In almost all sources I read so far, Brewster's angle is defined for light incident on an optically denser medium from an optically rarer medium. Is the concept of Brewster's angle applicable when light passes from an optically denser medium to an optically rarer medium? Will the reflected ray be completely polarized as in the usual case? In angles other than Brewster's angle, is the reflected or refracted rays partially polarized (composed predominantly of one direction of polarization over the other perpendicular direction) as in the regular case?
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2 Answers
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The Brewster angle is $\theta_B= \arctan (n_2/n_1)$, where $n_1$ is the refractive index for the light incident on an interface, which has a value whichever of $n_1$ or $n_2$ is greater. This angle is always less than the critical angle for total internal reflection, $\theta_c = \arcsin(n_2/n_1)$, where a critical angle exists ($n_1 > n_2$), since $\arcsin(x) > \arctan(x)$ for $0<x<1$.
The reflected wave is completely polarised at the Brewster angle. The transmitted wave is partially polarised. At other angles there is partial polarisation of both reflected and transmitted waves.
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The concept of the Brewster angle has no restriction on the ratio of the refractive index of the incident and transmitted medium. One can see it as follows.
Let $p$-polarized light is incident from a medium of refractive index $n_1$ to a medium with index $n_2$. We define the relative index $n:=\frac{n_1}{n_2}$. The reflection coefficient for $p$-polarized light reads $${r_p} = \frac{{\cos \left( {{\theta _1}} \right) - n\cos \left( {{\theta _2}} \right)}}{{\cos \left( {{\theta _1}} \right) + n\cos \left( {{\theta _2}} \right)}}.$$
To fulfill the Brewster condition, namely zero reflection coefficient, we demand $r_p=0$. Therefore
\begin{align} & \cos \left( {{\theta _1}} \right) - n\cos \left( {{\theta _2}} \right) = 0, \\ \Rightarrow~ &\cos \left( {{\theta _1}} \right) = n\sqrt {1 - {{\sin }^2}\left( {{\theta _2}} \right)} , \\ \Rightarrow~ &\cos \left( {{\theta _1}} \right) = n\sqrt {1 - \frac{1}{{n_2^2}} \cdot n_2^2 \cdot {{\sin }^2}\left( {{\theta _2}} \right)} , \\ \Rightarrow~ &\cos\left( {{\theta _1}} \right) = n\sqrt {1 - \frac{1}{{n_2^2}} \cdot n_1^2 \cdot {{\sin }^2}\left( {{\theta _1}} \right)},\\ &~~~~~~~~~~~~~~~~~~~~~~~~~~~\mathrm{Using~Snell's~law: {n_1}\sin \left( {{\theta _1}} \right) = {n_2}\sin \left( {{\theta _2}} \right).}\\ \\ \Rightarrow~ & \cos^2\left( {{\theta _1}} \right) = {n^2}\left( {1 - {n^2}{{\sin }^2}\left( {{\theta _1}} \right)} \right), \\ \\ \Rightarrow~ & 1 - {\sin ^2}\left( {{\theta _1}} \right) = {n^2}\left( {1 - {n^2}{{\sin }^2}\left( {{\theta _1}} \right)} \right), \\ \\ \therefore~ & \sin \left( {{\theta _1}} \right) = \sqrt {\frac{1}{{1 + {n^2}}}}. \tag{1} \label{ppolaBrw} \end{align}
As we can see from Eq. \eqref{ppolaBrw}, regardless of the value of relative index $n = \frac{n_1}{n_2}$, there is always an angle $\theta_1$ when the reflection coefficient is zero.
