Mie Scattering Function

Scott Prahl

1 Aug 2019, Version 6



In [1]:

    
%matplotlib inline

import numpy as np
import matplotlib.pyplot as plt

import miepython as mp

Mie scattering describes the special case of the interaction of light passing through a non-absorbing medium with a single embedded spherical object. The sphere itself can be non-absorbing, moderately absorbing, or perfectly absorbing.

Goals for this notebook:

show how to plot the phase function
explain the units for the scattering phase function
describe normalization of the phase function
show a few examples from classic Mie texts

Geometry

Specifically, the scattering function $p(\theta_i,\phi_i,\theta_o,\phi_o)$ describes the amount of light scattered by a particle for light incident at an angle $(\theta_i,\phi_i)$ and exiting the particle (in the far field) at an angle $(\theta_o,\phi_o)$. For simplicity, the scattering function is often assumed to be rotationally symmetric (it is, obviously, for spherical scatterers) and that the angle that the light is scattered into only depends the $\theta=\theta_o-\theta_i$. In this case, the scattering function can be written as $p(\theta)$. Finally, the angle is often replaced by $\mu=\cos\theta$ and therefore the phase function becomes just $p(\mu)$.

The figure below shows the basic idea. An incoming monochromatic plane wave hits a sphere and produces in the far field two separate monochromatic waves — a slightly attenuated unscattered planar wave and an outgoing spherical wave.

Obviously. the scattered light will be cylindrically symmetric about the ray passing through the center of the sphere.



In [2]:

    
t = np.linspace(0,2*np.pi,100)
xx = np.cos(t)
yy = np.sin(t)
fig,ax=plt.subplots(figsize=(10,8))
plt.axes().set_aspect('equal')
plt.plot(xx,yy)
plt.plot([-5,7],[0,0],'--k')

plt.annotate('incoming irradiance', xy=(-4.5,-2.3),ha='left',color='blue',fontsize=14)
for i in range(6):
    y0 = i -2.5
    plt.annotate('',xy=(-1.5,y0),xytext=(-5,y0),arrowprops=dict(arrowstyle="->",color='blue'))

plt.annotate('unscattered irradiance', xy=(3,-2.3),ha='left',color='blue',fontsize=14)
for i in range(6):
    y0 = i -2.5
    plt.annotate('',xy=(7,y0),xytext=(3,y0),arrowprops=dict(arrowstyle="->",color='blue',ls=':'))

plt.annotate('scattered\nspherical\nwave', xy=(0,1.5),ha='left',color='red',fontsize=16)
plt.annotate('',xy=(2.5,2.5),xytext=(0,0),arrowprops=dict(arrowstyle="->",color='red'))
plt.annotate(r'$\theta$',xy=(2,0.7),color='red',fontsize=14)
plt.annotate('',xy=(2,2),xytext=(2.7,0),arrowprops=dict(connectionstyle="arc3,rad=0.2", arrowstyle="<->",color='red'))

plt.xlim(-5,7)
plt.ylim(-3,3)
plt.axis('off')
plt.show()

Scattered Wave



In [3]:

    
fig,ax=plt.subplots(figsize=(10,8))
plt.axes().set_aspect('equal')
plt.scatter([0],[0],s=30)

m = 1.5
x = np.pi/3
theta = np.linspace(-180,180,180)
mu = np.cos(theta/180*np.pi)
scat = 15 * mp.i_unpolarized(m,x,mu)

plt.plot(scat*np.cos(theta/180*np.pi),scat*np.sin(theta/180*np.pi))
for i in range(12):
    ii = i*15
    xx = scat[ii]*np.cos(theta[ii]/180*np.pi)
    yy = scat[ii]*np.sin(theta[ii]/180*np.pi)
#    print(xx,yy)
    plt.annotate('',xy=(xx,yy),xytext=(0,0),arrowprops=dict(arrowstyle="->",color='red'))
    

