Wave OpticsMedium

Question

An unpolarized light is incident on the plane interface of air-dielectric medium shown in figure. If the incident angle is equal to Brewster angle, identify the expression representing reflected wave.

Options

A.$\left( E_{x}\widehat{i} + E_{y}\widehat{j} \right)sin(kx - kz - \omega t)$
B.$\left( E_{x}\widehat{i} + E_{z}\widehat{k} \right)sin(kx + ky - \omega t)$
C.$\left( E_{x}\widehat{j} + E_{y}\widehat{k} \right)sin(ky + kz - \omega t)$
D.$\left( E_{x}\widehat{i} + E_{y}\widehat{j} + E_{z}\widehat{k} \right)sin(kx + ky - kz - \omega t)$

Solution

Sol. [Key Concept: At Brewster's angle the reflected light is plane-polarised along $y$ and travels as $kx - kz - \omega\ t$ with no $z$-component.]

Is the light is propagating along the positive z -axis and its is reflected from plane then correct phase for reflected light is $kx - kz - \omega t$ and for electric field as the light incident on Brewster angle so the reflected light should be perfectly plane polarized that's why polarization direction should be y -axis so electric field of reflected light contain no zcomponent.

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