## Abstract

The trend to conduct volumetric particle tracking velocimetry (PTV) experiments with ever-increasing volumes, at a given particle density, poses increasing challenges on the design of such experiments in terms of the power of the laser source and the image analysis. These challenges, on one hand, require a reliable model to estimate the current signal from a pixel on a complementary metal-oxide semiconductor (CMOS) detector due to a Mie scattering particle. On the other hand, they require also a model for estimating the limiting factors upon the image resolution, where a large amount of particles within a three-dimensional (3D) volume are mapped into a two-dimensional (2D) image. Herein, we present a model that provides an analytical expression to estimate the current signal from a pixel of a CMOS detector due to a Mie scattering particle within an arbitrary large volume in a volumetric PTV experiments. We begin with a model for planar experiments and extend it into volumetric measurements. Our model considers the effect of the depth of field, particle density, Mie scattering signal and total Mie scattering loss, laser pulse energy, and other relevant optical parameters. Later, we investigate the consequence of the Rayleigh criterion upon the spatial resolution when it is applied to Mie particles within a volume of interest (VOI). Finally, we demonstrate how we applied our model to estimate the current signal and the limit upon the spatial resolution in three experiments carried out in our lab.

## 1 Introduction

The continual increase in computational power and the development of fast algorithms for particle tracking, such as shake the box (STB) [1], had been facilitating tracking of an increasing larger number of particles in tomographic particle imaging velocimetry (PIV) and PTV. Table 1 lists examples from the literature of several three dimensional-particle tracking velocimetry (3D-PTV) experiments using helium filled soap bubbles (HFSB), air-filled soap bubbles (AFSB), or Di-Ethyl-Hexyl-Sebacat (DEHS) particles. The values of the VOI, particle number *N*, particle diameter *d*, laser pulse energy *E _{p,}* and laser repetition rate

*f*

_{rep}are specified for each experiment. The last three rows in Table 1 are examples of 3D-PTV experiments (conducted in our lab) with small $1\u2009\mu m$ diameter DEHS particles, or with $15\u2009\mu m$ diameter AFSB particles. These experiments demonstrate that a high pulse-energy laser is required in order to obtain a signal in a relatively small VOI.

When one tries to work in an increasing VOI and with small tracer particles, one is confronted with the necessity to carefully optimize the light budget. A methodical study of 3D-PTV experiments shows that the current signal from a pixel of a CMOS camera depends on several factors: The initial laser pulse energy, the Fresnel losses due to guiding and shaping of the laser beam, scattering and absorption losses of the laser beam by the tracer particles and by the carrier fluid, the diameter of the tracer particle, the polar angle of the Mie scattering, the depth of field of the volume of interest, scattering and absorption losses of the Mie signal by the tracer particles and by the fluid; and the losses of the camera (transmission loss of the camera lens and the responsivity loss of the CMOS detector).

At the present, it is a common practice to study examples of existing experiments from the literature (c.f. Chapter 18 in Ref. [7]) and then conduct tests with existing equipment in order to estimate experimentally the attainable current signal and the spatial resolution for a given tracer and pulsed laser system in a given VOI. This type of experimental approach is inefficient in the design stage of 3D-PTV experiments. Since STB of volumetric PTV measurement are maturing (computationally) toward feasibility of tracking increasing amounts of particles in ever-increasing volumes, we recognize that there is a need to develop a sound and practical design method for volumetric experiments. In particular, it became evident to us that the dependence of the CMOS current signal and the spatial resolution upon the depth of field is not well understood. Thus, the main purpose of the present work is to develop an analytical (yet practical) model that can predict reliably the current signal from a CMOS pixel due to the imaged particles and the limit of the spatial resolution in 3D-PTV measurements.

## 2 Analytical Model for the Signal Level in Planar and Volumetric Particle Tracking Velocimetry

A simplified configuration of a 3D-PTV experiment is depicted in Fig. 1. A laser with pulse energy, *E _{p}*, irradiates a Mie particle, the particle scatters light into all directions according to Mie theory. Part of that scattered light passes through the camera lens and generates an image on the pixels of the CMOS detector. The image has an optical power

*P*. If the image fills the area of one pixel, the current signal

_{i}*i*equals to the incident optical power times the responsivity of the CMOS detector at the wavelength

_{S}*λ*of the laser, $iS=PiR(\lambda )$ (the responsivity has the units of Ampere/Watts).

*E*, pulse duration

_{p}*τ*and its Gaussian-elliptic beam profile. Let us approximate the shape of the laser pulse in the time domain to a square wave. Hence, $Ep=P\tau $, where the pulse power

*P*is now a constant during the pulse duration

*τ*. By incorporating this relation into the expression for an elliptic Gaussian laser beam with waists

*ω*and

_{a}*ω*[8], we obtain the intensity of the laser beam and its dependence on the pulse energy

_{b}*E*and pulse duration

_{p}*τ*

where the variables *r _{a}* and

*r*are the radial components along the semiminor and semimajor axes of the elliptic cross section of the laser beam, respectively.

_{b}Now, we are ready to calculate the current signal by tracking the path of the laser pulse from its output, until a signal is generated in a pixel of the CMOS detector due to scattering by a Mie particle. Since planar PTV is a special case of volumetric PTV with a minute depth of field, it is simpler to consider first the signal level in a planar PTV experiment and then extend the model to the case of a 3D-PTV experiment.

### 2.1 The Current Signal in Planar Particle Tracking Velocimetry.

The following six factors should typically be considered:

First, the Gaussian laser beam is shaped by lenses into an elliptical Gaussian beam and it is guided by reflecting mirrors into the volume of interest *V*, as depicted in Fig. 1. If the volume of interest is located within a glass chamber that is filled with a liquid, the beam will suffer Fresnel losses also from the glass chamber (or any other dielectric media). All the optical surfaces (of lenses, mirrors, and the chamber) will have an accumulative optical power-loss, which we designate by *L* (a factor that is smaller than 1). These losses will reduce the initial laser intensity, *I*, to $I(1\u2212L)$.

Second, when the laser beam propagates through a fluid (air or water, for example), it will suffer both scattering and absorption losses from the tracer particles and from the carrier fluid. The losses due to the tracer particles are characterized by scattering and absorption loss coefficients $\alpha s\u2212p$ and $\alpha a\u2212p$, respectively. The losses due to the fluid are characterized by scattering and absorption coefficients $\alpha s\u2212f$ and $\alpha a\u2212f$, respectively. For example, Mie particles with particle density *n* in air have a scattering loss coefficient $\alpha s\u2212p=\sigma sn$, where *σ _{s}* is the total Mie scattering cross section of one particle (It accounts for scattering into a full solid angle). In the air (the fluid), the scattering loss coefficient by molecules (known as Rayleigh scattering) and the absorption loss coefficient (due to molecular absorption) are negligible in the range of the visible spectrum. On the other hand, in water, Mie particles cause for scattering and absorption losses; these losses are characterized by $\alpha s\u2212p$ and $\alpha a\u2212p$. However, the water itself (the fluid) will show significant scattering losses due to density fluctuations and absorption losses, due to molecular absorption (see Chapter 43 in Ref. [9]). Let us assume that the laser beam will pass a distance

*z*through the seeded fluid until it reaches a particle that is located within the volume of interest. The laser intensity

_{l}*I*after a path

*z*will be $I(zl)=I(0)e\u2212\alpha zl$, where $\alpha =\alpha a\u2212p+\alpha s\u2212p+\alpha a\u2212f+\alpha s\u2212f$ accounts separately for the absorption and scattering losses by the tracer particles and by the fluid.