plt.annotate('incident irradiance', xy=(-4.5,-2.3),ha='left',color='blue',fontsize=14)
for i in range(6):
    y0 = i -2.5
    plt.annotate('',xy=(-1.5,y0),xytext=(-5,y0),arrowprops=dict(arrowstyle="->",color='blue'))

plt.annotate('unscattered irradiance', xy=(3,-2.3),ha='left',color='blue',fontsize=14)
for i in range(6):
    y0 = i -2.5
    plt.annotate('',xy=(7,y0),xytext=(3,y0),arrowprops=dict(arrowstyle="->",color='blue',ls=':'))

plt.annotate('scattered\nspherical wave', xy=(0,1.5),ha='left',color='red',fontsize=16)

plt.xlim(-5,7)
plt.ylim(-3,3)
#plt.axis('off')
plt.show()

Normalization of the scattered light

So the scattering function or phase function has at least three reasonable normalizations that involve integrating over all $4\pi$ steradians. Below $d\Omega=\sin\theta d\theta\,d\phi$ is a differential solid angle

$$ \begin{align} \int_{4\pi} p(\theta,\phi) \,d\Omega &= 1\\[2mm] \int_{4\pi} p(\theta,\phi) \,d\Omega &= 4\pi \\[2mm] \int_{4\pi} p(\theta,\phi) \,d\Omega &= a \qquad\qquad \mbox{Used by miepython}\\[2mm] \end{align} $$

where $a$ is the single scattering albedo,

$$ a = \frac{\sigma_s}{\sigma_s+\sigma_a} $$

and $\sigma_s$ is the scattering cross section, and $\sigma_a$ is the absorption cross section.

The choice of normalization was made because it accounts for light lost through absorption by the sphere.

If the incident light has units of watts, then the values from the scattering function $p(\theta,\phi)$ have units of radiant intensity or W/sr.

For example, a circular detector with radius $r_d$ at a distance $R$ will subtend an angle

$$ \Omega = \frac{\pi r_d^2}{R^2} $$

(assuming $r_d\ll R$). Now if $P_0$ of light is scattered by a sphere then the scattered power on the detector will be

$$ P_d = P_0 \cdot \Omega \cdot p(\theta,\phi) $$

Examples

Unpolarized Scattering Function

If unpolarized light hits the sphere, then there are no polarization effects to worry about. It is pretty easy to generate a plot to show how scattering changes with angle.



In [4]:

    
m = 1.5
x = np.pi/3
theta = np.linspace(-180,180,180)
mu = np.cos(theta/180*np.pi)
scat = mp.i_unpolarized(m,x,mu)

fig,ax = plt.subplots(1,2,figsize=(12,5))
ax=plt.subplot(121, projection='polar')
ax.plot(theta/180*np.pi,scat)
ax.set_rticks([0.05, 0.1,0.15]) 
ax.set_title("m=1.5, Sphere Diameter = $\lambda$/3")

plt.subplot(122)
plt.plot(theta,scat)
plt.xlabel('Exit Angle [degrees]')
plt.ylabel('Unpolarized Scattered light [1/sr]')
plt.title('m=1.5, Sphere Diameter = $\lambda$/3')
plt.ylim(0.00,0.2)

plt.show()

Rayleigh Scattering

Classic Rayleigh scattering treats small particles with natural (unpolarized) light.
The solid black line denotes the total scattered intensity. The red dashed line is light scattered that is polarized perpendicular to the plane of the graph and the blue dotted line is for light parallel to the plane of the graph. (Compare with van de Hult, Figure 10)



In [5]:

    
m = 1.3
x = 0.01
theta = np.linspace(-180,180,180)
mu = np.cos(theta/180*np.pi)
ipar = mp.i_par(m,x,mu)/2
iper = mp.i_per(m,x,mu)/2
iun  = mp.i_unpolarized(m,x,mu)

fig,ax = plt.subplots(1,2,figsize=(12,5))
ax=plt.subplot(121, projection='polar')
ax.plot(theta/180*np.pi,iper,'r--')
ax.plot(theta/180*np.pi,ipar,'b:')
ax.plot(theta/180*np.pi,iun,'k')

ax.set_rticks([0.05, 0.1,0.15]) 
plt.title('m=%.2f, Sphere Parameter = %.2f' %(m,x))

plt.subplot(122)
plt.plot(theta,iper,'r--')
plt.plot(theta,ipar,'b:')
plt.plot(theta,iun,'k')

plt.xlabel('Exit Angle [degrees]')
plt.ylabel('Normalized Scattered light [1/sr]')
plt.title('m=%.2f, Sphere Parameter = %.2f' %(m,x))
plt.ylim(0.00,0.125)
plt.text(130,0.02,r"$0.5I_{per}$",color="blue", fontsize=16)
plt.text(120,0.062,r"$0.5I_{par}$",color="red", fontsize=16)
plt.text(30,0.11,r"$I_{unpolarized}$",color="black", fontsize=16)


plt.show()

Differential Scattering Cross Section

Sometimes one would like the scattering function normalized so that the integral over all $4\pi$ steradians to be the scattering cross section

$$ \sigma_{sca} = \frac{\pi d^2}{4} Q_{sca} $$

The differential scattering cross section \frac{d\sigma_{sca}}{d\Omega}

$$ \sigma_{sca} = \int_{4\pi} \frac{d\sigma_{sca}}{d\Omega}\,d\Omega $$

Since the unpolarized scattering normalized so its integral is the single scattering albedo, this means that

$$ \frac{Q_{sca}}{Q_{ext}} = \int_{4\pi} p(\mu) \sin\theta\,d\theta d\phi $$

and therefore the differential scattering cross section can be obtained miepython using

$$ \frac{d\sigma_{sca}}{d\Omega} = \frac{\pi d^2 Q_{ext}}{4} \cdot p(\theta,\phi) $$

Note that this is $Q_{ext}$ and not $Q_{sca}$ because of the choice of normalization!

For example, here is a replica of figure 4



In [6]:

    
m = 1.4-0j
lambda0 = 532e-9  # m
theta = np.linspace(0,180,1000)
mu = np.cos(theta* np.pi/180)

d = 1700e-9        # m
x = 2 * np.pi/lambda0 * d/2
geometric_cross_section = np.pi * d**2/4 * 1e4  # cm**2
qext, qsca, qback, g = mp.mie(m,x)
sigma_sca = geometric_cross_section * qext * mp.i_unpolarized(m,x,mu)
plt.semilogy(theta, sigma_sca*1e-3, color='blue')
plt.text(15, sigma_sca[0]*3e-4, "%.0fnm\n(x10$^{-3}$)" % (d*1e9), color='blue')

d = 170e-9         # m
x = 2 * np.pi/lambda0 * d/2
geometric_cross_section = np.pi * d**2/4 * 1e4  # cm**2
qext, qsca, qback, g = mp.mie(m,x)
sigma_sca = geometric_cross_section * qext * mp.i_unpolarized(m,x,mu)
plt.semilogy(theta, sigma_sca, color='red')
plt.text(110, sigma_sca[-1]/2, "%.0fnm" % (d*1e9), color='red')

d = 17e-9                                                       # m
x = 2 * np.pi/lambda0 * d/2
geometric_cross_section = np.pi * d**2/4 * 1e4                 # cm**2
qext, qsca, qback, g = mp.mie(m,x)
sigma_sca = geometric_cross_section * qext * mp.i_unpolarized(m,x,mu)
plt.semilogy(theta, sigma_sca*1e6, color='green')
plt.text(130, sigma_sca[-1]*1e6, "(x10$^6$)\n%.0fnm" % (d*1e9), color='green')

plt.title("Refractive index m=1.4, $\lambda$=532nm")
plt.xlabel("Scattering Angle (degrees)")
plt.ylabel("Diff. Scattering Cross Section (cm$^2$/sr)")
plt.grid(True)
plt.show()

Normalization revisited

Evenly spaced $\mu=\cos\theta$

Start with uniformly distributed scattering angles that are evenly spaced over the cosine of the scattered angle.

