About partially polarized light and the degree of polariztion When I was taking Optics course, I found there were several questions about polarization of light. I use the textbook of Hecht.


*

*It seems that the definition of degree of polarization may be not so well-defined if $V=\frac{I_{max}-I_{min}}{I_{max}+I_{min}}$. For a elliptical polarized light, there is no natural polarized part, but still $V\neq 0$.

*I found it hard to deal with partially polarized light. First the definition. What is the definition of partially polarized light? Light with $0<V<1$?

*Is all partially polarized light can be presented by the superposition of a plane polarized light and a natural light? I suppose it is true but why? Is there any formal explanation?

*Consider a real problem. Suppose there is a beam of light incident on an air-glass interface with $n_{ti}=1.5$ at a certain degree, say $30°$. Then how to characterize the reflected light or the transmitted light since they are all partially polarized. What is $V$ for these lights? I gauss the answer may be $V=\frac{|r_p|^2-|r_s|^2}{|r_p|^2+|r_s|^2}$. But I can't convince my self why this correspondes to the definition above. (This is essentially a problem in Hecht.)
Thanks a lot! I am really confused with that.
 A: Here is some possibly useful information from Goodman's "Statistical Optics."  (Sorry about the lack of symbol quality --  so much for cut/paste from a PDF)

Light from a thermal source is regarded as unpolarized if two conditions are
      met. First, we require that the intensity of the light passed by a polarization
      analyzer, situated in a plane perpendicular to the direction of propagatof the wave, be independent of the rotational orientation of the analyzer.
      Second, we require that any two orthogonal field components $u_x( P, t) $ and
      $u_y(P, t)$ have the property that $(u_x(t + T)u_t(t))$ is identically zero for all
      rotational orientations of the $X- Y$ coordinate axes and for all delays $T$.

Later, in section 4.3.3, 

We define the degree of polarization of the wave as the ratio of
      the intensity of the polarized component to the total intensity,

I'd recommend reading section 4.3 in its entirety.
A: 
It seems that the definition of degree of polarization may be not so well-defined if $V=\frac{I_{max}-I_{min}}{I_{max}+I_{min}}$. For a
  elliptical polarized light, there is no natural polarized part, but
  still $V\neq 0$.

This definition is used in measurements.  Elliptical polarized light can be fully polarized, i.e. $V=1$ (theoretically).  I really don't like that Hecht uses V here, as that is also used for visibility which is an interference thing (but I understand why, see below).  
Specifically, we define the Stokes parameters 
$$\mathbf{s} = \begin{pmatrix}s_0 \\ s_1 \\ s_2 \\ s_3\end{pmatrix}$$.
Then the degree of polarization $$\mathscr{P} = \frac{\sqrt{s_1^2 + s_2^2 + s_3^2}}{s_0}$$
Partial polarization is a statistical quantity, it is the result of taking the ensemble average over some domain of the irradiance projection through polarization elements.  This means that it is really a coherence (i.e. interference) phenomena.

I found it hard to deal with partially polarized light. First the definition. What is the definition of partially polarized light? Light
  with $0<V<1$?

This is because for a deterministic ideal, mathematical point in space-time, partial polarization doesn't exist.  Partial polarization is the statistical time average of a time or space or space-time region.  Yes, this definition implies that $0<V<1$.
This is because optical detectors only measure irradiance, not optical field.  In longer wavelength regions, antennas can capture both phase and amplitude and the polarization can be reconstructed from those parameters, but this is not the case for optical detectors.
Mathematically :
$s_0 = \langle E_x^2 + E_y^2\rangle\\
s_1 = \langle E_x^2 - E_y^2\rangle\\
s_2 = 2\langle E_xE_y\cos(\delta)\rangle\\
s_3 = 2\langle E_xE_y\sin(\delta)\rangle
$
where $E_x$ and $E_y$ are the classical electric field components in the $x$ and $y$ directions (in optics we use the $z$-axis as the direction of energy flux), both functions of $t$ of course, and
$$\langle f(t) \rangle = lim_{T \to \infty}\frac{1}{T}\int_0^T f(t) dt $$
is the time average, and $\delta$ is the phase difference between $E_x$ and $E_y$.
Open access source aren't great for this explanation, but this paper may help.

Is all partially polarized light can be presented by the superposition
  of a plane polarized light and a natural light? I suppose it is true
  but why? Is there any formal explanation?

Not a super-position of natural light and polarized light, but a super-position of purely polarized light and of completely depolarized light.  It is obvious why if we look at the definition of $\mathscr{P}$, if $\mathscr{P} = 0$, then the light is completely depolarized and $s_1 = s_2 = s_3 = 0$. 
Then
$$\mathbf{s} = (1-\mathscr{P})\begin{pmatrix}s_0 \\ 0 \\ 0 \\ 0\end{pmatrix} + \mathscr{P}\begin{pmatrix}s_0 \\ s_1 \\ s_2 \\ s_3\end{pmatrix}$$
Assuming that the Stokes parameters are linear, (which follows from the linearity of optical fields in most dielectrics, air, etc.) then I'll leave it to you to verify this is correct.

Consider a real problem. Suppose there is a beam of light incident on
  an air-glass interface with $n_{ti}=1.5$ at a certain degree, say
  $30°$. Then how to characterize the reflected light or the transmitted
  light since they are all partially polarized. What is $V$ for these
  lights? I guess the answer may be
  $V=\frac{|r_p|^2-|r_s|^2}{|r_p|^2+|r_s|^2}$. But I can't convince my
  self why this corresponds to the definition above. (This is
  essentially a problem in Hecht.)

What you have written for $V$ here is called the diattenuation, not the depolarization.  This is the difference in amplitude between the the $s$ and $p$ polarizations reflected.  Remember that $s$ and $p$ are defined by the geometry and symmetries of the problem.
