Answer Happy

1. (a) Obtain the ratio of reflected (Ex) and incident (E.) and transmitted (Eur) and incident electric field amplitudes given below for light in normal incidence on a planar dielectric interface between medium 1 and medium 2 with refractive indices and wavenumbers, 1 k,and na ka ER E ET Nu - 12 1+12 2n =t 12 + 12 EI [4 marks] (b) A thin dielectric layer with refractive index, nz, is sandwiched between two semi- infinite dielectrics of refractive index, n. The thickness, d, of the layer is comparable to the wavelength of light, 2. None of the dielectrics absorbs radiation at this wavelength. Light with wavenumber, kr, is incident on the layer along the interface normal. The incident electric field amplitude is Eos. Find the electric field strength just above the lower interface at z = d between medium 2 and medium 1, for an incoming wave. (z = d' means z just less than d). ܐ n. [3 marks] n2 z=d Note: (c) Find the electric field strength just below the upper surface of the dielectric layer at z = 0 after propagation to z=d and a single reflection at z =d. (z = 0* means z just greater than o) [4 marks] (d) Explain how you would calculate the overall reflectivity of the thin layer for light incident from medium 1 along the interface normal. Illustrate your answer with a diagram. Hint consider multiple reflections by both interfaces with escape of some of the light into the two regions with refractive index nz. Show that this leads to the following expression for the ratio of the reflected and incident fields, Esk/Ex of the thin dielectric sheet embedded in medium 1 at z = 0. =r-trtek - trpetik-trtebiky 212 t' T: + 2 Note: r and were defined in part (a). (8 marks] (0) Show that the expression from part (d) can be rewritten as ER El r(1 - alk) 1 - 22 (6 marks) Note: t=1+r and t'=1- (1 - x) = 1 + x +