Photons per Pixel

The photons one pixel collects in one exposure:

N ≈ B × η(NA) × T × A_{pixel at sample} × t

where B is specimen emission per unit area per unit time, η is the fraction of emitted light the objective collects, T is the transmission of everything between objective and sensor, and A is the specimen area one pixel sees.

The trap is in that last term. A_{pixel at sample} = (pitch / M)², which already contains 1/M². Writing the expression with pixel area at the sample and a separate division by M² charges magnification twice and makes photons fall as M⁻⁴ instead of M⁻². At 60× against 25× that is an error of a factor of 5.8 — not a rounding problem, a different answer.

Write it one way and stay there: either pixel area at the specimen, or physical pixel area divided by M². Never both.

Collection efficiency is usually written as proportional to NA². That is the small-angle approximation, and high-NA objectives are not operating at small angles.

The fraction of a point emitter's isotropic output entering a cone of half-angle θ is (1 − cosθ)/2, with sinθ = NA/n. In oil at n = 1.515: NA 1.15 collects 17.4% of the full sphere, NA 1.25 collects 21.7%.

The ratio between them is 0.802 exactly, where NA² predicts 0.846. The approximation understates the penalty of dropping NA by about 5%, and it does so in the direction that flatters the weaker objective — which is the wrong direction for a purchasing decision.

For rough reasoning NA² is fine. For a written comparison between two objectives, use the exact form; it is one line of arithmetic.

A 3.7 µm pixel at 25× sees 148 nm on a side, or 0.0219 µm². A 6.5 µm pixel at 60× sees 108.3 nm, or 0.0117 µm². The first covers 1.87× the specimen area.

Now include the objectives, 1.1 NA against 1.2 NA. Exact collection efficiencies are 15.6% and 19.5%, a ratio of 0.802. So the photon ratio is 1.87 × 0.802 = 1.50×, not 1.87×.

Quoting the area ratio alone would have claimed a 1.9× advantage where the truth is 1.5×. Using NA² instead of the exact form would have given 1.57×. The lower-magnification, lower-NA configuration still wins on photons per pixel — the conclusion survives — but by half as much as the careless version claims, and the margin is now small enough that sampling and read noise decide the outcome.

Photons per pixel is worth computing because it tells you which noise regime you are in. Below roughly 5 detected photons, read noise dominates and a sub-electron sensor is genuinely decisive. Above about 10, shot noise dominates and every competent camera converges to within a few percent.

What it cannot do is rank two cameras. Pixel area is in it, so the answer partly describes how each sensor divides the field rather than how well it detects. Two cameras can differ by 37% in per-pixel SNR on one specimen and by about 2% per resolution element.

So use it to locate yourself on the noise curve, then convert to photons per unit specimen area or per resolution element before comparing anything to anything.