# What is the probability that a photon that enters the aperture is registered?

If I want to derive the probability that a photon that enters the aperture of a telescope is detected, how do I do that?

This should be closely related to the quantum efficiency of a telescope, but from what I read it is not quite the same thing. Quantum efficiency is the ratio between the number of incident photons and the number of charge carriers collected at the terminals of a photosensor. But does every charge carrier get registered? Is it possible to derive the probability of a photon getting registered from a telescope's specifications?

If a photon enters the aperture, then it has been captured. What you want to know is if that photon has been detected?

The calculation is relatively straightforward. Some fraction of the photons are lost in reflections from the mirrors or in passing through other optical elements or correctors that are in front of the detector - for example coloured filters or atmospheric dispersion correctors. These efficiencies are usually very quite high for a simple imaging cameras but can be much lower for complex spectrographs or fibre-fed systems.

The majority of photons will then impact the detector - usually a CCD if we are talking about optical telescopes. The CCD will have a well known detection efficiciency for photons that will be a function of wavelength. Modern CCD detectors on telescopes have detection efficiencies of 70+% across quite a broad wavelength range. These efficiencies are normally published/available and are usually what is meant when referring to the quantum efficiency of the detector - the fraction of incident photons that are detected.

Most "exposure time calculators" at big observatories will do these calculations for you, usually implicitly (i.e. they don't break up the calculation into its components for you). For example, I ran the ESO Imaging-mode calculator for the FORS2 instrument on the VLT, just accepting all the default values (a 20th magnitude A0V star observed for 100s through a V-band filter) and checked the "system efficiency (excluding atmosphere)" box. This returned the following graph, which I think is the number you are looking for, as a function of wavelength.

A more useful example might be the Isaac Newton Group exposure time calculator. I ran this with all the defaults set but selected WHT Prime Focus Imaging with the WHTWFC imaging CCD. The output gives the following useful information:

"Atm * tel * instr throughput 0.74 = 0.87 * 0.85 * 1.00 " "Detector efficiency 0.80"

Looking at the help page, we see that what this means is the numbers above account for losses (extinction) in the atmosphere, losses due to reflectivity, the throughput of the instrument (this is losses in the filter being used and there are apparently no losses for the V-bnd filter) and the average detector efficiency over the wavelengths being observed. The number you would want is $$0.85 \times 1.00 \times 0.80 = 0.68$$. i.e. 68% of the photons entering the telescope are detected. This is quite wavelength dependent. If we switch the calculator to a U-band filter we get $$0.85 \times 0.60 \times 0.55 = 0.28$$.

• "If a photon enters the aperture, then it has been captured? Surely what you want to know is if the photon is detected?" Indeed. Fixed. Dec 25, 2022 at 23:01
• Great answer. So the summary is that it can not easily be calculated from typical telescope specs I would find on e.g. wikipedia, but I'd have to use the telescope-specific calculation tools to find it. That is a bit of a bummer. But between 20% and 90% seems to be a safe bet for large astronomical telescopes, based on the examples you gave. Dec 26, 2022 at 21:03
• ... although it can get much lower depending on the type of instruments and other hardware used (e.g. interferometry). Dec 26, 2022 at 21:12
• @JanKanis the components of the calculation are not secret but differ for every telescope and instrument. Dec 26, 2022 at 22:48