Question

Consider the case of light incident from air (or vacuum) on a planar interface with diamond. Calculate, (a) the reflection and transmission coefficients at normal incidence. (b) the reflection and transmission coefficients at the Brewster angle. (c) the “crossover” angle where the transmission and reflection coefficients are equal. Referring back to part (b), we might have imagined that, as the reflection coefficient is zero at Brewster's angle, that the transmission coefficient would equal one. But as your answer should have showed, this is not the case. (d) Show that instead, this value of transmission coefficient corresponds to the condition that Si.n=St.n, where Si, St are the incident and transmitted Poynting vectors and n is the surface normal. What is the physical significance of this statement.

Answer 1

① For normal incidence Di=0 =) Ot=0 wing freuel kauctions nicosdi - ne cos Ot. 9 Sami nicos Oi- necos Oz retend ni ntr nit ne nicosat - Nicos oi than 6 H = Mi Cosat + nt ni- nt. nitnt ni casei nicos oi tulosot x x to = 2ni cosoi Ni Cosatt he cosi nicos 2 ni tt = nient z ni nitnt til = ( - )

VVV AN 111 for light incident and brewster's angle Di= Op Angle ware between reflected and transmitted = 30° =) Opt Ot = 90° tan (Op-De). T etom Opf Ot) ri tom (Optot) Vio 11 = & Sin Oz Cos oi Sin (Ot+ 0) Cos (Oi-Ot) 2 Sin Of Bos Op Sin (90) eos Op-ot) - 2 cos Op Cos of = cos(@p-0+) tl to . Op> Ot.

At (ross over angle rutu and cnicos di nicos Of Mt Cosot ni Cosot+ht Color - nicos q + htcore Giren Mi=1 nt = 2.42 diamont and, los ot - 2.42 Cosoi Cosat + 2-42 Cosoi - 2cos di. Caso +2:42 costi or Corot - 2.42 Cosoi - 2 Cos di 4.42 Col di cost - Qiz 863° & Sin - sen - = Eix Hinore - (Ext) on Ei ti cos oi = Er He cost Hi Coroi Et tu = HE Cosat M Cordi Cosot Misinde = Mesinot Mi=1 sin di = utsin Oz - ² sindi Sec²d-112 tam at Cos of B (Sun) = st.n) is a conservation of energy (initial final)