= Confocal excitation and detection point spread function
{title2=$H_a(\mathbf r)=E(\mathbf r)D_a(\mathbf r)$}
For an isolated fluorescent molecule at displacement $\mathbf r$ from the scanning <optical focus>, write $E(\mathbf r)$ for relative excitation <light intensity> and $D_a(\mathbf r)$ for the fraction of its collected emission passed by a centred detection pinhole of radius $a$. The expected response, apart from a common brightness and collection factor, is
$$
H_a(\mathbf r)=E(\mathbf r)D_a(\mathbf r),\qquad
D_a(\mathbf r)=\int_{|\mathbf u|\leq a}p_{\rm em}(\mathbf u;\mathbf r)\,d^2\mathbf u,\qquad
\int_{\mathbb R^2}p_{\rm em}(\mathbf u;\mathbf r)\,d^2\mathbf u=1.
$$
Here $\mathbf u$ is position in the detection-pinhole plane and $p_{\rm em}$ is the emitted <light intensity> distribution there conditional on collection, with units of inverse <area>. This <point spread function normalization> retains throughput: $D_a$ is a fraction, not a curve rescaled to have peak one. The product expresses excitation followed by detection; it is an intensity/probability product, not multiplication of coherent <wave amplitudes> from different molecules. <Confocal microscopy (chapter manuscript) by Brad Amos and others>[Manuscript p. 16, Fig. 4 discussion; Appendix §5, p. 37, Eq. A17: illumination-times-detection response]{ref}
Assume linear, unsaturated <fluorescence>, independent emitters whose <light intensities> add, a stationary specimen, fixed collection efficiency and <quantum efficiency>, and negligible changes from scattering, absorption or bleaching during a scan. Different excitation and emission <wavelengths> generally give different excitation and detection responses. For a sufficiently small pinhole, $D_a$ is approximately its <area> times $p_{\rm em}$ at the centre; this is the familiar excitation-times-detection <point spread function>. It is not universally the square of one <point spread function>. A spatially invariant response further allows the image to be expressed as a <convolution> with the emitter distribution; depth-dependent <optical aberrations> can invalidate that approximation.
For an explicitly synthetic axial example, put $t=z/z_0$ and choose an on-axis excitation $E(t)=1/(1+t^2)$, inspired by a <Gaussian beam>. Stipulate a circular <Gaussian function> for the detection-plane <light intensity>, with standard deviation $s(t)=s_0\sqrt{1+t^2}$. In dimensionless coordinates $\mathbf v=\mathbf u/s_0$, put $\alpha=a/s_0$. Then
$$
p(\mathbf v;t)=\frac{\exp[-|\mathbf v|^2/(2(1+t^2))]}{2\pi(1+t^2)},\qquad
D_\alpha(t)=1-\exp\!\left[-\frac{\alpha^2}{2(1+t^2)}\right],\qquad
H_\alpha(t)=\frac{D_\alpha(t)}{1+t^2}.
$$
The expression for $D_\alpha$ follows by integrating $p$ over a <disk> using <polar coordinates>. The assumptions specify a toy model with matched axial scales; its parameters are not a calibration for a real objective <optical lens>. It omits <Airy disk> rings, vector <diffraction>, and unequal excitation and emission <wavelengths>.
\Image[/confocal-synthetic-axial-pinhole-response.png]
{title=Synthetic axial response: narrower pinholes reduce throughput and suppress <defocus>}
{id=confocal-synthetic-axial-pinhole-response}
{description=Original Gaussian toy model, not measurements. The left panel keeps a common signal scale; the right divides each response by its own peak to compare widths. Here $\alpha=a/s_0$ is a radius in units of detection-plane Gaussian standard deviation, not an Airy-unit diameter. The dashed reference is the same scanned excitation with all collected emission accepted, not conventional uniform widefield illumination. Both panels retain nonzero out-of-focus tails.}
For a vanishingly small pinhole, the peak-normalized synthetic response tends to $(1+t^2)^{-2}$, whose <full width at half maximum> is $2\sqrt{\sqrt2-1}\,z_0\simeq1.2872z_0$. With all collected emission accepted it tends to $(1+t^2)^{-1}$, of width $2z_0$. Meanwhile $H_\alpha(0)=1-e^{-\alpha^2/2}\to0$ as the pinhole closes. The narrower normalized curve therefore accompanies a vanishing detected signal; peak normalization must not conceal that loss.
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