_{l}*I*into a differential unit of a solid angle by one particle is

_{s}where *I* is the intensity of the incoming laser and *σ* is the scattering cross section of a particle into a differential solid angle $\u2202\Omega $. The integration over $4\pi \u2009[str]$ will give the total Mie scattering cross section *σ _{s}*. We note that due to the symmetry of scattering from a sphere, the scattering cross section depends only on the polar angle

*θ*. The polar Mie scattering $\sigma (\theta )$ can be calculated by using openly available resources, e.g., (see Footnote

^{2}).

*θ*over the whole solid angle. However, in typical planar (and volumetric PTV), the solid angle extended from particles to the camera lens is very small. Therefore, taking $\sigma (\theta )$ to be a constant withing the small solid angle is a good approximation. This small solid angle is given by the ratio between the area of the camera lens to the surface area of a sphere that its center is at the location of the particle and its radius is

*z*(the working distance in planar PTV). For example, in planar PTV, the solid angle is given by

_{o}where *D _{f}* is the diameter of the lens. The intensity of Mie scattering into the camera lens will be approximately $Is=I\sigma (\theta )\Omega par$.

Fourth, the Mie scattering signal from a particle that is located at a distance *z _{o}* from the camera lens will suffer scattering and absorption losses similar to those suffered by the incoming laser pulse (as explained in the second factor). Accordingly, the intensity of the signal will fall as $Is(zo)=Is(0)e\u2212\alpha zo$.

Fifth, the camera lens usually consists of a combination of several types of lenses (to compensate for lens aberrations). If these lenses are not specifically coated by Anti-Reflection (AR) coatings, a considerable Fresnel loss will ensue. Let us designate the optical transmission of the camera lens at the laser wavelength, *λ*, by $T(\lambda )$.

Sixth (and final), one needs to consider the ratio of the area of one pixel to the area of the image of one particle. In order to avoid peak locking [7], it is advised that the particle image will have an area of at least four pixels. Therefore, the amount of optical power incident onto one pixel will be about $14$ of the total power of the image. We designate this ratio by fill factor ratio (*FFR*).

*τ*is used in a planar PTV experiment, the amount of incident optical power that arrives into the area of one pixel from one scattering Mie particle is found to be

_{par}, and multiplying it by the CMOS responsivity, $R(\lambda )$, we obtain an explicit expression for the current signal in one pixel for a planar PTV experiment

Observe that particles that are located at the center of the elliptical laser beam (where $ra=rb=0$) will have the maximal laser intensity (and therefore the maximum signal). While particles that are located where $ra=\omega a,\u2009rb=\omega b$ will have the lowest laser intensity and therefore the lowest current signal.

### 2.2 The Current Signal in Volumetric Particle Tracking Velocimetry.

In 3D-PTV, we adjust the diameter *D _{a}* of the camera aperture to obtain a depth of field $\Delta zo$ that equals to the thickness of our rectangular cuboid VOI. By applying two minor and judicious changes to Eq. (5), we can obtain an expression that estimates the current signal,

*i*, for volumetric PTV with a large depth of field.

_{S}First, we need to identify the particles within the VOI that would have the faintest signal. We note that particles at the far face (the plane at $zo+$) of the VOI will have a smaller solid angle, Ω_{par}, compared with particles located at the object plane or at the $zo\u2212$ plane (see Fig. 1). More specifically, considering the configuration of the camera and of the laser as is depicted in Fig. 1, we note that particles at the vertices of the far face of the VOI will have the smallest solid angle, Ω_{par}. The two vertices at the right hand with respect to the view seen by the camera will experience a laser beam with the lowest power due to scattering and absorption losses over the longest path. Furthermore, the polar Mie scattering by particles in these two vertices will be the weakest (generally, Mie scattering is maximal in forward scattering and falls down with increasing polar angle). Finally, these particles have the greatest distance to the camera lens and thus their Mie-signal will suffer the greatest scattering and absorption losses as it propagates toward the camera lens.

For simplicity, we will consider the signal from particles at the distance $zo+$ (the explicit expression for $zo+$ is found from (A2) and (A8) in the Appendix). Then, we will replace the lens diameter, *D _{f}*, with the aperture diameter,

*D*, in Eq. (5). By doing so, we account for the signal from particles with almost the faintest Mie scattering signal. If we need to detect a signal from

_{a}*all*particles within the rectangular cuboid VOI, we need to consider the signal from particles that are located at the vertices of the far face of the volume of interest, to the right with respect to the viewing camera.

Second, when the diameter of the camera aperture is decreased in order to match the required depth of field, Mie particles that are not in the object plane will appear blurred on the image plane. An ideal point-like light source at the object plane will appear in the image plane as a disk with diameter $2.44\lambda Da$ [11]. An ideal point-like light source at the far (or near) face of the VOI (relative to the camera view) will appear larger and blurred with diameter *CoC* (the Circle of Confusion, see explanation in the Appendix).

*FFR*with $FFR(CoC)$ in (5), yielding an expression for estimating the current signal in volumetric PTV

## 3 The Spatial Resolution in Planar and Volumetric Particle Tracking Velocimetry

To achieve the highest spatial resolution $\Delta V$ in a planar PTV experiment (or 3D-PTV), we would like to work with the smallest possible trace particles at the highest density. Therefore, let us consider two Mie scattering particles. What would be the image of such two adjacent point-like, light-source particles, which are located at the object plane?

Considering the wavelength of the illuminating laser and the distance of the camera lens from the particles, the scattered light that reaches the lens of the camera is as good as a coherent plane wave. Since our particles are point-like particles, we cannot use geometric optics to predict their image. The “image” of these two adjacent particles is a diffraction pattern.

According to diffraction theory, when a circular aperture (i.e., the lens or the camera aperture) is illuminated by a plane wave, one will observe a diffraction pattern that is called an Airy pattern (Ref. [11]). A single Airy pattern is made of a central disk (the Airy disk) that is surrounded by a series of dark and lighted concentric rings (Airy rings). When two adjacent point-like, light sources, are imaged by a lens with diameter *D _{f}*, we will get two adjacent Airy patterns on the image plane.

