Image of the day

From the
ATWB Customer Gallery

NGC 6960, Veil Nebula

My Account

New to Astromart?

Register an account...

Need Help?

Image Resolution calculations.

Started by benkokes, 01/19/2005 09:27AM
Posted 01/19/2005 09:27AM Opening Post
Hey there.
I was asked the other day if I would be able to seee the Lunar Landers, and didnt know. But I whipped out my trusty HP48G and figured out that my eyepiece has a 4000foot viewing image at what I assumed the highest power.

How do I figure out what size objects in the sky that I can resolve? ( straight math, not taking atmospheric conditions into play)

Posted 01/19/2005 09:33AM #1
Hi Ben,

The theoretical resolution of a telescope is given by:

a" = (0.25 x wavelength in microns) / mirror diam in metres

Green light has a wavelength of about 500 nm or abour 0.4 microns, so using a 1m telescope (don't we wish!!!) the resolution would be 0.1 arc seconds.

Hope this helps.


* Teach a man to fish and you feed him for a day.
* Teach a man to use the internet and he won't bother you for weeks.
Posted 01/19/2005 10:28AM #2
No the lunar landers can't be viewed from earth. I'm quite certain about this.

I used to know another formula for calculating the resolving power of telescope but I can't recall it right now.

I do remember from an APOD (Astronomy Picture of the Day) regarding the Hubble telescope that its 1.2 meter(?) mirror can only resolve lunar objects down to around 80 meters. That further supports my earlier statement regarding earth based scopes.

Charles Packard
Posted 01/19/2005 03:15PM #3
I have heard that the smallest thing you can resolve on the moon from the earth is about the size of the Colliseum in Los Angeles (Or other large athletic facility).

However, if you want to see the landers, it is much easier to just go to wherever the soundstage is that they used to film the landings. :S

Posted 01/19/2005 03:24PM | Edited 01/19/2005 03:49PM #4
Hi Ben.

From a land surveyor's perspective: roughly (very) assuming a distance of 250,000mi to the moon (402,250Km or 402,250,000m) using basic trig for a 3m object at that distance:

[arctan(1.5m/402,250,000m)]*2 would equal resolving power in terms of "arcsec" of 0.00154" (or considering that the difference is negligible in this case: arctan(3m/402,250,000m) would suffice).

That is to say: if you were to be capable of detecting the "speck" on the moon that would be the 3m base of the LEM, you would require a telescope with a resolving power of about 0.00154 arcseconds.

0.68"/0.00154" = 442. You would require a telescope with roughly 442 times the resolving power of what you are currently using to detect a 3m object on the moon (assuming no atmosphere).

Posted 01/19/2005 04:43PM | Edited 01/20/2005 06:23AM #5
Well, Dawes limit is the most common referred to formula that most people are aware of for the calculation of some type of resolution, but it is not the basis for the measurement of angular resolution in optics. That is another of the mis-informations. Rayleigh Limit is the correct formula to use, as I described in detail above.

Dawes Limit is an empirical measure of the double stars that can be "seen" with a telscope, not resolved. In Dawes, you will always have overlap. Of course there will be those that can say, well if I can see it I have resolved it. But that would not be accurate. Seeing overlapped double stars, seeing resolved double stars and seeing an object that has not been enlarged by your telescope Airy disk are very different things.

NO object, no matter how small, can form an image in the telescope smaller than the Airy disk. Resolution is the determination of what size object can be seen to a dimension in the telescope image that is larger than the diameter of the central bright spot of the Airy disk. Any object smaller than that will be bloated up to the size of the Airy disk by nature of the telscope optics.

Dawes Limit has nothing to do with determining the size of the Airy disk in telescope optics, even yet another of the mis-informations spread on the internet and in text books.

And once again I will repeat, Rayleigh Limit is the radius of the Airy disk. Because only a percentage of the light is put into the central bright spot, you can see faint stars cleanly split at approx one half the diameter of the Airy disk, convieniently that is RL for faint stars. However you must double RL and calculate the diameter of the central bright spot to determine the size of a bright object that will form an image larger than the Airy disk. I gave a little more specific information in the previous post. see that post

Posted 01/19/2005 11:56PM #6
The math here is pretty easy. As mentioned before, the diffraction limit of a telescope is given by the wavelength of light divided by the aperture. Take an 8" telescope and 500nm light, and you have a resolution of 0.52". You get this by 500e-9/0.2 = 2.5e-6. Note that first I converted everything into meters, and the answer comes out in radians. Multiply the answer by 57.296 to convert from radians to degrees, then multiply by 3600 to convert to arcseconds. Actually, the answer in radians is more useful, as we shall see.
The distance to the moon is roughly 233,000 miles, which is 3.9e8 meters. Using the definition of the radian, the length of an arc (the edge of a circle) is the radius times the angle in radians. If we multiply the distance to the moon (as the radius) by the diffraction limit (as the angle), we get the linear distance on the surface of the moon that you can resolve. In this case, it's 970m, or a little less than a kilometer. The Hubble Space Telescope has a 2.4m mirror, which is about 10 time bigger than the case above, so the resolution is 10 times better. But this is still about 100m, or roughly the size of a football field. Even in the photos taken by the Apollo command module from lunar orbit, the LEM is just a tiny speck, and all you really see is the shadow it casts.

Mike Connelley
Posted 01/20/2005 08:28AM #7
And this is the beginners forum??... Man I'd hate to see the intermediate or advanced grin You've all left me way behind ( grin )

Mark Visser

Posted 01/21/2005 03:50PM #8
I saw this question posted either here, or at another site a while back. According to the person who wrote the reply, it was interesting to discover the only telescope which could possibly resolve the lunar lander (and only if modified) was the Hubble.

What you're attempting to resolve is an object that is approximately 3 meters wide, in a crater (if I remember correctly, and 225,000 miles away.

I'm not sure about the calculations behind this, but he definitely seemed to know what he was talking about. If anybody out there remembers seeing this post, share the info with us.

Clear Skies,

David Eccles
Lompoc. CA