= Integrating-sphere escape probability
{title2=$P_{\rm esc}$}
The \b[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 $q=\rho(1-f)$.<Integrating Sphere Theory and Applications by Labsphere>[§1.2, printed p. 2, equations 6–12; §1.2.2, zero port reflectance versus general average reflectance]{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 $n\ge1$ has probability $q^{n-1}e$. Summing the <geometric series> gives
$$
P_{\rm esc}=\sum_{n=1}^{\infty}q^{n-1}e
=\frac{\rho f}{1-\rho(1-f)},\qquad q<1.
$$
Similarly, $P_{\rm abs}=(1-\rho)/(1-q)$ and $P_{\rm esc}+P_{\rm abs}=1$. These are an original probability derivation of the guide's repeated-reflection bookkeeping, not an additional source of optical energy.
\Image[/integrating-sphere-escape-and-survival.png]
{title=Ideal photon accounting from the first wall encounter: eventual escape versus port fraction, and cumulative escape, absorption and survival for a stated reflectance and port fraction}
For an original numerical example, $\rho=0.96$ and $f=0.04$ give $q=0.9216$, so
$$
P_{\rm esc}=\frac{0.0384}{0.0784}=\frac{24}{49}\approx0.4898,
\qquad P_{\rm abs}=\frac{25}{49}\approx0.5102.
$$
Although 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 $P_{\rm esc}(1-q^N)$, the absorbed fraction is $P_{\rm abs}(1-q^N)$, 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.<Integrating Sphere Theory and Applications by Labsphere>[§1.2.2, average reflectance; §2.3, baffles and detector field of view]{ref} The probability above is therefore a useful ideal-model prediction, not a universal detector efficiency.
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