Verifying normalization numerically

Specifically, to ensure proper normalization, the integral of the scattering function over all solid angles must be unity

$$ a = \int_0^{2\pi}\int_0^\pi \, p(\theta,\phi)\,\sin\theta\,d\theta\,d\phi $$

or with a change of variables $\mu=\cos\theta$ and using the symmetry to the integral in $\phi$

$$ a = 2\pi \int_{-1}^1 \, p(\mu)\,d\mu $$

This integral can be done numerically by simply summing all the rectangles

$$ a = 2\pi \sum_{i=0}^N p(\mu_i)\,\Delta\mu_i $$

and if all the rectanges have the same width

$$ a = 2\pi\Delta\mu \sum_{i=0}^N p(\mu_i) $$

Case 1. n=1.5, x=1

The total integral total= in the title should match the albedo a=.

For this non-strongly peaked scattering function, the simple integration remains close to the expected value.



In [7]:

    
m = 1.5
x = 1
mu = np.linspace(-1,1,501)
intensity = mp.i_unpolarized(m,x,mu)
qext, qsca, qback, g = mp.mie(m,x)
a = qsca/qext

#integrate over all angles
dmu = mu[1] - mu[0]
total = 2 * np.pi * dmu * np.sum(intensity)

plt.plot(mu,intensity)
plt.xlabel(r'$\cos(\theta)$')
plt.ylabel('Unpolarized Scattering Intensity [1/sr]')
plt.title('m=%.3f%+.3fj, x=%.2f, a=%.3f, total=%.3f'%(m.real,m.imag,x,a, total))
plt.show()

Case 2: m=1.5-1.5j, x=1

Aagin the total integral total= in the title should match the albedo a=.

For this non-strongly peaked scattering function, the simple integration remains close to the expected value.



In [8]:

    
m = 1.5 - 1.5j
x = 1
mu = np.linspace(-1,1,501)
intensity = mp.i_unpolarized(m,x,mu)
qext, qsca, qback, g = mp.mie(m,x)
a = qsca/qext

#integrate over all angles
dmu = mu[1] - mu[0]
total = 2 * np.pi * dmu * np.sum(intensity)

plt.plot(mu,intensity)
plt.xlabel(r'$\cos(\theta)$')
plt.ylabel('Unpolarized Scattering Intensity [1/sr]')
plt.title('m=%.3f%+.3fj, x=%.2f, a=%.3f, total=%.3f'%(m.real,m.imag,x,a, total))
plt.show()

Normalization, evenly spaced $\theta$

The total integral total in the title should match the albedo $a$.

For this non-strongly peaked scattering function, even spacing in $\theta$ improves the accuracy of the integration.



In [9]:

    
m = 1.5-1.5j
x = 1
theta = np.linspace(0,180,361)*np.pi/180
mu = np.cos(theta)

intensity = mp.i_unpolarized(m,x,mu)
qext, qsca, qback, g = mp.mie(m,x)
a = qsca/qext

#integrate over all angles
dtheta = theta[1]-theta[0]
total = 2 * np.pi * dtheta * np.sum(intensity* np.sin(theta))

plt.plot(mu,intensity)
plt.xlabel(r'$\cos(\theta)$')
plt.ylabel('Unpolarized Scattering Intensity [1/sr]')
plt.title('m=%.3f%+.3fj, x=%.2f, a=%.3f, total=%.3f'%(m.real,m.imag,x,a, total))
plt.show()

Comparison to Wiscombe's Mie Program

Wiscombe normalizes as

$$ \int_{4\pi} p(\theta,\phi) \,d\Omega = \pi x^2 Q_{sca} $$

where $p(\theta)$ is the scattered light.

Once corrected for differences in phase function normalization, Wiscombe's test cases match those from miepython exactly.

