can understand that it
decreases as the aperture increases. But how can the
sharpness of objects at high magnification be
proportionally related to the aperture? In fact,
intuition might even tell us that it should be
exactly the opposite -- the "perfect focus" camera
is one that has "infinitesimal" aperture, and so
any defect on the lens or mirror surface would not
affect the projected image. (of course, I do
understand that with tiny apertures, telescopes
would be useless, since one would not be able to
see objects that are too faint -- you have to gather
light together to "amplify" the visual brightness).

Any kind soul could help me understand the "physical"
reasons for this phenomenon? (BTW, I'm an Electrical
Engineer, so feel free to get technical and use any
maths or physics needed to explain it)

Thanks!

Carlos
--

Hi Carlos,
My comments in line

Carlos Moreno wrote:

Hi,

For quite a while, I've accepted the informal rule for
maximum useful magnification of a telescope -- roughly
50 times the aperture in inches.


The eye, a good young one can resolve to about 2 arc minutes.
At 50X this is 2.4 arc seconds which about what the distortion
of the atmosphere, called seeing, produces near the ground.
Seeing can get better than 1 arc second and so more magnification is
needed.

A 1 inch aperture produces a spot about 4.5 arc seconds in diameter.
(fwhm)
and so 50X is twice the spatial sampling frequency of that spot.

This is similar to the optics and atmosphere being treated as a low pass
filter.
The cut off frequency of the optical filter is inversely proportional to
the diameter.
so 1 inch is one .22 cycles per arc second and 4.5 inches is 1 cycle per
arc second etc.

for a low pass filter, as signal frequency goes up the the amplitude
will be reduced.

This is equivalent to what is observes in a optical system where the
contrast
decreases as a function of spatial frequency. The effective cutoff
spatial frequency of
the eye, telescope combination, is the point where the contrast drops
below the
detectable level for the eye. Some observers like the image diameter
larger as it
matches the spatial frequency contrast function of their eye/brain
better.



After thinking calmly about it and trying to explain it
to someone else (someone even less experienced than I am),
I find myself struggling to understand the physical/
mathematical/geometric reasons why this is so.

The more I think about it, the less it makes sense.
Limiting visual magnitude I can understand that it
decreases as the aperture increases.

Mag~ -2.5 log brightness and so limiting magnitude goes up with
aperture.

But how can the
sharpness of objects at high magnification be
proportionally related to the aperture? In fact,
intuition might even tell us that it should be
exactly the opposite -- the "perfect focus" camera
is one that has "infinitesimal" aperture, and so
any defect on the lens or mirror surface would not
affect the projected image. (of course, I do
understand that with tiny apertures, telescopes
would be useless, since one would not be able to
see objects that are too faint -- you have to gather
light together to "amplify" the visual brightness).


The pin hole camera has a large depth of field and so you don't need to
focus
but the resolution is a function of diameter, see above,
while light gathering is a function of the square of the diameter



Any kind soul could help me understand the "physical"
reasons for this phenomenon? (BTW, I'm an Electrical
Engineer, so feel free to get technical and use any
maths or physics needed to explain it)

Thanks!

Carlos
--

How was that ? (signal to noise wise)
Dan McKenna a fellow EE

I always thought it out this way.

You use higher magnifications to see finer details in an object. Although
there are other secondary reasons, for this discussion this is the reason
for using higher magnifictions. But the ability of any optics to see finer
details is directly related to the aperture of the primary lens / mirror.
The larger the aperture, the closer together two features can be and you
can still resolve them. The famous but overly used and often mis- used
Dawes limit is based on this.

The rule of thumb relating maximum useable magnification to inches of
aperture is also related to this fact of life, but the ruel is not hard and
invariable. For instance, it depends on all factors that can affect seeing
finer details through a scope, with seeing and the type of optical
configuration being the primary ones. Anything with large secondary
obstruction pushes the number down. For bad to fair seeing the number
goes down. Hence most of the time, through the most commonly found
scopes (SCTs and Newtonians), the figure is around 20x to 30x per inch
of aperture.

