Analyzing Reflection from an Absorbing Medium

[876d34f]

Homework Statement

3. Consider the reflection from the boundary between air and an absorbing medium. We
can use the usual Fresnel Equations, but with a complex refractive index for the
absorbing material. For various angles of incidence, use the real parts of the refractive
index to determine the angle of refraction from Snell’s Law. Let the absorbing medium
have nR=2.1 and nI=0.5.
a) Find the percentage reflected power for s and p-polarized reflections for incident
angles of 0° to 90° in 10° increments and plot the results.

Homework Equations

The Attempt at a Solution

 

[jbsteele]

The Fresnel Equations are: Rs = (nI2-sin2Θ)/(nI2+sin2Θ) and Rp = (nR2cos2Θ-sin2Θ)/(nR2cos2Θ+sin2Θ) Where nR is the real part of the refractive index and nI is the imaginary part. So for nR=2.1 and nI=0.5: For 0° incidence Θ = 0 Rs = (0.5^2-sin^2(0))/(0.5^2+sin^2(0)) = 0.5 and Rp = (2.1^2cos^2(0)-sin^2(0))/(2.1^2cos^2(0)+sin^2(0)) = 1 For 10° incidence Θ = 10° Rs = (0.5^2-sin^2(10°))/(0.5^2+sin^2(10°)) = 0.4996 and Rp = (2.1^2cos^2(10°)-sin^2(10°))/(2.1^2cos^2(10°)+sin^2(10°)) = 0.9995 For 20° incidence Θ = 20° Rs = (0.5^2-sin^2(20°))/(0.5^2+sin^2(20°)) = 0.4978 and Rp = (2.1^2cos^2(20°)-sin^2(20°))/(2.1^2cos^2(20°)+sin^2(20°)) = 0.9991 For 30° incidence Θ = 30° Rs = (0.5^2-sin^2(30°))/(0.5^2+sin^2(30°)) = 0.4945 and Rp = (2.1^2cos^2(30°)-sin^2(