Optical Considerations for Designing Laser-Based Volumetric Particle Tracking Velocimetry

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 $1 μm$ diameter DEHS particles, or with $15 μm$ 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.

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_i. If the image fills the area of one pixel, the current signal i_S equals to the incident optical power times the responsivity of the CMOS detector at the wavelength λ of the laser, $iS=PiR(λ)$ (the responsivity has the units of Ampere/Watts).

where the variables r_a and r_b are the radial components along the semiminor and semimajor axes of the elliptic cross section of the laser beam, respectively.

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.

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 $I(1−L)$⁠.

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 $αs−p$ and $αa−p$⁠, respectively. The losses due to the fluid are characterized by scattering and absorption coefficients $αs−f$ and $αa−f$⁠, respectively. For example, Mie particles with particle density n in air have a scattering loss coefficient $αs−p=σsn$⁠, 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 $αs−p$ and $αa−p$⁠. 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_l through the seeded fluid until it reaches a particle that is located within the volume of interest. The laser intensity I after a path z_l will be $I(zl)=I(0)e−αzl$⁠, where $α=αa−p+αs−p+αa−f+αs−f$ accounts separately for the absorption and scattering losses by the tracer particles and by the fluid.

where I is the intensity of the incoming laser and σ is the scattering cross section of a particle into a differential solid angle $∂Ω$⁠. The integration over $4π [str]$ 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 $σ(θ)$ can be calculated by using openly available resources, e.g., (see Footnote²).

where D_f is the diameter of the lens. The intensity of Mie scattering into the camera lens will be approximately $Is=Iσ(θ)Ωpar$⁠.

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 $Is(zo)=Is(0)e−αzo$⁠.

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 $T(λ)$⁠.

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).

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 $ra=ωa, rb=ωb$ will have the lowest laser intensity and therefore the lowest current signal.

In 3D-PTV, we adjust the diameter D_a of the camera aperture to obtain a depth of field $Δzo$ 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_S, for volumetric PTV with a large depth of field.

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 $zo−$ 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_a, 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 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 $2.44λDa$ [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).

To achieve the highest spatial resolution $ΔV$ 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 sets a physical limit (due to diffraction) upon the optical resolution between two particles. $ΔθRzo$ 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 $ΔV$⁠:

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 $Δzo$ may take any value. Overshadowing will appear when $Δzo≥Dav$⁠. 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 $H=2ωb$⁠, thickness $Δzo=2ωa$⁠, and width W that equals to the Rayleigh range of a Gaussian beam, i.e., $W=2πωa2λ$⁠, the area of the field of view (FoV) with a proper laser intensity will be $4πωa2ωbλ$⁠. 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 $Dav=Δzo=2ωa$⁠. Meaning that the more we attempt to improve the spatial resolution of our measurement, by decreasing the waist ω_a to mitigate overshadowing, the smaller the area with proper light intensity we will get (within the field of view).

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 $Is(zo)=Is(0)e−σszon$⁠, where n is the density of the Mie particles and σ_s 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, 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.

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 $ΔVR$ 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 $Npix(2+Δalgo)2$⁠. In addition, the image of the VOI needs to fill the whole area of the CMOS detector.

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: $(2+2CoCΔpix+Δalgo)2$⁠. The spatial resolution that is limited by the number of pixels of the CMOS detector will be $ΔV=Dav3=WHΔzoNpix(2+2CoCΔpix+Δalgo)2$⁠. 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.

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.

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 (⁠$300×300 mm2$⁠) corresponds to that of the CMOS detector (⁠$20×20 mm2$⁠). 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) $≈0.066$⁠. 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 $H=300 mm$⁠, hence $zo>882 mm$⁠. From the thin lens Eq. (A1) and the magnification $M=zizo$ we find $zo=fM+1M≈16f$⁠. Thus, when the focal length $f=60 mm$⁠, the working distance $zo=960 mm>882 mm$ and our image at $zi=7.2 cm$ 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_o 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).