For high quality refractors with no secondary obstructions under
dead-steady seeing conditions, the figure can be pushed as high as
80x to perhaps 90x per inch of aperture, as I personally witnessed
using a 6- inch Jaegars refractor during another Mars opposition many
years ago. High quality Casssegrains and long focal length Newtonians
with small secondary mirrors also can reach these figures if the optics
are high quality and the seeing conditions warrent it.

Hoping this less technical overview of where this Rule of Thumb comes
from will assist everyone.

Clear and Steady Nights !
--
----------------------------------------------------------------------
Never be afraid of trying something new for the love of it.
Remember... amateurs built the Ark.
Professionals built the Titanic!


----------------------------------------------------------------------

"Carlos Moreno" wrote in message
...

Hi,

For quite a while, I've accepted the informal rule for
maximum useful magnification of a telescope -- roughly
50 times the aperture in inches.

After thinking calmly about it and trying to explain it
to someone else (someone even less experienced than I am),
I find myself struggling to understand the physical/
mathematical/geometric reasons why this is so.

The more I think about it, the less it makes sense.
Limiting visual magnitude I can understand that it
decreases as the aperture increases. But how can the
sharpness of objects at high magnification be
proportionally related to the aperture? In fact,
intuition might even tell us that it should be
exactly the opposite -- the "perfect focus" camera
is one that has "infinitesimal" aperture, and so
any defect on the lens or mirror surface would not
affect the projected image. (of course, I do
understand that with tiny apertures, telescopes
would be useless, since one would not be able to
see objects that are too faint -- you have to gather
light together to "amplify" the visual brightness).

Any kind soul could help me understand the "physical"
reasons for this phenomenon? (BTW, I'm an Electrical
Engineer, so feel free to get technical and use any
maths or physics needed to explain it)

Thanks!

Carlos
--

The telescope picks up a limited amount of fine detail because of
diffraction.

If you magnify the image beyond a certain point, you no longer make
additional detail visible.

It's like looking at your computer screen with a magnifying glass -- you
don't actually see any pixels that you couldn't already see.

Excessive magnification does not bring out detail but does make the image
fainter by spreading the light over a larger area.

Carlos,
Pure EE/optics answer, programmable with a little effort using a 2-D
fast Fourier transform (FFT), so you can experiment with it and teach
yourself:

Treat wavefront at telescope pupil as a complex phasor:
T(x,y) = A(x,y) exp[jP(x,y)]
where A(x,y) is the amplitude and P(x,y) is the combined optical path
difference (OPD) phase error from telescope aberrations and time-varying
atmospheric propagation:

P(x,y) = 2(pi) OPD(x,y) / wavelength.

Two arrays holding the real and imaginary pupil values are initialized.
The real array contains A(x,y)cos[P(x,y)] values, and the imaginary
array contains A(x,y)sin[P(x,y)] values.

2-D complex Fourier transform of T(x,y) = complex-valued point spread
function h(fx,fy)

h(fx,fy) and T(x,y) are 2-D Fourier transform pairs. Using the 2-D
in-place FFT, the scaling in the frequency domain is given by

(x)(fx) = (y)(fy) = (wavelength)(focal length)/N

where N is the FFT row or column size (1024, 2048, etc.)

However, your eye, film or CCD cannot see h(fx,fy), but instead responds
to intensity. The 2-D intensity I(fx,fy) is calculated as the squared
modulus of the complex point spread function, hh*. In the computer,

I(fx,fy) = Re[h(fx,fy)]² + Im[h(fx,fy)]².

This is a one-way operation - you can't recover h(fx,fy) from I(fx,fy)
because values are squared and sign information is lost. I(fx,fy) is
more generally known as the Airy diffraction pattern.

The 2-D Fourier transform of I(fx,fy) gives H(1/x,1/y), the
complex-valued optical transfer function. This is the low-pass
filtering that Dan McKenna refers to. H(1/x,1/y) is also the complex
2-D autocorrelation function of the original telescope pupil function
T(x,y), but it is MUCH faster computationally to calculate using the 2D
FFT.

