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The Brewster angle is also called the polarization angle because the light that reflects from a surface at this angle is entirely polarized perpendicular to the plane of incidence.

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Analytically, the focal length is described by the lens maker's equation: 1/f = (n - 1)(1/R1 + 1/R2), where R1 and R2 are the radii of curvature, f is the focal ...

Hope so , this article has given you a valid reason why the Brewster angle is not defined for perpendicular polarization.

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sin^{2}\theta {B\perp }= \frac{1-\frac{\mu {1}\varepsilon {2}}{\mu {2}\varepsilon {1}}}{1-\left ( \mu {1}/\mu _{2} \right )^{2}} —(1)

Brewster angle is the angle of incidence for which the reflection coefficient is zero. But, as we know, we have two cases of polarizations viz. perpendicular and parallel.

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sin^{2}\theta {B||}= \frac{1-\frac{\mu {2}\varepsilon {1}}{\mu {1}\varepsilon {2}}}{1-\left ( \epsilon {1}/\epsilon _{1} \right )^{2}} —(2)

Medium 1 will have intrinsic impedance \eta_{1} and medium 2 will have intrinsic impedance as \eta_{2}. So, due to impedance mismatching, some power gets reflected back we called it a reflected wave and the remaining power which gets transmitted back into another medium is called a transmitted wave.

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We know that an EM wave carries electromagnetic power with it. It depends on the intrinsic impedance of the medium it travels.

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When the electromagnetic wave travels from one medium to another then due to impedance mismatching some power gets transmitted into one medium while some power gets reflected back into another medium.

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Let us see the definition of the reflection coefficient, it is the ratio of the peak value of E in the reflected wave to the peak value of E in the incident wave.

Mathematical reason :- Let us see mathematical proof, that Why the Brewster angle is not defined for perpendicular polarization.

And, you must be aware that Brewster angle is not practically defined for the perpendicular polarization. Let’s see, why?

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