Total Internal Reflection — Question 4

Question 4

Solid angle gives the geometrical fraction initially inside an escape cone

An isotropic point source in index n emits total power P. Above it is an infinite flat interface to air. Count rays inside the upward critical cone and idealize every such ray as transmitted, neglecting Fresnel losses. Count only the first surface encounter.

1. Integrate over solid angle to find the escaping fraction of total emitted power.

2. Evaluate the fraction for n=4/3 and state how real Fresnel reflection changes it.

Question 4 - Solution

Solid angle gives the geometrical fraction initially inside an escape cone

(a) Method and calculation

$$dP/P=d\Omega/(4\pi),\quad\Omega_c=2\pi\int_0^{\theta_c}\sin\theta\,d\theta=2\pi(1-\cos\theta_c).$$$$\boxed{P_{\mathrm{cone}}/P=\frac{1-\sqrt{1-1/n^2}}{2}}.$$

(b) Interpretation and further calculation

$$n=4/3\Rightarrow\boxed{P_{\mathrm{cone}}/P=(1-\sqrt7/4)/2=0.169281}.$$$$P_{\mathrm{real}}=\frac{P}{4\pi}\int_{\mathrm{cone}}T(\theta,\text{polarization})\,d\Omega\le P_{\mathrm{cone}}.$$

Physical check

The factor of one-half accounts for normalization to the full sphere, including downward radiation; the result is a geometric upper bound.