The modulation transfer function is the modulus of H(1/x,1/y):

MTF(1/x,1/y) = SQRT( Re[H(1/x,1/y)]² + Im[H(1/x,1/y)]² )

Image contrast is (1+M)/(1-M) to one.

The MTF cutoff frequency in cycles/millimeter is 1/(wavelength)/(focal
ratio), wavelength in mm. For an f/5 telescope at 0.55 micron
wavelength, the cutoff frequency is 1/(0.00055)/(5) = 363 cycles/mm at
the focal plane. Angular details in the image smaller than this are not
resolved. As you increase magnification you begin to visually or
instrumentally resolve image details near this cutoff frequency. In
practice the time-varying atmospheric turbulence phase error swamps
image detail well before cutoff frequency, so the resolution of a highly
magnified image is primarily limited by turbulence.

The rolloff in telescope optical resolution and image contrast is as
exactly modeled by this methodology as one can express the phase
function in T(x,y). Good beginning references are Goodman's
"Introduction to Fourier Optics", Andrews and Phillips "Laser Beam
Propagation through Random Media", and Dick Suiter's "Star Testing
Astronomical Telescopes". I hope you can take the time to program this
for yourself and experiment with it, you will learn much from that
little program. I first programmed this back in 1975 or thereabouts,
and it is still one of the most useful and used programs I've ever
written.

Mike

On 2003-09-07, Carlos Moreno wrote:

Any kind soul could help me understand the "physical"
reasons for this phenomenon? (BTW, I'm an Electrical
Engineer, so feel free to get technical and use any
maths or physics needed to explain it)

It's due to the wave nature of light. A textbook on physical optics
will explain it, although the mathematics of computing the diffraction
pattern of a circular aperture can be "interesting". The wave front
is disturbed by the edges of the aperture and you get constructive
and destructive interference that produce a pattern that depends on
the shape of the aperture. For circular apertures you get a circular
pattern that consists of a central disc surrounded by diffraction rings.
The angular dimensions of the diffraction pattern depend on the size of
the aperture and the wavelength of the electromagnetic radiation. The
diffraction blurs the image and puts a limit on the resolution of the
telescope. It's the same thing that limits the resolution of radar
antennas.

"Mike Jones" wrote in message
...
Carlos,
Pure EE/optics answer, programmable with a little effort using a 2-D
fast Fourier transform (FFT), so you can experiment with it and teach
yourself:

Treat wavefront at telescope pupil as a complex phasor:
T(x,y) = A(x,y) exp[jP(x,y)]
where A(x,y) is the amplitude and P(x,y) is the combined optical path
difference (OPD) phase error from telescope aberrations and time-varying
atmospheric propagation:

P(x,y) = 2(pi) OPD(x,y) / wavelength.

Two arrays holding the real and imaginary pupil values are initialized.
The real array contains A(x,y)cos[P(x,y)] values, and the imaginary
array contains A(x,y)sin[P(x,y)] values.

2-D complex Fourier transform of T(x,y) = complex-valued point spread
function h(fx,fy)

h(fx,fy) and T(x,y) are 2-D Fourier transform pairs. Using the 2-D
in-place FFT, the scaling in the frequency domain is given by

(x)(fx) = (y)(fy) = (wavelength)(focal length)/N

where N is the FFT row or column size (1024, 2048, etc.)

However, your eye, film or CCD cannot see h(fx,fy), but instead responds
to intensity. The 2-D intensity I(fx,fy) is calculated as the squared
modulus of the complex point spread function, hh*. In the computer,

I(fx,fy) = Re[h(fx,fy)]² + Im[h(fx,fy)]².

This is a one-way operation - you can't recover h(fx,fy) from I(fx,fy)
because values are squared and sign information is lost. I(fx,fy) is
more generally known as the Airy diffraction pattern.

The 2-D Fourier transform of I(fx,fy) gives H(1/x,1/y), the
complex-valued optical transfer function. This is the low-pass
filtering that Dan McKenna refers to. H(1/x,1/y) is also the complex
2-D autocorrelation function of the original telescope pupil function
T(x,y), but it is MUCH faster computationally to calculate using the 2D
FFT.

The modulation transfer function is the modulus of H(1/x,1/y):

MTF(1/x,1/y) = SQRT( Re[H(1/x,1/y)]² + Im[H(1/x,1/y)]² )

Image contrast is (1+M)/(1-M) to one.

The MTF cutoff frequency in cycles/millimeter is 1/(wavelength)/(focal
ratio), wavelength in mm. For an f/5 telescope at 0.55 micron
wavelength, the cutoff frequency is 1/(0.00055)/(5) = 363 cycles/mm at
the focal plane. Angular details in the image smaller than this are not
resolved. As you increase magnification you begin to visually or
instrumentally resolve image details near this cutoff frequency. In
practice the time-varying atmospheric turbulence phase error swamps
image detail well before cutoff frequency, so the resolution of a highly
magnified image is primarily limited by turbulence.

The rolloff in telescope optical resolution and image contrast is as
exactly modeled by this methodology as one can express the phase
function in T(x,y). Good beginning references are Goodman's
"Introduction to Fourier Optics", Andrews and Phillips "Laser Beam
Propagation through Random Media", and Dick Suiter's "Star Testing
Astronomical Telescopes". I hope you can take the time to program this
for yourself and experiment with it, you will learn much from that
little program. I first programmed this back in 1975 or thereabouts,
and it is still one of the most useful and used programs I've ever
written.

Mike


As a beginner, and a liberal arts graduate from long ago, I gotta tell you,
I love it when you engineer guys talk like that. Didn't understand a word
of it but it sure is fine to listen to. Or, read, in this case.


--

----

Joe S.

Carlos Moreno wrote:
After thinking calmly about it and trying to explain it
to someone else (someone even less experienced than I am),
I find myself struggling to understand the physical/
mathematical/geometric reasons why this is so.

I've got a short bit on this at

http://astro.isi.edu/notes/magnify.html

Brian Tung
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

David Nakamoto wrote:

Anything with large secondary
obstruction pushes the number down. For bad to fair seeing the number
goes down. Hence most of the time, through the most commonly found
scopes (SCTs and Newtonians), the figure is around 20x to 30x per inch
of aperture.

Not necessarily. This opposition, I have regularly used 446x to as high as
588x on Mars using my 10 inch f/5.6 Newtonian (22% central obstruction). On
some deep-sky objects such as planetary nebulae, I have gone as high as 720x
to tease out that extra bit of detail, although 500x to 600x is more typical.
All it requires is reasonably-good optical quality and good seeing. Clear
skies to you.
--
David W. Knisely
Prairie Astronomy Club: http://www.prairieastronomyclub.org
Hyde Memorial Observatory: http://www.hydeobservatory.info/

**********************************************
* Attend the 10th Annual NEBRASKA STAR PARTY *
* July 27-Aug. 1st, 2003, Merritt Reservoir *
* http://www.NebraskaStarParty.org *
**********************************************

"Joe S." writes:

"Mike Jones" wrote in message
...
Carlos,
Pure EE/optics answer, programmable with a little effort using a 2-D
fast Fourier transform (FFT), so you can experiment with it and teach
yourself:

Treat wavefront at telescope pupil as a complex phasor:
T(x,y) = A(x,y) exp[jP(x,y)]
where A(x,y) is the amplitude and P(x,y) is the combined optical path
difference (OPD) phase error from telescope aberrations and time-varying
atmospheric propagation:
snip

As a beginner, and a liberal arts graduate from long ago, I gotta tell you,
I love it when you engineer guys talk like that. Didn't understand a word
of it but it sure is fine to listen to. Or, read, in this case.


Joe S.
Damn. I'm an EE and I haven't been able to talk like that in years. I
feel incredibly average now, thank you very much.

-Al