According to geometric optics, a $1 μm$ diameter particle will have an image with a diameter of $dM=0.066 μm$⁠. 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 $1 μm$ diameter particle with this magnification. Instead, we can calculate the diffraction image from a $1 μm$ 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 $2.44λDff=1.55 μm$⁠. By viewing the $1 μm$ 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 $1+1.55=2.55 μm$⁠. 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 $20×20 μm2$⁠. In such a case, the signal in one pixel would represent the number of particles in a voxel with a volume of $1000×300×300 μm2$⁠. 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 $umax=10 m/s$ within the object plane, the STB algorithm requires particle image displacement of about $∼5$ pixels. The time between pulses is given by $Δt=5ΔpixumaxM=5·20 μm10 m/s×0.066=150 μs$⁠. The laser repetition rate is determined by the required temporal resolution, $frep=1Δt=10.151×10−3 s=6.6 kHz$⁠.

Laser Pulse Width, τ: To avoid streaking, $τ<Δpix2 Mumax=20×10−6 m2·0.066×10 m/s=15 μs$⁠. The Nanio laser has a pulse duration of $τ=0.1 μs$⁠. Hence, this laser is suitable.

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 $0.3×0.3×0.3 mm3$⁠. This corresponds to a particle density of $n=3.7×104/cm3$⁠. In this case, the loss due to total Mie scattering and absorption by DEHS particles is negligible. The electron count from a pixel for $Ep=1 mJ$ is found by dividing i_S with the electron charge $e=1.6×10−19$ Coulomb. For a $1 mJ$ 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 $1 mJ$ 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 $rb=ωb,ra=0$ 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$ΔV$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): $Dav=1.22λDfzo=1.220.532×10−6 m0.05×0.96=12.4 μm$⁠. This corresponds to a distance of $MDav=0.066×12.4 μm=0.83 μm$ in the image plane. This imaged distance is smaller than the $20 μm$ pixel size of the CMOS detector. Hence, with the present CMOS resolution (of $20 μm$⁠) 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 $Δpix/M=20 μm/0.066=300 μm$⁠. 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 $1 mm$⁠. If we work with an average distance between the particles of $Dav=0.3 mm$⁠, 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 $ωa=0.15 mm$⁠. 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 $W<14πωa2λ=33 mm$⁠. We see that although we speak of planar PTV, a more accurate understanding of planar PTV can be achieved in terms of volumetric PTV.

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 $∼10.000−100.000$⁠. 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 $zo>80 mm$⁠. Due to experimental constraints, a minimum working distance of $1000 mm$ is required, which fulfills the aberration constraint. The volume of interest has a FoV of $27×2.7 mm2$⁠, 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 $zo=1000 mm$⁠, the focal length of the lens should be $f=500 mm$ and $zi=1000 mm$⁠, 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 $300 mm$ Telelphoto lens is basically an elaborated Galilean telescope (Ref. [11]). It has a Front Effective Focal Length $FEFL=300 mm$⁠. When it is continued with a Nikkor teleconverter 2 M lens, its $FEFL=600 mm$⁠. 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 $umax=3 m/s$⁠, the required time between pulses is given by $Δt=5ΔpixumaxM=5·13.5 μm3 m/s×1=22.5 μs$⁠. The corresponding laser repetition rate is thus $frep=1Δt=10.0225×10−3 s=44.4 kHz$

Laser Pulse Width, τ: To avoid streaking $τ<13.5 μm2·1·3 m/s=2.25 μs$⁠. The Blizz laser from InnoLas has a pulse duration of $τ=0.01 μs$⁠, which is suitable.

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 $e=1.6×10−19$ Coulomb to obtain 187 electron counts from an image of one particle per $1 mJ$ 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 $zo=1 m, f/N=11,f=300 mm,$ and a depth of field $Δzo=2.7 mm$⁠, one can use Eq. (A9) to estimate that the circle of confusion, $CoC=3.4 μm$⁠. and from Eq. (A8) we find that $zo−=998.5 mm$ and $zo+=1001.5 mm$⁠. Using Eq. (8), we find that the Rayleigh distance: $Dav=1.22λDazo=1.220.532×10−6 m0.027×1=24 μm$⁠. The camera sensor has a pixel pitch of $13.5 μm$⁠. The Rayleigh distance corresponds to a distance of two pixels.

As was explained in Sec. 3, the images of point-like light sources at $zo−$ 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 $2.44λf/N=14.2 μm$ and $CoC=3.4 μm$⁠. Hence, the image diameter of a particle at $zo−$ and a particle at $zo+$ will be $14.2+3.4=17.5 μm$⁠. The Rayleigh distance for particles at $zo+$ plane is $24+3.4=27.4 μm$⁠. Hence, the best spatial resolution would be $(27.4)3 μm3$⁠.

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 $1 cm$ 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 $1 mJ$ pulse-energy as in the first example (a cross section of $1×300 mm2$⁠), we cannot expect to have a detectable Mie-signal when the cross section of the VOI is $100×300 mm2$⁠. 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 $∼15 μm$ air-filled soap bubbles (their Mie scattering signal is two orders of magnitude larger than the Mie scattering of $1 μm$ 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_o should be larger than $882 mm$⁠. 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 z_o that are as small as possible. Therefore, we use a Nikkor $35 mm$ lens (from Nikon Corporation) which has aberration-free imaging down to a distance of $300 mm$ (this lens consists of 8 lenses that together decrease lens aberrations and facilitate an angle of view of $42°$⁠).

The dimensions of our CMOS sensor are $27×27 mm2$ and the dimensions of the FoV are $300×300 mm2$⁠. 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 $zo=350 mm$ with M = 0.1. By using Eq. (A1) and the magnification, we find that for $f=35 mm$⁠, the required working distance is $zo=350 mm$⁠. Since $f/N=fDa$⁠, for $f/N=11$ the aperture diameter is $Da=3.18 mm$⁠.

From Fig. 2, one can study the variations of $zo+$ and $zo−$ as a function of CoC for $zo=350 mm,f=35 mm$ and $Da=3.18 mm$⁠. For a depth of field $zo+−zo−=Δzo=101 mm$⁠, the circle of confusion is $CoC=50 μm$ (where $zo−=306 mm$ and $zo+=407 mm$⁠).

Laser Repetition Rate, f_rep: For a maximum velocity of $umax=3 m/s$⁠, the time between pulses is found from $Δt=5ΔpixumaxM−=5·13.5 μm3 m/s×0.11=204.5 μs$ (note that $M−=zizo−=35306=0.11$⁠). Since $frep=1Δt$⁠, the laser pulse repetition rate should be above $frep=1204×10−6=4.88 kHz$

Laser Pulse Width, τ: Since the ratio of the depth of field to the working distance is not negligible, one should use $M−=35306=0.11$ to estimate the pulse width $τ<13.5×10−6 m2·0.11×3 m/s=20 μs$⁠. Our laser has $τ=0.01 μs$ and is therefore suitable.

Laser Pulse Energy, E_p: We assume an elliptic Gaussian laser beam with $ωa=50 mm$ and $ωb=150 mm$ 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 $zo+=407 mm$⁠). We used Philip Laven's free-ware² to calculate the differential Mie scattering $σ(90°)=10−16Wcm2$ at $0.407 m$ for $d=15 μm$ and $λ=532 nm$⁠. The solid angle of this particle with the aperture $Da=3.18 mm$ is calculated from Eq. (3) to be $Ωpar=(3.18)216×4072=3.8·10−6 str$⁠. The image diameter of this particle on the image plane is $(dM++CoC)=14+50=64 μm$⁠, where the diameter of the particle due to diffraction (⁠$14 μm$⁠) 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 $e=1.6×10−19$ 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 $1 mJ$ 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 $zo=407 mm$⁠) will have a signal 8 times smaller (the Mie signal is proportional to the intensity of the laser).