The diffraction limit of the Hubble ST

[TheMan112]

I'm trying to calculate the diffraction limit/angular resolution for the Hubble Space Telescope. I know this can be found using the formula:

\theta = 1.22 \frac{\lambda}{D}

Where \lambda is the wavelength of the light being observed and D is the diameter of the objective lens (2.5 m on Hubble).

Now since Hubble is able to observe a wade range of wavelenghts from the ultraviolet to the visible to the infrared spectrum I would describe the diffraction limit as an interval depeding on this range of wavelenghts.

However, in all examples I've been able to find, even http://www.nasa.gov/missions/highlights/webcasts/shuttle/sts109/hubble-qa.html" , the diffraction limit of optical telescopes is calculated using a single wavelength of 500 nm (~cyan). What's the justification for using this particular wavelength for Hubble (and other optical telescopes).

500 nm coincides with:

1. The irradience top of sunlight (and thus any G-type star). See: http://en.wikipedia.org/wiki/File:EffectiveTemperature_300dpi_e.png
2. The sensitivity top of the human eye.
3. The approximate middle wavelength of the visible spectrum.

However (respectively):

1. G-type stars aren't all that common, comprising only 7.6% of all stars, thus the usage is limited and this point cannot be applied to all visible starlight.
2. The HST employs CCD sensors and not direct human observation, thus point 2 is irrelevant.
3. This is the only point that seems relevant. However the HST would theoretically achieve greater angular resolution by using say the 380-450 nm range.

So what's the point of using 500 nm when calculating the diffraction limit?

 

[mgb_phys]

For optics you tradiationaly use the Sodium D line (532nm) because before lasers it was the easiest bright monochromatic source.
The diffraction limit of the HST is different at each wavelength, the optics aren't necessarily diffraction limited at all wavelengths or the detectors have enough resolution to sample the diffraction limit.

 

[TheMan112]

Yeah, it apprears the detectors on WFPC2 can resolve down to 0.04 arcsec, given that lamda = 500 nm we get an angular resolution for the telescope itself at roughly 0.05 arcsec. The other camera appears to have much worse resolutions. Thanks!