Calculating the Resolving power of a Telescope

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Summary:
I'm trying to work out what "resolution" my telescope will have at various magnifications. And what the limit in magnification is based on how much light there is and the size of the object.
Lets assume I have a telescope with an aperture of 200 mm, if we also assume λ has a wavelength of 500 nm

According to;
http://labman.phys.utk.edu/phys222core/modules/m9/resolving_power.htm
θ = 1.22 λ/d, where θmin is the angular separation that can be resolved.

1.22 * 500 nm / 200 mm = 3.05

Assuming Jupiter has an apparent diameter of 49 arcseconds, and we can resolve differences in light down to 3 arc seconds which seems very low.
49 / 3.05 = 16

assuming nm needs to be converted to mm???
16 * 1,000,000 = approx 16,000,000 which seems way to high.

Lets now assume I have a telescope with a focal length of 2000 mm and a 5 mm eyepiece with an apparent field of view of 68° this gives a true field of view of 0.17° (68 / 2000 / 5). At 49 arc seconds Jupiter has an apparent field of view of 0.0136° and would therefore fill 8% of the eyepiece.

I can get a Barlow, a lens which is placed between the objective lens and the focal point, changing the angle of light, effectively simulating a longer focal length, this will increase magnification. However as the aperture remains the same, the telescope will resolve with the same power. For instance if I get a 4x Barlow to increase the focal length of the telescope to 8,000 mm, with a magnification of 1,600x, Jupiter will fill 33% of the eyepiece. (At this magnification the exit pupil will be too low for the eye to physically see)

Ultimately I want to add some additional formulas to my astronomy spreadsheet that gives a number I can compare with all my Telescopes / Eyepieces and Barlows, I can then use certain combinations to understand what that number means from a practical point of view. And then when I want to buy a new telescope/eyepiece/barlow, I can plug those values in and find out what I would expect to see. I can then optimise the hardware for every deep sky object I wish to view.

Does this make sense? Can anyone see a better way to understand, how blurry / faint an objective will be perceived?
 

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  • #2
davenn
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Summary:: I'm trying to work out what "resolution" my telescope will have at various magnifications. And what the limit in magnification is based on how much light there is and the size of the object.

assuming nm needs to be converted to mm???
16 * 1,000,000 = approx 16,000,000 which seems way to high.

indeed ... where did you get the 1,000,000 from ?

1nm = 0.000001mm
 
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Is that not the same as 1mm = 1,000,000nm ?
 
  • #4
Tom.G
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The formula you are using, θ = 1.22 λ/d, is for calculating the size of an Airy disc due to diffraction thru a circular aperture.

There are several other things to consider for ultimate resolving power of a telescope. For instance:

The accuracy of each lens and mirror surface,
any uncorrected astigmatism in your eye,
atmospheric stability & clarity,
and this from:
https://www.britannica.com/science/optical-telescope/Light-gathering-and-resolution
The resolving power of a telescope can be calculated by the following formula: resolving power = 11.25 seconds of arc/d, where d is the diameter of the objective expressed in centimetres.

Your question is perfectly valid, unfortunately with no simple answers.
You might try some of the posts listed in the Related Threads..., at the bottom of the page.

Cheers,
Tom
 
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Hi:

The number you are getting for resolving power (r) is in Radians.
r = 1.22 * .0005 mm/200mm
= 3.05 e-6 radians
to convert to arc seconds...
= 3.05 e-6 * 180/Pi* 3600 = Arc seconds
= 0.63 arc seconds

Cheers. Murray
 
  • #6
sophiecentaur
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I have two mutually confusing comments about this. The Rayleigh criterion is a very pessimistic one and makes assumptions about our vision (and not the potential of modern image processing). Otoh, the atmosphere seems to be the limiting factor for Earth based telescopes.

But, on the other other hand, the processing in modern large telescope arrays does a great job of overcoming many of the effects of simple 'Seeing'. So much so that there is an opinion that large terrestrial arrays have much greater potential than even the (imminent - hurrah) James Webb telescope for image gathering. (And you can open the box and mend them too)
 
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Vanadium 50
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. So much so that there is an opinion that large terrestrial arrays have much greater potential than even the (imminent - hurrah) James Webb telescope for image gathering.
Except that the atmosphere isn't so transparent in the wavelengths where JWST will observe.
 
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  • #8
sophiecentaur
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Except that the atmosphere isn't so transparent in the wavelengths where JWST will observe.
Haha - a point that I failed to make.
 

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