Wiscombe's Test Case 14



In [10]:

    
"""
MIEV0 Test Case 14:  Refractive index:  real    1.500  imag  -1.000E+00,  Mie size parameter =     1.000
  Angle    Cosine           S-sub-1                     S-sub-2               Intensity  Deg of Polzn
   0.00  1.000000   5.84080E-01   1.90515E-01   5.84080E-01   1.90515E-01   3.77446E-01        0.0000
  30.00  0.866025   5.65702E-01   1.87200E-01   5.00161E-01   1.45611E-01   3.13213E-01       -0.1336
  60.00  0.500000   5.17525E-01   1.78443E-01   2.87964E-01   4.10540E-02   1.92141E-01       -0.5597
  90.00  0.000000   4.56340E-01   1.67167E-01   3.62285E-02  -6.18265E-02   1.20663E-01       -0.9574
"""

x=1.0
m=1.5-1.0j
mu=np.cos(np.linspace(0,90,4) * np.pi/180)

qext, qsca, qback, g = mp.mie(m,x)
albedo = qsca/qext
unpolar = mp.i_unpolarized(m,x,mu) # normalized to a
unpolar /= albedo                  # normalized to 1

unpolar_miev = np.array([3.77446E-01,3.13213E-01,1.92141E-01,1.20663E-01])
unpolar_miev /= np.pi * qsca * x**2                         # normalized to 1
ratio = unpolar_miev/unpolar

print("MIEV0 Test Case 14:  m=1.500-1.000j,  Mie size parameter =     1.000")
print()
print("             %9.1f°%9.1f°%9.1f°%9.1f°"%(0,30,60,90))
print("MIEV0        %9.5f %9.5f %9.5f %9.5f"%(unpolar_miev[0],unpolar_miev[1],unpolar_miev[2],unpolar_miev[3]))
print("miepython    %9.5f %9.5f %9.5f %9.5f"%(unpolar[0],unpolar[1],unpolar[2],unpolar[3]))
print("ratio        %9.5f %9.5f %9.5f %9.5f"%(ratio[0],ratio[1],ratio[2],ratio[3]))









    



MIEV0 Test Case 14:  m=1.500-1.000j,  Mie size parameter =     1.000

                   0.0°     30.0°     60.0°     90.0°
MIEV0          0.18109   0.15027   0.09218   0.05789
miepython      0.18109   0.15027   0.09218   0.05789
ratio          1.00000   1.00000   1.00000   1.00000

Wiscombe's Test Case 10



In [11]:

    
"""
MIEV0 Test Case 10:  Refractive index:  real    1.330  imag  -1.000E-05,  Mie size parameter =   100.000
  Angle    Cosine           S-sub-1                     S-sub-2               Intensity  Deg of Polzn
   0.00  1.000000   5.25330E+03  -1.24319E+02   5.25330E+03  -1.24319E+02   2.76126E+07        0.0000
  30.00  0.866025  -5.53457E+01  -2.97188E+01  -8.46720E+01  -1.99947E+01   5.75775E+03        0.3146
  60.00  0.500000   1.71049E+01  -1.52010E+01   3.31076E+01  -2.70979E+00   8.13553E+02        0.3563
  90.00  0.000000  -3.65576E+00   8.76986E+00  -6.55051E+00  -4.67537E+00   7.75217E+01       -0.1645
"""

x=100.0
m=1.33-1e-5j
mu=np.cos(np.linspace(0,90,4) * np.pi/180)

qext, qsca, qback, g = mp.mie(m,x)
albedo = qsca/qext
unpolar = mp.i_unpolarized(m,x,mu) # normalized to a
unpolar /= albedo                  # normalized to 1

unpolar_miev = np.array([2.76126E+07,5.75775E+03,8.13553E+02,7.75217E+01])
unpolar_miev /= np.pi * qsca * x**2                         # normalized to 1
ratio = unpolar_miev/unpolar

print("MIEV0 Test Case 10:  m=1.330-0.00001j,  Mie size parameter =     100.000")
print()
print("             %9.1f°%9.1f°%9.1f°%9.1f°"%(0,30,60,90))
print("MIEV0        %9.5f %9.5f %9.5f %9.5f"%(unpolar_miev[0],unpolar_miev[1],unpolar_miev[2],unpolar_miev[3]))
print("miepython    %9.5f %9.5f %9.5f %9.5f"%(unpolar[0],unpolar[1],unpolar[2],unpolar[3]))
print("ratio        %9.5f %9.5f %9.5f %9.5f"%(ratio[0],ratio[1],ratio[2],ratio[3]))