*The Rayleigh Criterion*for optical angular resolution is an ad hoc criterion, yet, empirically verified [11]. The criterion tells us that when the maximum of the first Airy disk (from one point-source) falls on the first dark ring of the second Airy pattern (from the second point-source), one can just barely deduce that the diffraction pattern is due to two different point-like light sources. In that case, the angular distance between the two light sources is found to be

The Rayleigh criterion sets a physical limit (due to diffraction) upon the *optical resolution* between two particles. $\Delta \theta Rzo$ is the minimal distance between two particles in the object plain that could be resolved in the image plane as two separate light points.

Apart from the physical limit set by the Rayleigh criterion, we identify four technical factors that may limit the spatial resolution $\Delta V$:

*Particles in the Front Overshadow Particles That are in the Rear:* If the depth of field is equal to, or smaller than, the average distance between the particles, *D _{av}*, we will avoid overshadowing of particles. In planar PTV experiments, we tend to shape the laser beam to be thin (a laser sheet), hence we can mitigate overshadowing by reducing the thickness of the laser sheet. In volumetric PTV, the depth of field $\Delta zo$ may take any value. Overshadowing will appear when $\Delta zo\u2265Dav$. In that case, it is possible to mitigate overshadowing by using several cameras to view simultaneously a VOI from different angles. This allows for the algorithm of the image analysis [12] to identify particles that are overshadowed in one viewing camera but are distinguished when viewed by the other cameras.

*Illumination by a Gaussian Elliptic Laser Beam*: When a Gaussian elliptic laser beam is used to illuminate a rectangular cuboid VOI with height $H=2\omega b$, thickness $\Delta zo=2\omega a$, and width *W* that equals to the Rayleigh range of a Gaussian beam, i.e., $W=2\pi \omega a2\lambda $, the area of the field of view (FoV) with a proper laser intensity will be $4\pi \omega a2\omega b\lambda $. Thus, we see that if we use a Gaussian elliptic laser beam, the area of the field of view (with proper light intensity) depends quadratically on its waist *ω _{a}*. Typically, in order to avoid overshadowing, we would like that $Dav=\Delta zo=2\omega a$. Meaning that the more we attempt to improve the spatial resolution of our measurement, by decreasing the waist

*ω*to mitigate overshadowing, the smaller the area with proper light intensity we will get (within the field of view).

_{a}*CMOS Cameras Have a Fixed Number of Pixels*: Since the number of pixels,

*N*

_{pix}, of a CMOS detector is fixed, one cannot detect more than

*N*

_{pix}particles. In practice, in order to avoid Peak-Locking (c.f. Sec. 6.2.3 in Ref. [7]) in image analysis, the area of the images of the particles needs to be at least 2 × 2 pixels on the CMOS detector. Also, the algorithms of particle tracking requires a minimal distance of Δ

_{algo}pixels between the images of the particles. Thus, the maximum number of particles we can image is $Npix(2+\Delta algo)2$ and thus the spatial resolution is limited by the CMOS pixel number

*N*

_{pix}

*Signal Loss Due to Scattering and Absorption Losses*: When a particle within the VOI generates a Mie signal, the signal needs to reach the camera lens at a distance *z _{o}* (or $zo+$ in 3D-PTV). However, this signal will be rescattered and absorbed by other Mie particles and by the fluid as it propagates toward the camera lens. It was argued previously that the Mie scattering from a particle in the VOI is reduced according to $Is(zo)=Is(0)e\u2212\sigma szon$, where

*n*is the density of the Mie particles and

*σ*is the total Mie scattering cross section (a similar factor will appear if the Mie particle has a significant absorption). To achieve the best spatial resolution, we will tend to increase the tracer density,

_{s}*n*, toward the density that is limited by the Rayleigh spatial resolution. However, as we increase the tracer density

*n*, the signal intensity will drop exponentially. Thus, we could face a situation where the density of the tracer will limit the optimal spatial resolution.

### 3.1 The Highest Spatial Resolution in Planar Particle Tracking Velocimetry.

*D*, views two adjacent point-like light sources that are located on the object plane. The distance between the particles is

_{f}*D*. According to the Rayleigh Criterion, the image of two particles is resolvable if

_{av}where the subscript, *R*, reminds us that the limit is due to the Rayleigh Criterion. This is the best possible spatial resolution, limited by physical optics.

In order to reach the Rayleigh spatial resolution $\Delta VR$ with a CMOS detector that has *N*_{pix} pixels and an average particle distance as given by (10), we will need to reduce the dimensions of the VOI until the amount of particles left within the VOI is below $Npix(2+\Delta algo)2$. In addition, the image of the VOI needs to fill the whole area of the CMOS detector.

### 3.2 The Highest Spatial Resolution in Three-Dimensional-Particle Tracking Velocimetry.

*D*that corresponds to the depth of field $\Delta zo$. Hence, in order to resolve the image, the distance between these two adjacent point-like, light sources, cannot exceed

_{a}*CoC*($CoC\Delta pix$ in pixel number), the distance between their boundaries will decrease by

*CoC*($CoC\Delta pix$ in pixel number). To compensate for that, we require that the particles will be $1.22\lambda Dazo++CoC$ apart. This means that the best spatial resolution in 3D-PTV will be

This is the best possible spatial resolution for particles in a VOI. It is limited by physical optics and by the circle of confusion.

When we have a CMOS detector with *N*_{pix} pixels, the area (in pixel number) that should be reserved to the image of each particle needs to account for the *CoC*: $(2+2CoC\Delta pix+\Delta algo)2$. The spatial resolution that is limited by the number of pixels of the CMOS detector will be $\Delta V=Dav3=WH\Delta zoNpix(2+2CoC\Delta pix+\Delta algo)2$. In addition, the image of the VOI needs to fill the whole area of the CMOS detector.

In order to reach the Rayleigh spatial resolution of Eq. (13) with a CMOS detector that has *N*_{pix} pixels, we would need to reduce the dimensions of the VOI until the amount of particles left within the VOI matches the limit value due to the CMOS detector.

## 4 Experimental Examples

In this section, the practical design steps for a planar PTV, a small and a large volumetric PTV experiments are analyzed. We use the models and the relevant equations that were deduced in Secs. 2 and 3 along with experimental parameters which are specified in Tables 2–4.

