= Solution
\b[Normalization and angular pattern.] A <scattering phase function> describes the distribution of single-scattered power over outgoing <solid angle>, for a specified incident direction, wavelength and <polarization> convention. Use the common normalization
$$
\frac1{4\pi}\int_{4\pi}P(\Theta)\,d\Omega=1,\qquad
\boxed{g=\frac1{4\pi}\int_{4\pi}P(\Theta)\cos\Theta\,d\Omega}.
$$
The <scattering asymmetry parameter> is the mean cosine of the <scattering angle>: forward scattering has positive $g$, backward scattering negative $g$, and a forward–backward symmetric pattern has $g=0$.
Let $C=E_0(2\pi/\lambda)^4[(n^2-1)/(n^2+2)]^2r^6$. The given small-sphere angular intensity is $I=C(1+\mu^2)/2$, where $\mu=\cos\Theta$. Its total scattered power is
$$
\int I\,d\Omega=2\pi C\int_{-1}^1\frac{1+\mu^2}{2}\,d\mu=\frac{8\pi C}3.
$$
Normalizing gives
$$
\boxed{P(\Theta)=\frac{4\pi I}{\int I\,d\Omega}=\frac34(1+\cos^2\Theta)}.
$$
Dimensionally the supplied $I$ is <radiant intensity>, power per <solid angle>. If interpreted as detector <irradiance> at a distance, the appropriate inverse-square distance factor is also needed; it cancels out of the normalized angular pattern.
The polar plot has equal forward and backward lobes, each with radius $3/2$, and lateral values $3/4$. A small nonabsorbing particle behaves as an induced <electric dipole>. Radiation transverse to the dipole and the average over two orthogonal incident <polarizations> produce the $1+\cos^2\Theta$ form. Its odd angular moment vanishes:
$$
\boxed{g=\frac38\int_{-1}^1\mu(1+\mu^2)\,d\mu=0}.
$$
As the size parameter $2\pi r/\lambda$ grows, <Rayleigh scattering> in the electric-dipole approximation no longer suffices. Higher multipoles and interference produce structure in the angular pattern, and large particles generally develop a strong narrow forward <diffraction> peak. Their $g$ is normally positive, although detailed values and angular features depend on <refractive index>, size distribution and shape; they are not fixed by radius alone. Large spheres may show rainbows and backscattering structure, while nonspherical ice introduces further angular features.
\Image[/past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2016-rayleigh-scattering-phase.png]
{title=Normalized Rayleigh phase function for an unpolarized incident beam}
\b[Optically thick transmission.] In a nonabsorbing two-stream or transport approximation, it is the transport <optical depth> $(1-g)\tau$ that controls the loss of forward directional memory. Using the same idealized albedo closure as in option 1 gives total forward-hemisphere transmittance
$$
T\simeq1-A=\frac2{2+(1-g)\tau}.
$$
For the same large but finite extinction <optical depth>,
$$
\boxed{T_{\rm water}>T_{\rm ice}>T_{\rm small\ smoke}}.
$$
The respective transport-depth factors are $0.15$, $0.30$ and $1$. The water <cloud> scatters most strongly forward and returns the least light backward. All transmittances tend to zero in the infinite-depth limit. The strictly unscattered direct-beam transmittance is instead $e^{-\tau}$, identical in the three cases: the ranking concerns total transmitted radiation, including forward-scattered light, not that direct component alone. Full <radiative transfer> can also depend on higher angular moments beyond $g$.
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