    



MIEV0 Test Case 10:  m=1.330-0.00001j,  Mie size parameter =     100.000

                   0.0°     30.0°     60.0°     90.0°
MIEV0        419.22116   0.08742   0.01235   0.00118
miepython    419.22168   0.08742   0.01235   0.00118
ratio          1.00000   1.00000   1.00000   1.00000

Wiscombe's Test Case 7



In [12]:

    
"""
MIEV0 Test Case  7:  Refractive index:  real    0.750  imag   0.000E+00,  Mie size parameter =    10.000
  Angle    Cosine           S-sub-1                     S-sub-2               Intensity  Deg of Polzn
   0.00  1.000000   5.58066E+01  -9.75810E+00   5.58066E+01  -9.75810E+00   3.20960E+03        0.0000
  30.00  0.866025  -7.67288E+00   1.08732E+01  -1.09292E+01   9.62967E+00   1.94639E+02        0.0901
  60.00  0.500000   3.58789E+00  -1.75618E+00   3.42741E+00   8.08269E-02   1.38554E+01       -0.1517
  90.00  0.000000  -1.78590E+00  -5.23283E-02  -5.14875E-01  -7.02729E-01   1.97556E+00       -0.6158
"""

x=10.0
m=0.75
mu=np.cos(np.linspace(0,90,4) * np.pi/180)

qext, qsca, qback, g = mp.mie(m,x)
albedo = qsca/qext
unpolar = mp.i_unpolarized(m,x,mu) # normalized to a
unpolar /= albedo                  # normalized to 1

unpolar_miev = np.array([3.20960E+03,1.94639E+02,1.38554E+01,1.97556E+00])
unpolar_miev /= np.pi * qsca * x**2                          # normalized to 1
ratio = unpolar_miev/unpolar

print("MIEV0 Test Case 7:  m=0.75,  Mie size parameter =     10.000")
print()
print("             %9.1f°%9.1f°%9.1f°%9.1f°"%(0,30,60,90))
print("MIEV0        %9.5f %9.5f %9.5f %9.5f"%(unpolar_miev[0],unpolar_miev[1],unpolar_miev[2],unpolar_miev[3]))
print("miepython    %9.5f %9.5f %9.5f %9.5f"%(unpolar[0],unpolar[1],unpolar[2],unpolar[3]))
print("ratio        %9.5f %9.5f %9.5f %9.5f"%(ratio[0],ratio[1],ratio[2],ratio[3]))









    



MIEV0 Test Case 7:  m=0.75,  Mie size parameter =     10.000

                   0.0°     30.0°     60.0°     90.0°
MIEV0          4.57673   0.27755   0.01976   0.00282
miepython      4.57673   0.27755   0.01976   0.00282
ratio          1.00000   1.00000   1.00000   1.00000

Comparison to Bohren & Huffmans's Mie Program

Bohren & Huffman normalizes as

$$ \int_{4\pi} p(\theta,\phi) \,d\Omega = 4 \pi x^2 Q_{sca} $$

Bohren & Huffmans's Test Case 14



In [13]:

    
"""
BHMie Test Case 14, Refractive index = 1.5000-1.0000j, Size parameter = 1.0000

Angle  Cosine              S1                        S2             
0.00   1.0000  -8.38663e-01 -8.64763e-01  -8.38663e-01 -8.64763e-01 
0.52   0.8660  -8.19225e-01 -8.61719e-01  -7.21779e-01 -7.27856e-01
1.05   0.5000  -7.68157e-01 -8.53697e-01  -4.19454e-01 -3.72965e-01
1.57   0.0000  -7.03034e-01 -8.43425e-01  -4.44461e-02  6.94424e-02
"""

x=1.0
m=1.5-1j
mu=np.cos(np.linspace(0,90,4) * np.pi/180)

qext, qsca, qback, g = mp.mie(m,x)
albedo = qsca/qext
unpolar = mp.i_unpolarized(m,x,mu) # normalized to a
unpolar /= albedo                  # normalized to 1

s1_bh = np.empty(4,dtype=np.complex)
s1_bh[0] = -8.38663e-01 - 8.64763e-01*1j
s1_bh[1] = -8.19225e-01 - 8.61719e-01*1j
s1_bh[2] = -7.68157e-01 - 8.53697e-01*1j
s1_bh[3] = -7.03034e-01 - 8.43425e-01*1j

s2_bh = np.empty(4,dtype=np.complex)
s2_bh[0] = -8.38663e-01 - 8.64763e-01*1j
s2_bh[1] = -7.21779e-01 - 7.27856e-01*1j
s2_bh[2] = -4.19454e-01 - 3.72965e-01*1j
s2_bh[3] = -4.44461e-02 + 6.94424e-02*1j

# BHMie seems to normalize their intensities to 4 * pi * x**2 * Qsca
unpolar_bh = (abs(s1_bh)**2+abs(s2_bh)**2)/2
unpolar_bh /= np.pi * qsca * 4 * x**2                          # normalized to 1
ratio = unpolar_bh/unpolar

print("BHMie Test Case 14:  m=1.5000-1.0000j, Size parameter = 1.0000")
print()
print("             %9.1f°%9.1f°%9.1f°%9.1f°"%(0,30,60,90))
print("BHMIE        %9.5f %9.5f %9.5f %9.5f"%(unpolar_bh[0],unpolar_bh[1],unpolar_bh[2],unpolar_bh[3]))
print("miepython    %9.5f %9.5f %9.5f %9.5f"%(unpolar[0],unpolar[1],unpolar[2],unpolar[3]))
print("ratio        %9.5f %9.5f %9.5f %9.5f"%(ratio[0],ratio[1],ratio[2],ratio[3]))
print()
print("Note that this test is identical to MIEV0 Test Case 14 above.")
print()
print("Wiscombe's code is much more robust than Bohren's so I attribute errors all to Bohren")









    



BHMie Test Case 14:  m=1.5000-1.0000j, Size parameter = 1.0000

                   0.0°     30.0°     60.0°     90.0°
BHMIE          0.17406   0.14780   0.09799   0.07271
miepython      0.18109   0.15027   0.09218   0.05789
ratio          0.96118   0.98353   1.06296   1.25600

Note that this test is identical to MIEV0 Test Case 14 above.

Wiscombe's code is much more robust than Bohren's so I attribute errors all to Bohren

Bohren & Huffman, water droplets

Tiny water droplet (0.26 microns) in clouds has pretty strong forward scattering! A graph of this is figure 4.9 in Bohren and Huffman's Absorption and Scattering of Light by Small Particles.

A bizarre scaling factor of $16\pi$ is needed to make the miepython results match those in the figure 4.9.



In [14]:

    
x=3
m=1.33-1e-8j

theta = np.linspace(0,180,181)
mu = np.cos(theta*np.pi/180)

scaling_factor = 16*np.pi
iper = scaling_factor*mp.i_per(m,x,mu)
ipar = scaling_factor*mp.i_par(m,x,mu)

P = (iper-ipar)/(iper+ipar)

plt.subplots(2,1,figsize=(8,8))
plt.subplot(2,1,1)
plt.semilogy(theta,ipar,label='$i_{par}$')
plt.semilogy(theta,iper,label='$i_{per}$')
plt.xlim(0,180)
plt.xticks(range(0,181,30))
plt.ylabel('i$_{par}$ and i$_{per}$')
plt.legend()
plt.title('Figure 4.9 from Bohren & Huffman')
plt.subplot(2,1,2)
plt.plot(theta,P)
plt.ylim(-1,1)
plt.xticks(range(0,181,30))
plt.xlim(0,180)
plt.ylabel('Polarization')
plt.plot([0,180],[0,0],':k')
plt.xlabel('Angle (Degrees)')
plt.show()

van de Hulst Comparison

This graph (see figure 29 in Light Scattering by Small Particles) was obviously constructed by hand. In this graph, van de Hulst worked hard to get as much information as possible