Max. velocity | $umax=10\u2009m/s$ |

Temporal resolution | $\Delta t=0.5\u2009ms$ |

Spatial resolution | $\Delta V=?$ |

Laser: | Nanio from InnoLas |

Wavelength | $\lambda =532\u2009nm$ |

Laser waist | $\omega a=0.5\u2009mm,\u2009\omega b=150\u2009mm$ |

Repetition rate | $frep=?$ |

Pulse duration | $\tau =?$ |

Pulse energy | $Ep=?$ |

Particles: | DEHS |

Average diameter | $d=1\u2009\mu m$ |

Refractive index | 1.45 at $532\u2009nm$ |

CMOS camera(s): | 1 × Photron Nova S9 |

Image sensor dimensions | $20.48\xd720.48\u2009mm2$ |

Pixels | 1024 × 1024 |

Pixel pitch | $\Delta pix=20\u2009\mu m$ |

Camera frame rate | $9000\u2009fps$ |

Responsivity | $R(\lambda =532\u2009nm)=0.78$ |

Optics | |

Beam guiding loss | L = 0.2 |

Signal transmission of the lens | $T(\lambda =532\u2009nm)=0.8$ |

Objective lens diameter | $Df=50\u2009mm$ |

Max. velocity | $umax=10\u2009m/s$ |

Temporal resolution | $\Delta t=0.5\u2009ms$ |

Spatial resolution | $\Delta V=?$ |

Laser: | Nanio from InnoLas |

Wavelength | $\lambda =532\u2009nm$ |

Laser waist | $\omega a=0.5\u2009mm,\u2009\omega b=150\u2009mm$ |

Repetition rate | $frep=?$ |

Pulse duration | $\tau =?$ |

Pulse energy | $Ep=?$ |

Particles: | DEHS |

Average diameter | $d=1\u2009\mu m$ |

Refractive index | 1.45 at $532\u2009nm$ |

CMOS camera(s): | 1 × Photron Nova S9 |

Image sensor dimensions | $20.48\xd720.48\u2009mm2$ |

Pixels | 1024 × 1024 |

Pixel pitch | $\Delta pix=20\u2009\mu m$ |

Camera frame rate | $9000\u2009fps$ |

Responsivity | $R(\lambda =532\u2009nm)=0.78$ |

Optics | |

Beam guiding loss | L = 0.2 |

Signal transmission of the lens | $T(\lambda =532\u2009nm)=0.8$ |

Objective lens diameter | $Df=50\u2009mm$ |

Note: ? is to be calculated.

Max. velocity | $umax=3\u2009m/s$ |

Temporal resolution | $\Delta t=0.25\u2009ms$ |

Spatial resolution | $\Delta V=?$ |

Laser: | Blizz InnoLas |

Wavelength | $\lambda =532\u2009nm$ |

Laser waist | $\omega a=\omega b=\omega 0=1.35\u2009mm$ |

Repetition rate | $frep=?$ |

Pulse duration | $\tau =?$ |

Pulse energy | $Ep=?$ |

Particles: | DEHS |

Average diameter | $d=1\u2009\mu m$ |

Refractive index | 1.45 at $532\u2009nm$ |

CMOS camera(s): | 4 × Phantom V2640 |

Image sensor dimensions | $27.6\xd726.3\u2009mm2$ |

Pixels | 2048 × 1952 |

Pixel pitch | $\Delta pix=13.5\u2009\mu m$ |

Camera framerate | $4855\u2009fps$ |

Responsivity | $R(\lambda =532\u2009nm)=0.23$ |

Optics | |

Beam guiding loss | L = 0.04 |

Signal transmission | $T(\lambda =532\u2009nm)=0.8$ |

Objective lens diameter | $Df=70\u2009mm$ |

Max. velocity | $umax=3\u2009m/s$ |

Temporal resolution | $\Delta t=0.25\u2009ms$ |

Spatial resolution | $\Delta V=?$ |

Laser: | Blizz InnoLas |

Wavelength | $\lambda =532\u2009nm$ |

Laser waist | $\omega a=\omega b=\omega 0=1.35\u2009mm$ |

Repetition rate | $frep=?$ |

Pulse duration | $\tau =?$ |

Pulse energy | $Ep=?$ |

Particles: | DEHS |

Average diameter | $d=1\u2009\mu m$ |

Refractive index | 1.45 at $532\u2009nm$ |

CMOS camera(s): | 4 × Phantom V2640 |

Image sensor dimensions | $27.6\xd726.3\u2009mm2$ |

Pixels | 2048 × 1952 |

Pixel pitch | $\Delta pix=13.5\u2009\mu m$ |

Camera framerate | $4855\u2009fps$ |

Responsivity | $R(\lambda =532\u2009nm)=0.23$ |

Optics | |

Beam guiding loss | L = 0.04 |

Signal transmission | $T(\lambda =532\u2009nm)=0.8$ |

Objective lens diameter | $Df=70\u2009mm$ |

Note: ? is to be calculated.

Max. velocity | $umax=3\u2009m/s$ |

Temporal resolution | $\Delta t=0.25\u2009ms$ |

Spatial resolution | $\Delta V=?$ |

Laser: | Green elliptic |

Wavelength | $532\u2009nm$ |

Laser waist | $\omega a=50\u2009mm,\u2009\omega b=150\u2009mm$ |

Repetition rate | $frep=?$ |

Pulse duration | $\tau =?$ |

Pulse energy | $Ep=?$ |

Particles: | Air-filled soap bubbles |

Average diameter | $d=15\u2009\mu m$ |

Refractive index | 1.33 at $532\u2009nm$ (soap membrane) |

CMOS camera(s): | 4 × Phantom V2640 |

Image sensor dimensions | $27.6\xd726.3\u2009mm2$ |

Pixels | 2048 × 1952 |

Pixel pitch | $\Delta pix=13.5\u2009\mu m$ |

Camera framerate | $4855\u2009fps$ |

Responsivity | $R(\lambda =532\u2009nm)=0.23$ |

Optics | |

Beam guiding loss | L = 0.04 |

Signal transmission | $T(\lambda =532\u2009nm)=0.8$ |

Objective lens diameter | $Df=50\u2009mm$ |

Max. velocity | $umax=3\u2009m/s$ |

Temporal resolution | $\Delta t=0.25\u2009ms$ |

Spatial resolution | $\Delta V=?$ |

Laser: | Green elliptic |

Wavelength | $532\u2009nm$ |

Laser waist | $\omega a=50\u2009mm,\u2009\omega b=150\u2009mm$ |

Repetition rate | $frep=?$ |

Pulse duration | $\tau =?$ |

Pulse energy | $Ep=?$ |

Particles: | Air-filled soap bubbles |

Average diameter | $d=15\u2009\mu m$ |

Refractive index | 1.33 at $532\u2009nm$ (soap membrane) |

CMOS camera(s): | 4 × Phantom V2640 |

Image sensor dimensions | $27.6\xd726.3\u2009mm2$ |

Pixels | 2048 × 1952 |

Pixel pitch | $\Delta pix=13.5\u2009\mu m$ |

Camera framerate | $4855\u2009fps$ |

Responsivity | $R(\lambda =532\u2009nm)=0.23$ |

Optics | |

Beam guiding loss | L = 0.04 |

Signal transmission | $T(\lambda =532\u2009nm)=0.8$ |

Objective lens diameter | $Df=50\u2009mm$ |

Note: ? is to be calculated.

### 4.1 Planar Particle Imaging Velocimetry/Particle Tracking Velocimetry: $300\u2009mm\xd7300\u2009mm\xd71\u2009mm$.

This experiment concerns with measuring variations in the concentration of Mie-scattering particles [13]. We will demonstrate how the current signal and the spatial resolution are estimated by using the experimental parameters listed in Table 2.

*Minimal Working Distance, z _{o}:* The aspect ratio of the FoV ($300\xd7300\u2009mm2$) corresponds to that of the CMOS detector ($20\xd720\u2009mm2$). In order to use the full area of the CMOS detector, we would like to work with a demagnification $M=$ (Width of CMOS)/(Width of FoV) $\u22480.066$. We find the minimal working distance with a negligible lens aberration by requiring, $zo>H0.34$ (due to the paraxial ray approximation [11]). The height of the FoV is $H=300\u2009mm$, hence $zo>882\u2009mm$. From the thin lens Eq. (A1) and the magnification $M=zizo$ we find $zo=fM+1M\u224816f$. Thus, when the focal length $f=60\u2009mm$, the working distance $zo=960\u2009mm>882\u2009mm$ and our image at $zi=7.2\u2009cm$ will have negligible lens aberrations. This estimation is for a single thin lens. If we use an objective lens that consists of several lenses to reduce spherical aberrations, the working distance could be made shorter and without lens aberrations. Adjusting the focal lens

*f*and the working distance

*z*is a practical approach in the design of PTV experiments (lenses are cheaper than CMOS detectors, and adjusting a working distance is usually a practical option).

_{o}According to geometric optics, a $1\u2009\mu m$ diameter particle will have an image with a diameter of $dM=0.066\u2009\mu m$. This image diameter is smaller than the predicted diameter of a point-like light source using diffraction theory. Therefore, we cannot use geometric optic to predict the image diameter of the $1\u2009\mu m$ diameter particle with this magnification. Instead, we can calculate the diffraction image from a $1\u2009\mu m$ diameter particle that consists of a continuous distribution of point-like light sources. The diffraction image of one point-like light source would be an airy pattern. Its airy disk would have the diameter $2.44\lambda Dff=1.55\u2009\mu m$. By viewing the $1\u2009\mu m$ diameter particle as if it consists of continuous point-like light sources, we can approximate (instead of a rigorous complex integration) that the diameter of the image would be the sum of all Airy disks from each point. The total diameter would be about $1+1.55=2.55\u2009\mu m$. The area of the image of one particle is smaller than the area of one pixel. With a proper particle density, the images of several particles could fit into one pixel with an area of $20\xd720\u2009\mu m2$. In such a case, the signal in one pixel would represent the number of particles in a voxel with a volume of $1000\xd7300\xd7300\u2009\mu m2$. Hence, we will image the density of the particles (which is indeed the objective of this PIV experiment), rather than imaging the individual particle as in a PTV experiment.

*Laser Repetition Rate, f*_{rep}*:* For a maximum velocity $umax=10\u2009m/s$ within the object plane, the STB algorithm requires particle image displacement of about $\u223c5$ pixels. The time between pulses is given by $\Delta t=5\Delta pixumaxM=5\xb720\u2009\mu m10\u2009m/s\xd70.066=150\u2009\mu s$. The laser repetition rate is determined by the required temporal resolution, $frep=1\Delta t=10.151\xd710\u22123\u2009s=6.6\u2009kHz$.

*Laser Pulse Width, τ:* To avoid streaking, $\tau <\Delta pix2\u2009Mumax=20\xd710\u22126\u2009m2\xb70.066\xd710\u2009m/s=15\u2009\mu s$. The Nanio laser has a pulse duration of $\tau =0.1\u2009\mu s$. Hence, this laser is suitable.

*Laser Pulse Energy, E*The laser beam has an elliptic cross section profile: $\omega a=0.5\u2009mm$ and $\omega b=150\u2009mm$. The elliptical cross section of the laser beam overlaps, approximately, with the cross section of $1\xd7300\u2009mm$ of our rectangular cuboid VOI. We use the free-ware of Philip Laven

_{p}:^{2}to calculate that the angular Mie scattering $\sigma (90\xb0)=5\xd710\u221218Wcm2$ at a distance of $0.96\u2009m$ and that the total Mie scattering cross section $\sigma s\u224810\u221216\u2009cm2$. We use a lens with a diameter $Df=50\u2009mm$ and a working distance $zo=960\u2009mm$, yielding from Eq. (3) a solid angle of $\Omega par=Df216zo2=1.7\xd710\u22124\u2009str$. The

*FFR*=

*1 (since the area of the image of one particle falls into one pixel in this experiment). Equation (5) for the current signal is rewritten here for convenience*

Let us consider a rarefied particle density so that only one particle is imaged into one pixel. For this to happen, there should be 1 particle in one voxel with dimensions of $0.3\xd70.3\xd70.3\u2009mm3$. This corresponds to a particle density of $n=3.7\xd7104/cm3$. In this case, the loss due to total Mie scattering and absorption by DEHS particles is negligible. The electron count from a pixel for $Ep=1\u2009mJ$ is found by dividing *i _{S}* with the electron charge $e=1.6\xd710\u221219$ Coulomb. For a $1\u2009mJ$ pulse energy, we obtain 19 electron counts from an image of one particle. The Photron Nova S9 camera has an electron noise-count of about 20 per pixel. Hence, we will have

*SNR*=

*1 at $1\u2009mJ$ laser pulse energy. If we will increase the particle density to above 10 particles per voxel, we will see a signal that corresponds to the density of the particles and our*

*SNR*will be workable.

The laser intensity at $rb=\omega b,ra=0$ falls by about 87%. Since the Mie signal is proportional to the power density of the illumination source, the signal will fall exponentially, outward, from the center of the beam.

*Spatial Resolution*$\Delta V$*and Particle Density n:* According to the Rayleigh criterion, the minimum discernible distance between adjacent particles in the object plane is given by Eq. (8): $Dav=1.22\lambda Dfzo=1.220.532\xd710\u22126\u2009m0.05\xd70.96=12.4\u2009\mu m$. This corresponds to a distance of $MDav=0.066\xd712.4\u2009\mu m=0.83\u2009\mu m$ in the image plane. This imaged distance is smaller than the $20\u2009\mu m$ pixel size of the CMOS detector. Hence, with the present CMOS resolution (of $20\u2009\mu m$) and the given demagnification, we would *not* be able to reach the Rayleigh resolution. We could only distinguish between two particles with a distance larger than $\Delta pix/M=20\u2009\mu m/0.066=300\u2009\mu m$. The limit on the optical resolution is due to the pixel size (called the pixel resolution) and the demagnification. Also, we note that the thickness of the laser sheet is $1\u2009mm$. If we work with an average distance between the particles of $Dav=0.3\u2009mm$, we would experience some overshadowing (the laser sheet is 3.3 times larger than the average distance *D _{av}* between the particles). In order to mitigate overshadowing, we could reduce the laser waist to $\omega a=0.15\u2009mm$. However, then our field of view (with useful laser light) will be reduced due to the Rayleigh Range of a Gaussian laser beam. The new width will be $W<14\pi \omega a2\lambda =33\u2009mm$. We see that although we speak of planar PTV, a more accurate understanding of planar PTV can be achieved in terms of volumetric PTV.

