Question

2. A block of plastic with refractive index np-1.68 is immersed in oil (no-1.516). Randomly polarized light (ie, 50% s-polarization and 50% p-polarization) is incident from the oil towards the plastic at an angle of 65" to the normal. a) What percentage of the p-polarized component of the light is reflected off the plastic? b) What percentage of the total incident power of the light is transmitted into the block? c) What is the ratio of the s-polarized reflected power to the p-polarized reflected power?

Answer 1

a] Use Snell's law to find the refracted angle

The reflectance for p polarized light is:

$R_p=\frac{tan^2(\theta_i-\theta_t)}{tan^2(\theta_i+\theta_t)}$

=> R_p = 5.74 x 10^-5 = 0.00574 % of the p-polarized light is reflected.

b]

Transmittance for p polarization is:

T_p = 1 - R_p = 0.9999425

and for s polarization,

$R_s=\frac{sin^2(\theta_i-\theta_t)}{sin^2(\theta_i+\theta_t)} =1.42\times 10^{-4}$ = 0.0142%

so, T_s = 0.999858

therefore the percentage of incident power is transmitted is:

T = [(0.9999425) + (0.999858)] x 100 x 0.5 = 99.986%.

c] R_s/R_p = 2.474.