Total Internal Reflection — Question 6

Question 6

Evanescent field and squared-amplitude penetration lengths differ by two

At a lossless total-reflection boundary, the low-index-side electric field envelope varies as $E(z)=E(0)e^{-\kappa z}$, where z points away from the interface and $\kappa=(2\pi/\lambda_0)\sqrt{n_1^2\sin^2\theta-n_2^2}$. Take n1=1.50, n2=1.00, theta=60 degrees, and vacuum wavelength 600 nm.

1. Find the field-amplitude 1/e penetration depth.

2. Find the 1/e depth of the squared field and explain why this does not imply an average normal transmitted power.

Question 6 - Solution

Evanescent field and squared-amplitude penetration lengths differ by two

(a) Method and calculation

$$\kappa=\frac{2\pi}{600\,\mathrm{nm}}\sqrt{2.25(3/4)-1}=\frac{2\pi}{600\,\mathrm{nm}}\frac{\sqrt{11}}4.$$$$\boxed{\ell_E=\kappa^{-1}=115.169\,\mathrm{nm}}.$$

(b) Interpretation and further calculation

$$|E(z)|^2/|E(0)|^2=e^{-2\kappa z}\Rightarrow\boxed{\ell_{E^2}=1/(2\kappa)=57.5844\,\mathrm{nm}}.$$$$\boxed{\langle S_z\rangle=0}\quad\text{for a single isolated lossless TIR interface}.$$

Physical check

The evanescent field can carry energy parallel to the interface; its nonzero amplitude is not a propagating normal ray.