### 4.2 Volumetric Particle Tracking Velocimetry: $V=27\u2009mm\xd72.7\u2009mm\xd72.7\u2009mm$.

This experiment is designed to measure the full instantaneous velocity gradient tensor at sub-Kolmogorov scales in a turbulent round jet. The measurements are carried out at a high spatial resolution in order to obtain the full dissipation rate tensor at Reynolds numbers ranging between $\u223c10.000\u2212100.000$. More details on the laboratory setup can be found in [6] and [14]. The parameters of this experiment are listed in Table 3.

*Minimal Working Distance, z _{o}:* To avoid lens aberrations, the paraxial ray condition give a $zo>H0.34$, thus $zo>80\u2009mm$. Due to experimental constraints, a minimum working distance of $1000\u2009mm$ is required, which fulfills the aberration constraint. The volume of interest has a FoV of $27\xd72.7\u2009mm2$, which matches the CMOS image sensor area. Hence, the required magnification is

*M*=

*1. The thin lens equation with the required magnification, gives that for $zo=1000\u2009mm$, the focal length of the lens should be $f=500\u2009mm$ and $zi=1000\u2009mm$, which is an unusually large number for a CMOS camera (the sensor will need to be 1 meter away from the lens). The solution is to use a Galilean telescope. The Nikkor $300\u2009mm$ Telelphoto lens is basically an elaborated Galilean telescope (Ref. [11]). It has a Front Effective Focal Length $FEFL=300\u2009mm$. When it is continued with a Nikkor teleconverter 2 M lens, its $FEFL=600\u2009mm$. By turning the focusing cylinder knob, we tune the Back Effective Focal Lens*

*BEFL*so that the field of view is in focus at a distance of 240 mm from the edge of the Nikkor teleconverter 2 M lens.

*Laser Repetition Rate, f*_{rep}*:* For a maximum velocity of $umax=3\u2009m/s$, the required time between pulses is given by $\Delta t=5\Delta pixumaxM=5\xb713.5\u2009\mu m3\u2009m/s\xd71=22.5\u2009\mu s$. The corresponding laser repetition rate is thus $frep=1\Delta t=10.0225\xd710\u22123\u2009s=44.4\u2009kHz$

*Laser Pulse Width, τ:* To avoid streaking $\tau <13.5\u2009\mu m2\xb71\xb73\u2009m/s=2.25\u2009\mu s$. The Blizz laser from InnoLas has a pulse duration of $\tau =0.01\u2009\mu s$, which is suitable.

*Laser Pulse Energy, E*The laser has a circular Gaussian laser beam with $\omega a=\omega b=1.35\u2009mm$ that approximately overlaps with the rectangular cuboid VOI. The depth of field is relatively small, and we can use

_{p}:*z*instead of $zo+$ in Eq. (6). The Mie scattering factor at $90\u2009deg$ is calculated by using Philip Laven free-ware.

_{o}^{2}The software supplies a Mie factor of $\sigma (90\xb0)=5\xd710\u221218Wcm2$ at 1 meter distance. For $f/N=11$, the aperture $Da=300\u2009mm11=27.3\u2009mm$ and from Eq. (3), $\Omega par=Da216zo2=4.6\xd710\u22125$. Since the diameter of the DEHS particles is of the order of the wavelength of the laser light, the diameter of the image of the particle must be estimated using diffraction theory. For an f-number (

*f*/

*N*) of 11, the airy disk has a diameter of $2.44\lambda f/N=14.3\u2009\mu m$. This means that the area of the image of a particle is slightly larger than the area of one pixel. The $FFR\u22641$ for particles located in the object plane and decreases for particles that are out of focus. Thus, we take FFR

*=*

*0.25. Plugging these numbers and those from Table 3 into Eq. (6), we obtain*

Let us assume that the density of the tracer particles is low and thus the total Mie scattering and the absorption by the tracer particles is negligible. We calculate the electron-count from a pixel that images particles, which are located at the center of the Gaussian beam where *r *=* *0 (the radial distance from the center of the beam outward). We divide *i _{S}* by the electron charge $e=1.6\xd710\u221219$ Coulomb to obtain 187 electron counts from an image of one particle per $1\u2009mJ$ pulse energy. According to the manual of the V2640 camera, the read out noise per pixel in the standard mode is 7.2 electrons. Hence, the

*SNR*=

*26 for particles located at the center of the laser beam. Since the intensity of the cross section of the laser beam drops exponentially with*

*r*, the intensity of the laser at the waists will be about 8 times smaller. This means that the

*SNR*for particles located at the faces of the VOI will be 8 times smaller.

*Optical Resolution and Density:* For $zo=1\u2009m,\u2009f/N=11,f=300\u2009mm,$ and a depth of field $\Delta zo=2.7\u2009mm$, one can use Eq. (A9) to estimate that the circle of confusion, $CoC=3.4\u2009\mu m$. and from Eq. (A8) we find that $zo\u2212=998.5\u2009mm$ and $zo+=1001.5\u2009mm$. Using Eq. (8), we find that the Rayleigh distance: $Dav=1.22\lambda Dazo=1.220.532\xd710\u22126\u2009m0.027\xd71=24\u2009\mu m$. The camera sensor has a pixel pitch of $13.5\u2009\mu m$. The Rayleigh distance corresponds to a distance of two pixels.

As was explained in Sec. 3, the images of point-like light sources at $zo\u2212$ and $zo+$ will have a diameter that is approximately the diameter of the airy disk plus the circle of confusion. The diameter of the Airy disk is $2.44\lambda f/N=14.2\u2009\mu m$ and $CoC=3.4\u2009\mu m$. Hence, the image diameter of a particle at $zo\u2212$ and a particle at $zo+$ will be $14.2+3.4=17.5\u2009\mu m$. The Rayleigh distance for particles at $zo+$ plane is $24+3.4=27.4\u2009\mu m$. Hence, the best spatial resolution would be $(27.4)3\u2009\mu m3$.

### 4.3 Volumetric Particle Tracking Velocimetry: $300\u2009mm\xd7300\u2009mm\xd7100\u2009mm$.

In this experimental design, the idea is to globally measure the turbulent flow from a jet at Reynolds numbers ranging between ∼10.000 and 100.000. The $1\u2009cm$ diameter of the nozzle of the jet is 10 times smaller than the nozzle in the previous example. Therefore, one can cover a large portion of the flow-field with proper illumination. If we use a laser with $1\u2009mJ$ pulse-energy as in the first example (a cross section of $1\xd7300\u2009mm2$), we cannot expect to have a detectable Mie-signal when the cross section of the VOI is $100\xd7300\u2009mm2$. In fact, we would need a laser pulse that is 100 times more powerful in order to keep on the same power density as in the first example. However, if we will work with a shorter working distance *z _{o}* and use $\u223c15\u2009\mu m$ air-filled soap bubbles (their Mie scattering signal is two orders of magnitude larger than the Mie scattering of $1\u2009\mu m$ diameter particles), the Mie signal might be detectable. We need to investigate if we can detect particles from other parts of the volume, in particular particles at the far face of the volume of interest.

*Minimal Working Distance, z _{o}:* The spherical aberration of a simple lens are negligible when $zo>H0.34$. Hence, the working distance

*z*should be larger than $882\u2009mm$. As can be seen from Eq. (6), the current signal depends inversely on the second power of $zo+$. To enhance the signal, it seems sensible to work with values of

_{o}*z*that are as small as possible. Therefore, we use a Nikkor $35\u2009mm$ lens (from Nikon Corporation) which has aberration-free imaging down to a distance of $300\u2009mm$ (this lens consists of 8 lenses that together decrease lens aberrations and facilitate an angle of view of $42\xb0$).

_{o}The dimensions of our CMOS sensor are $27\xd727\u2009mm2$ and the dimensions of the FoV are $300\xd7300\u2009mm2$. Thus, for optimal imaging of the field of view, the magnification should be $M=$(Width of CMOS)/(Width of the FoV)$=0.09$. In order to maximize the signal, we consider first a working distance $zo=350\u2009mm$ with *M *=* *0.1. By using Eq. (A1) and the magnification, we find that for $f=35\u2009mm$, the required working distance is $zo=350\u2009mm$. Since $f/N=fDa$, for $f/N=11$ the aperture diameter is $Da=3.18\u2009mm$.

From Fig. 2, one can study the variations of $zo+$ and $zo\u2212$ as a function of *CoC* for $zo=350\u2009mm,f=35\u2009mm$ and $Da=3.18\u2009mm$. For a depth of field $zo+\u2212zo\u2212=\Delta zo=101\u2009mm$, the circle of confusion is $CoC=50\u2009\mu m$ (where $zo\u2212=306\u2009mm$ and $zo+=407\u2009mm$).

*Laser Repetition Rate, f*_{rep}*:* For a maximum velocity of $umax=3\u2009m/s$, the time between pulses is found from $\Delta t=5\Delta pixumaxM\u2212=5\xb713.5\u2009\mu m3\u2009m/s\xd70.11=204.5\u2009\mu s$ (note that $M\u2212=zizo\u2212=35306=0.11$). Since $frep=1\Delta t$, the laser pulse repetition rate should be above $frep=1204\xd710\u22126=4.88\u2009kHz$

*Laser Pulse Width, τ:* Since the ratio of the depth of field to the working distance is not negligible, one should use $M\u2212=35306=0.11$ to estimate the pulse width $\tau <13.5\xd710\u22126\u2009m2\xb70.11\xd73\u2009m/s=20\u2009\mu s$. Our laser has $\tau =0.01\u2009\mu s$ and is therefore suitable.

*Laser Pulse Energy, E _{p}:* We assume an elliptic Gaussian laser beam with $\omega a=50\u2009mm$ and $\omega b=150\u2009mm$ that approximately fills the VOI. We recall that since the depth of field is not negligible, we need to calculate the signal level for a particle at the far edge of the depth of field (at $zo+=407\u2009mm$). We used Philip Laven's free-ware

^{2}to calculate the differential Mie scattering $\sigma (90\xb0)=10\u221216Wcm2$ at $0.407\u2009m$ for $d=15\u2009\mu m$ and $\lambda =532\u2009nm$. The solid angle of this particle with the aperture $Da=3.18\u2009mm$ is calculated from Eq. (3) to be $\Omega par=(3.18)216\xd74072=3.8\xb710\u22126\u2009str$. The image diameter of this particle on the image plane is $(dM++CoC)=14+50=64\u2009\mu m$, where the diameter of the particle due to diffraction ($14\u2009\mu m$) was used instead of

*dM*

^{+}. Thus, the area of a blurred image of a particle will cover about 25 pixels(!) and the $FFR=125$. This example emphasizes a situation where the

*FFR*becomes a dominant limiting factor. For particles in the object plane, the

*FFR*is 0.25. For particles at the far face of the volume of interest, the

*FFR*is smaller by a factor of 6.25.

We assume that the losses due to total Mie scattering or absorption are negligibly small.

We divide *i _{S}* by the electron charge $e=1.6\xd710\u221219$ Coulomb and obtain 0.014 electron counts in one pixel per one

*mJ*of pulse energy. The read out noise in the Phantom camera is 7.2 electrons (in the standard mode). Hence, we would not be able to detect a signal with $1\u2009mJ$ pulse energy. Our estimate is for a laser intensity at the center of the beam (at $ra=rb=0$). The laser intensity at the waist will be at least eight times smaller than its intensity at its center. This means that our estimation for the current signal from particles at the far face of the depth of field (at $zo=407\u2009mm$) will have a signal 8 times smaller (the Mie signal is proportional to the intensity of the laser).

The above estimation shows that our current signal will be very poor. In order to improve the electron count, we should consider the following:

Change the angle of the Mie scattering toward forward scattering. For example: Mie scattering at $40\u2009deg$ is 18 times larger than Mie scattering at right angle,

Increase the pulse energy by an order of magnitude,

Decrease the volume of interest, in particular, by shortening of the depth of field so that the

*CoC*decreases and the*FFR*increases.Increase the particle diameter. A particle with diameter of $40\u2009\mu m$ will have 10 times more Mie signal.

*Optical Resolution and Density:* When $f/N=11\u2009mm$ and $f=35\u2009mm,$ the camera aperture diameter is $Da=3.18\u2009mm$. Figure 2 shows that for a working distance $zo=350\u2009mm$ and a depth of field $zo=101\u2009mm$, the far face of the VOI is at $zo+=407\u2009mm$ and the $CoC=50\u2009\mu m$. The Rayleigh distance at $zo+=407\u2009mm$ is $Dav=1.22\lambda Dazo++CoC=1.220.532\xd710\u22126\u2009m0.003,18\xd70.406+50=122.8\u2009\mu m$. The spatial resolution is the third power of this value. Due to the large depth of field, a point-like particle at $zo+$ has a circle of confusion $CoC=50\u2009\mu m$, which compromises the current signal, the optical resolution and as a consequence, the spatial resolution $\Delta V$.

## 5 Conclusions

This report presents an analytical model that estimates the current signal in a pixel of a CMOS camera in planar PTV and 3D-PTV experiments. Additionally we found expressions for estimating the spatial resolution in 2D and 3D-PTV experiments.

The key idea behind our approach is to identify the particles within the VOI which will have the faintest current signal and the worst optical resolution. The location of these particles depends on the configuration of the experiment. In an experimental configuration, as is shown in Fig. 1, the particles at the vertices of the far face will have the faintest signal and the worst optical resolution.

In typical 3D-PTV experiments, one uses 2–4 identical cameras to view the volume of interest with 2–4 corresponding viewing angles. To employ our model, one needs to use for each camera view: the appropriate polar Mie scattering, the specific scattering and absorption losses of the laser and the specific scattering and absorption losses of the signal.

Our analysis assumes the paraxial approximation of geometric optics, i.e., there are no spherical aberration to our images. However, when one is working with four cameras that look at the same VOI, it could happen that the field of view would have to be wide and then the effect of spherical aberration: astigmatism, coma and distortion (see Ref. [11]) will creep into the images. As a result, the current signal will fall and the image resolution will degrade. It is thus worth investing in high-quality lenses that compensate for spherical aberration at large viewing angles.

*μ*m HFSB in air, the effect of diffraction is washed away. The intensity of the signal is due to reflection of the light. This intensity is given by the Fresnel equation for reflection

where, $nw=1.44$ is the refractive index of water, *θ _{i}* is the angle of incidence of the laser beam, and

*θ*is the angle of refraction, found from Snell law. The calculation of the current signal due to the reflection of a laser light from the surface of HFSB in a volumetric PTV experiment would result in a similar expression to Eq. (6). The same six optical consideration that were put forth in Sec. 2 would have to appear. However, the Mie scattering factor should be replaced by the Fresnel reflection, while considering the spherical surface of the HFSB and its effect on the angle of incident upon the Fresnel reflection. Additionally, the scattering cross section and absorption cross section by large HFSB need to be worked out (by using Fresnel reflection and transmission equations).

_{t}The presented model considers a Gaussian elliptic laser source as the light source, which is the most prevalent illumination source at present when using Mie particles. Light emitting diodes (LEDs) have been proven to be able to generate high power densities suitable for illuminating HFSB in large volumes [4]. LEDs have a wide spectral distribution and more complex beam profiles, such as, Lambertian, top-hat profile or a complex beam profile emanating from an array of LEDs. Therefore, a model for the current signal and the image resolution of HFSB within a VOI, which are illuminated by LEDs, could be of interest in future work.

## Acknowledgment

H. A. and C. M. V. acknowledge financial support from the Poul Due Jensen Foundation: financial support from the Poul Due Jensen Foundation (Grundfos Foundation) for this research is gratefully acknowledged.

C. M. V., S. L. R. and Y. Z. acknowledge financial support from the European Research Council: this project has received funding from the European Research Council (ERC) under the European Unions Horizon 2020 research and innovation program.

The authors would like to thank Benny Edelsten and Jakob Skov Nielsen for proofreading the paper.

## Funding Data

Poul Due Jensen Foundation (Grundfos Foundation) (No. 2018-03).

European Research Council (ERC) under the European Unions Horizon 2020 Research and Innovation Program (Grant No. 803419; Funder ID: 10.13039/501100000781).

## Data Availability Statement

The datasets generated and supporting the findings of this article are obtainable from the corresponding author upon reasonable request.

## Nomenclature

*d*=particle diameter (

*μ*m)*D*=_{a}aperture diameter (mm)

*D*=_{av}averaged distance between particles (mm)

*D*=_{f}lens diameter (mm)

*E*=_{p}pulse energy (J)

*f*_{rep}=laser repetition rate (Hz)

*i*=_{s}current signal (Ampere)

*I*=laser intensity (W/cm

^{2})*I*=_{s}signal intensity (W/cm

^{2})*n*=particle density (cm)

^{−3}*P*=_{i}optical power (W)

*z*=_{l}distance between laser to VOI (mm)

*z*=_{o}distance of a lens to the object plane (mm)

*α*=loss coefficient (cm)

^{−1}- $\Delta V$ =
spatial resolution (

*μ*m)- $\Delta zo$ =
Depth of field (mm)

- $\Delta \theta $ =
angular rayleigh resolution (rad)

*θ*=polar angle (rad)

*λ*=wavelength (nm)

*σ*=cross section (cm)

^{2}*τ*=laser pulse duration (S)

- $\omega a,\u2009\omega b$ =
laser waists (mm)

- Ω =
solid angle (str)

### Appendix: Depth of Field and Circle of Confusion

The thin lens equation is essential for understanding the image generation of the Mie scattering particles in planar and volumetric PTV. Figure 3 illustrates the main concepts involved with a thin lens. Three planes are defined in perpendicular to the horizontal optical axis: The object plane (the plane where the object lies), the lens plane (the plane where the lens is positioned) and the image plane (the plane where the image is formed).

According to geometric optics, the finite-sized objects (seeding particles) consist of a distribution of point-like light sources that lie in the object plane. Each point source of the object emits light rays in all directions. The rays of each point source define a cone, where its apex is the point-like light source and its base is the area of the lens. Ideally, all the rays within that cone will be imaged into a single point on the image plane.

*z*from the lens plane will be refracted at the lens and meet at a point on the other side of the lens at distance

_{o}*z*from the lens plane (the conjugated point). The thin lens equation gives a relation between the focal length of the lens

_{i}*f*and the distances between the lens plane to the object,

*z*, and to the image,

_{o}*z*, respectively:

_{i}A digital camera consists of a CMOS detector, a lens and an aperture. In order to view an object at a distance *z _{o}* away from the lens “in focus,” the CMOS detector must be at a distance

*z*away from the lens on the opposite side of the lens. Then all the point sources of the object will be mapped into image points in the image plane.

_{i}The point-like light source of the object at *z _{o}* is imaged into a point-like light source at

*z*on the CMOS detector (the green rays). However, the other two point-like light sources, $zo+$ and $zo\u2212$, are imaged as blurred circles on the image plane

_{i}*z*, since their foci lie in front of and behind the image plane. The resulting blurred circles due to the lack of focus are known as the circle of confusion with diameter

_{i}*CoC*.

*CoC*as the base and the other has the aperture diameter

*D*as the base. The following relation is easily deduced from the geometry:

_{a}*D*as its base. We can find the following relation from their similarity:

_{a}The depth of focus is obtained from the difference $\Delta zi=zi\u2212\u2212zi+$. The depth of focus is thus the segment along the optical axis where a point source from the object plane is imaged into a circle that is smaller than the circle of confusion. Correspondingly, the depth of field, $\Delta zo=zo+\u2212zo\u2212$, is a segment along the optical axis where any point source within the depth of field will appear on the image plane as a circle of light with a diameter smaller than *CoC*.

*f*, and $zo\u2212$ in terms of $zi\u2212$ and

*f*:

*D*,

_{a}*CoC*,

*f*, and

*z*. One is then able to use $\Delta zo=zo+\u2212zo\u2212$ to calculate the depth of field. Sometimes, it can be more useful to invert (A7) and express

_{o}*CoC*as a function of the depth of field, working distance, aperture diameter and focal length

where $zi=fzozo\u2212f$.