In [15]:

    
x=5
m=10000

theta = np.linspace(0,180,361)
mu = np.cos(theta*np.pi/180)

fig, ax = plt.subplots(figsize=(8,8))

x=10
s1,s2 = mp.mie_S1_S2(m,x,mu)
sone = 2.5*abs(s1)
stwo = 2.5*abs(s2)
plt.plot(theta,sone,'b')
plt.plot(theta,stwo,'--r')
plt.annotate('x=%.1f  '%x,xy=(theta[-1],sone[-1]),ha='right',va='bottom')

x=5
s1,s2 = mp.mie_S1_S2(m,x,mu)
sone = 2.5*abs(s1) + 1
stwo = 2.5*abs(s2) + 1
plt.plot(theta,sone,'b')
plt.plot(theta,stwo,'--r')
plt.annotate('x=%.1f  '%x,xy=(theta[-1],sone[-1]),ha='right',va='bottom')

x=3
s1,s2 = mp.mie_S1_S2(m,x,mu)
sone = 2.5*abs(s1) + 2
stwo = 2.5*abs(s2) + 2
plt.plot(theta,sone,'b')
plt.plot(theta,stwo,'--r')
plt.annotate('x=%.1f  '%x,xy=(theta[-1],sone[-1]),ha='right',va='bottom')

x=1
s1,s2 = mp.mie_S1_S2(m,x,mu)
sone = 2.5*abs(s1) + 3
stwo = 2.5*abs(s2) + 3
plt.plot(theta,sone,'b')
plt.plot(theta,stwo,'--r')
plt.annotate('x=%.1f  '%x,xy=(theta[-1],sone[-1]),ha='right',va='bottom')

x=0.5
s1,s2 = mp.mie_S1_S2(m,x,mu)
sone = 2.5*abs(s1) + 4
stwo = 2.5*abs(s2) + 4
plt.plot(theta,sone,'b')
plt.plot(theta,stwo,'--r')
plt.annotate('x=%.1f  '%x,xy=(theta[-1],sone[-1]),ha='right',va='bottom')

plt.xlim(0,180)
plt.ylim(0,5.5)
plt.xticks(range(0,181,30))
plt.yticks(np.arange(0,5.51,0.5))
plt.title('Figure 29 from van de Hulst, Non-Absorbing Spheres')
plt.xlabel('Angle (Degrees)')
ax.set_yticklabels(['0','1/2','0','1/2','0','1/2','0','1/2','0','1/2','5',' '])
plt.grid(True)
plt.show()

Comparisons with Kerker, Angular Gain

Another interesting graph is figure 4.51 from The Scattering of Light by Kerker.

The angular gain is $$ G_1 = \frac{4}{x^2} |S_1(\theta)|^2 \qquad\mbox{and}\qquad G_2 = \frac{4}{x^2} |S_2(\theta)|^2 $$



In [16]:

    
## Kerker, Angular Gain

x=1
m=10000

theta = np.linspace(0,180,361)
mu = np.cos(theta*np.pi/180)

fig, ax = plt.subplots(figsize=(8,8))

s1,s2 = mp.mie_S1_S2(m,x,mu)

G1 = 4*abs(s1)**2/x**2
G2 = 4*abs(s2)**2/x**2

plt.plot(theta,G1,'b')
plt.plot(theta,G2,'--r')
plt.annotate('$G_1$',xy=(50,0.36),color='blue',fontsize=14)
plt.annotate('$G_2$',xy=(135,0.46),color='red',fontsize=14)

plt.xlim(0,180)
plt.xticks(range(0,181,30))
plt.title('Figure 4.51 from Kerker, Non-Absorbing Spheres, x=1')
plt.xlabel('Angle (Degrees)')
plt.ylabel('Angular Gain')
plt.show()



In [ ]: