Astronomy Picture of the Day

If you can find it the American Scientific guide to telescope making is a good resource. I read it like 50 years ago and it had some good information.It shows you how to check the figure on your mirror with simple equipment and even how to build a rough in machine for initial grinding. Never did make a scope however.

@Thalassokrator said in #7:

A telescope at Lagrange point L2 (en.wikipedia.org/wiki/Lagrange_point#L2_point), about 1.5 million km (≈930,000 miles) away from mother earth, is ill-suited as a spy satellite. Espionage is done from low earth orbit (altitudes of 250 to 1000 km or 155 to 620 miles), for example:
en.wikipedia.org/wiki/KH-11_KENNEN

This makes sense, because in espionage, small surface features really matter. So you want to be as close as possible (for your limited angular resolution to translate into optimal ground sample distance) without deorbiting due to atmospheric drag. Travelling 1% of the Earth-Sun distance away from the earth (that's about 6,000 times farther away than typical spy satellites) makes absolutely no sense whatsoever for this purpose.

Thank you, thank you, for taking the time to detail why the idea the Webb is a spy satellite is plainly misguided, to say the least, simply due to where they put it. But, hey, at least it would be harder to shoot down from Earth with lasers, lol.

@JCee62 said in #10:

Makes me wonder if anyone here has ever ground their own telescope mirror? I want to try it someday. I even have a record of H.G. Wells War of the Worlds broadcast to use as bearings. Can't find glass thick enough to do it, and am too cheap to buy it. Would be an interesting project.

My father ground a telescope lens by hand once. It's a long, tedious and very painstaking exercise. He had to literally keep count of every revolution, and you have to keep the same pressure all the time on both sides. And take into account temperature and humidity, and adjust accordingly.
You need special glass.

Don't grind your own, just go to Edmund Scientific ...

@Eyvazova_2009 said in #1:

It's magnificent, isn't it?

That's gotta be my new background.

Dude i like astronomy

@boilingFrog said in #9:

Dude, CGI bro

Whatever you say, bro. The Phantom Galaxy (Messier 74) was discovered 242 years ago and has been imaged thousands of times since then. Yet here you are, claiming that the new JWST image that looks just like all other images of this galaxy (just better resolution than previous infrared images) is "CGI".
Do you seriously think that hundreds of professional astronomers would be fooled by CGI when they have known and studied this object for the past 242 years?

But anyways, let's have some fun, shall we? Let's assume the official technical specifications of the JWST to be correct and let's estimate the angular resolution that the JWST should have according to basic physics (i.e. the Rayleigh criterion). Consequently I will compare that resolution to what we see in the new JWST picture of M74.
If it's a real image and a real telescope the theoretical prediction should match what we see in this image (up to measurement uncertainty), shouldn't it?

HYPOTHESIS: The predicted angular resolution θ ≈ 1.22*(λ/D) should only be slightly smaller than the smallest angular separation between two distinguishable light sources θ_s observed in the image.

How JWST can take colour images:

First we have to know how astronomical colour images like this are produced: Space telescopes like the JWST take black and white pictures only (light intensity on a grayscale). They are not alone in this, cameras do something similar.

How then can we come to know what colour something is? Well, the JWST doesn't just take one picture. It takes multiple pictures through multiple filters. Bandpass filters (https://en.wikipedia.org/wiki/Band-pass_filter) that let certain EM-wave frequencies (light colours) through while attenuating (or blocking) other frequencies.
So you might have a filter that only passes orange light while blocking all other colours. Or one that only passes red light while blocking all other colours. And many more such filters. Take a grayscale picture through each of these filters and you can retrieve the colour information by colouring the individual grayscale pictures in the colour associated with their respective filters and overlaying them.

Collecting relevant information:

So I looked up which colours were associated with which instrument filter. Robert Eder, who processed the raw data into the above APOD, kindly provided this information:
https://www.astrobin.com/3782ws/0/

I picked a MIRI (Mid-Infrared Instrument) filter, the F2100W filter in particular. Robert Eder decided to colour-code it in red in his false colour image (human eyes cannot see infrared light, so a true colour image would be pointless).

The F2100W bandpass filter passes mid-infrared light with wavelengths between 18.5 μm and 23.5 μm (according to the first diagram in https://jwst-docs.stsci.edu/jwst-mid-infrared-instrument/miri-instrumentation/miri-filters-and-dispersers). Let's take the arithmetic mean of those and say that things colour-coded in red in Robert Eder's image should correspond to observations made by MIRI at a wavelength of 21.0 μm. This figure can also be found as the λ_0 of the F2100W filter in Table 1.

Calculating a theoretical prediction:

According to the theoretical Rayleigh criterion (https://en.wikipedia.org/wiki/Angular_resolution#The_Rayleigh_criterion), the angular resolution θ of any given telescope should be given by

θ ≈ 1.22*(λ/D),

where θ (theta) is the angular resolution in radians, λ (lambda) is the wavelength of light and D is the diameter of the telescope's primary mirror. The angular resolution θ is the minimal angle of separation at which two light sources can still be distinguished.

In our case, λ = 21 μm and D = 6.5 m. This gives a theoretical prediction of:

θ ≈ 1.22*([21 μm]/[6,500,000 μm]) = 3.9415*10^(-6)

Multiplying by 180º/π, the angular resolution should be about θ ≈ 0.00022583º. There are 60 arc minutes (') in a degree and 60 arc seconds ('') in an arc minute. So this angle corresponds to θ ≈ 0.81'' (arc seconds)

Finding suitable light sources in Eder's APOD image:

Then I picked two red light sources northeast (in Eder's image) of the centre of the galaxy. They are located between the galaxy's two main spiral arms. In case you want to find them yourself and check my work, they are to be found at pixel coordinates (x, y) = (1369, 788), where the x-axis goes from left to right and the y-axis goes from bottom to top. You can download the image I used from https://apod.nasa.gov/apod/ap220722.html

I picked those two particular red light sources because they were imaged using MIRI's F2100W filter (otherwise they wouldn't be colour-coded in red) and because they nearly meet the Rayleigh criterion (like the middle Airy diffraction pattern in this figure: https://commons.wikimedia.org/wiki/File:Airy_disk_spacing_near_Rayleigh_criterion.png).

They can still be distinguished as two different light sources. The pixel separation of their maxima is 4 pixels in the x-direction and 3 pixels in the y-direction. Using the Pythagorean theorem their pixel separation comes out as √(4^2 + 3^2) = √25 = 5 pixels.

Estimating the angular width of the field of view:

The apparent angular size (in visual light) of M74 on the night sky is 10.5 arc minutes by 9.5 arc minutes.
Robert Eder's APOD picture has a width of 2500 pixels. That means that 2500 pixels should correspond to about 10 arc minutes (probably a bit less because we don't see the entire galaxy in the picture). One pixel should therefore at the very most correspond to 10/2500 = 0.004 arc minutes, or 0.24 arc seconds. Again, probably a bit less because the field of view having an angular width of 10 arc minutes is likely an overestimation.

Putting it all together:

With one pixel at most corresponding to 0.24 arc seconds, the angular separation θ_s of the two red light sources's maxima should be at most five times that:

(angular separation) ≤ (pixel separation)*([angular width of the field of view]/[pixel width of the field of view])

θ_s ≤ (5 pixels)*(0.24 arc seconds/pixel)
θ_s ≤ 1.2 arc seconds

This is in excellent agreement with the theoretical prediction of θ ≈ 0.81 arc seconds arrived at via the Rayleigh criterion.

Potential sources of statistical error include a very small uncertainty in the exact wavelength of light (I've simply assumed λ_0 = 21 μm, but wavelengths could range between 18.5 μm and 23.5 μm) and a larger uncertainty in the angular size of the field of view (I've assumed 10 arc seconds). I've roughly accounted for the latter by writing the formula for θ_s as an inequality. A more thorough account would have to resort to propagation of uncertainty (https://en.wikipedia.org/wiki/Propagation_of_uncertainty) in order to calculate the statistical uncertainties precisely, but I've omitted this process here for (relative) brevity's sake.

IN SUMMARY:

I have tested the hypothesis formulated in the beginning of this post. Indeed the theoretically possible angular resolution θ ≈ 0.81 arc seconds is only slightly smaller than the smallest observed angular separation between red (F2100W filtered) light sources in the image θ_s ≤ 1.2 arc seconds.

0.81 arc seconds = θ ≤ θ_s ≤ 1.2 arc seconds

The physics all checks out as we would expect it to given that Eder's APOD is an authentic image of a known galaxy processed from publicly available data by the James-Webb Space Telescope.