Integrating-sphere escape probability ($P_{\mathrm{esc}}$)

The integrating-sphere escape probability is the chance that a photon in a specified integrating sphere model eventually leaves through a port rather than being absorbed by the wall. It depends on both the port area fraction $f$ and wall reflectance $\rho$, as well as where the model begins. In the ideal uniformly diffuse model with nonreflecting ports, the integrating-sphere port-loss factor is $1-f$ and the survival probability between successive wall encounters is $q=\rho (1-f)$.^[ref]

Start immediately before a wall encounter. The photon is absorbed there with probability $a=1-\rho$, reflects and escapes before the next wall encounter with probability $e=\rho f$, or returns to the wall with probability $q$. These satisfy $a+e+q=1$. Under the same well-mixed encounter rule each time, eventual escape at encounter $n\ge 1$ has probability $q^{n-1}e$. Summing the geometric series givesSimilarly, $P_{\mathrm{abs}}=(1-\rho )/(1-q)$ and $P_{\mathrm{esc}}+P_{\mathrm{abs}}=1$. These are an original probability derivation of the guide's repeated-reflection bookkeeping, not an additional source of optical energy.

For an original numerical example, $\rho =0.96$ and $f=0.04$ give $q=0.9216$, soAlthough only four percent of the spherical area is open, almost half the initial photon ensemble eventually exits: surviving photons get repeated opportunities. After $N$ completed encounters, the escaped fraction is $P_{\mathrm{esc}}(1-q^N)$, the absorbed fraction is $P_{\mathrm{abs}}(1-q^N)$, and the fraction still inside is $q^N$. The three fractions sum to one for every $N$.

The stopping encounter has a geometric distribution with termination probability $1-q$. Its expected count is $1/(1-q)$, about 12.76 in this example. This counts wall encounters; it does not specify elapsed time without a model of path lengths and propagation.

Boundary cases clarify the assumptions. With $\rho <1$ and $f=0$, every photon is eventually absorbed. With $\rho =1$ and $f>0$, every photon eventually escapes. If both $\rho =1$ and $f=0$, the ideal photon never terminates and the formula's $0/0$ is inapplicable. Starting after a successful first reflection instead would also give a different conditional escape probability; the starting point must accompany the result.

A measurement can vary a known aperture while injecting a stable radiant flux that first strikes the wall. A monitored detector then tests the apparatus response against a model including its viewing geometry. A single detector signal is not automatically the total energy escaping through all ports. Real coatings, reflective port fixtures and direct paths require corrections; baffles prevent the receiver from directly viewing an unusually bright first-illumination patch.^[ref] The probability above is therefore a useful ideal-model prediction, not a universal detector efficiency.