Someone argues: "Aluminium's refractive index at is

PHYSICS · Multi-part · Fermi University Physics · Optics & Photonics · Wave propagation and light in media · Wave propagation and light in media

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Problem

Someone argues: "Aluminium's refractive index at 633 nm633\ \mathrm{nm} is n~=1.37+7.62i\tilde n = 1.37+7.62i. Since Re(n~)=1.37>1\mathrm{Re}(\tilde n)=1.37>1, aluminium must have a true Brewster angle where Rp=0\mathcal{R}_p=0, found the same way as for a dielectric: tan⁡θB=Re(n~)=1.37\tan\theta_B = \mathrm{Re}(\tilde n) = 1.37, giving θB=53.9∘\theta_B = 53.9^\circ." ((a)) Using rp=nicos⁡θt−n~cos⁡θinicos⁡θt+n~cos⁡θir_p = \dfrac{n_i\cos\theta_t - \tilde n\cos\theta_i}{n_i\cos\theta_t + \tilde n\cos\theta_i} with ni=1n_i=1 (air) and sin⁡θt=sin⁡θi/n~\sin\theta_t=\sin\theta_i/\tilde n, compute Rp=∣rp∣2\mathcal{R}_p=|r_p|^2 at θi=53.9∘\theta_i=53.9^\circ for aluminium. ((b)) State what this shows about the claim, and quote the true minimum of Rp\mathcal{R}_p for aluminium and the angle at which it occurs.

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