by VictorBorun » Mon Jul 18, 2022 3:08 pm
Ann wrote: ↑Thu Jul 14, 2022 5:18 am
JWST User Documentation wrote:
JWST NIRCam offers 29 bandpass filters in the short wavelength (0.6–2.3 μm) and long wavelength (2.4–5.0 μm) channels.
The JWST NIRCam is the camera detecting the shortest infrared wavelengths. There is another camera too, which I
think is called MIRI (Mid Infrared Instrument) (but you'll have to google, because I'm too lazy), which detects longer, mid infrared wavelengths.
So anyway. If I understand things correctly, and 0.6 μm is the same thing as 600 nm, then the shortest wavelength that JWST can detect corresponds to this visible color:
███
So this is the shortest wavelength that JWST can detect, and any emission at this wavelength will be mapped as blue
███. (The color sample I just showed you corresponds to a wavelength of 441 nm, which probably qualifies as the height of "blue-ness" that the human eye can detect.) The difference in wavelength between 600 nm and 441 nm is not so large, only some "160 steps on the wavelength meter" (or how do you put that in words?).
All right. But please note that 2.3 μm, 2300 nm,
also counts as a short wavelength to JWST, and this wavelength will, perhaps, be mapped as this visible color:
███ (This color sample corresponds to 550 nm, which is a wavelength that is sometimes used by Hubble as a green filter.)
Let's assume that 2300 nm is mapped as "green", or 550 nm, by JWST. The difference between 550 nm and 2300 nm corresponds to "1750 steps on the wavelength meter". That's a lot.
2300 nm is way, way off the optical scale of wavelengths that the human eye can detect. After all, when Hubble takes pictures in the infrared, it is usually no more infrared than - if I remember correctly - 814 nm. 2300 nm is almost three times longer than the longest infrared wavelength than Hubble can see.
Ann
How do we compress a waveband uniformly if it is wide? And even wider than Hubble + Webb NIR, if you look at
today's APOD: — which combine those with Hubble UF and Webb MIR.
It's as simple as this: we use logarithmic scale. Like tones in a melody, like decibels in a loudness and like pH in an acid.
You take your target band where we humans are sensitive to the wavelength: from violet 435 nm to red 625 nm.
You ignore the margins that we humans can see but can not tell the colour difference, < 435 nm or > 625 nm.
You take your source band which is from
UV 200 nm to
MIR 26000 nm..
And here is your mapping: present any source wavelength λ_source with base 10 logarithms as
10^(log(435) + (log(625) - log(435))(log(λ_source) - log(200))/(log(26000) - log(200))).
PS Coding in Wolfram's Mathematica, made it base e logarithms, to the same effect:
- Attachments
-
[quote=Ann post_id=324234 time=1657775910 user_id=129702]
[quote][url=https://jwst-docs.stsci.edu/jwst-near-infrared-camera/nircam-instrumentation/nircam-filters]JWST User Documentation[/url] wrote:
JWST NIRCam offers 29 bandpass filters in the short wavelength (0.6–2.3 μm) and long wavelength (2.4–5.0 μm) channels.[/quote]
The JWST NIRCam is the camera detecting the shortest infrared wavelengths. There is another camera too, which I [b][i]think[/i][/b] is called MIRI (Mid Infrared Instrument) (but you'll have to google, because I'm too lazy), which detects longer, mid infrared wavelengths.
So anyway. If I understand things correctly, and 0.6 μm is the same thing as 600 nm, then the shortest wavelength that JWST can detect corresponds to this visible color: [color=#ffbb00]███[/color]
So this is the shortest wavelength that JWST can detect, and any emission at this wavelength will be mapped as blue [color=#000bff]███[/color]. (The color sample I just showed you corresponds to a wavelength of 441 nm, which probably qualifies as the height of "blue-ness" that the human eye can detect.) The difference in wavelength between 600 nm and 441 nm is not so large, only some "160 steps on the wavelength meter" (or how do you put that in words?).
All right. But please note that 2.3 μm, 2300 nm, [b][i]also[/i][/b] counts as a short wavelength to JWST, and this wavelength will, perhaps, be mapped as this visible color:[color=#a3ff00] ███[/color] (This color sample corresponds to 550 nm, which is a wavelength that is sometimes used by Hubble as a green filter.)
Let's assume that 2300 nm is mapped as "green", or 550 nm, by JWST. The difference between 550 nm and 2300 nm corresponds to "1750 steps on the wavelength meter". That's a lot.
2300 nm is way, way off the optical scale of wavelengths that the human eye can detect. After all, when Hubble takes pictures in the infrared, it is usually no more infrared than - if I remember correctly - 814 nm. 2300 nm is almost three times longer than the longest infrared wavelength than Hubble can see.
Ann
[/quote]
How do we compress a waveband uniformly if it is wide? And even wider than Hubble + Webb NIR, if you look at [url="https://asterisk.apod.com/viewtopic.php?t=42532"]today's APOD:[/url] — which combine those with Hubble UF and Webb MIR.
It's as simple as this: we use logarithmic scale. Like tones in a melody, like decibels in a loudness and like pH in an acid.
You take your target band where we humans are sensitive to the wavelength: from violet 435 nm to red 625 nm.
You ignore the margins that we humans can see but can not tell the colour difference, < 435 nm or > 625 nm.
You take your source band which is from [url="https://www.researchgate.net/figure/Band-passes-of-the-HST-STIS-and-WFPC2-filters-The-STIS-F25SRF2-and-FCN182NM-and-the_fig1_2234777"]UV 200[/url] nm to [url="https://jwst-docs.stsci.edu/jwst-mid-infrared-instrument"]MIR 26000 nm[/url]..
And here is your mapping: present any source wavelength λ_source with base 10 logarithms as
10^(log(435) + (log(625) - log(435))(log(λ_source) - log(200))/(log(26000) - log(200))).
PS Coding in Wolfram's Mathematica, made it base e logarithms, to the same effect: