The Light-Emitting Diode (LED) — Question 5

Question 5

Total internal reflection restricts a planar escape cone

Photons are emitted isotropically inside a semiconductor of refractive index $n=2.50$ adjoining air through one planar face. Ignore Fresnel reflection, absorption, and repeated encounters. Only rays within the critical-angle cone toward that face escape on the first encounter.

1. Find the critical angle and the escaping fraction of all emission directions.

2. Use the large-n approximation to estimate that fraction and compare with the exact value.

Question 5 - Solution

Total internal reflection restricts a planar escape cone

(a) Method and calculation

$$\boxed{\theta_c=\sin^{-1}(1/n)=23.58^\circ}.$$$$\boxed{f_{\mathrm{esc}}=\frac{2\pi\int_0^{\theta_c}\sin\theta\,d\theta}{4\pi}=\frac{1-\cos\theta_c}{2}=.04174}.$$

(b) Interpretation and further calculation

$$\cos\theta_c=\sqrt{1-1/n^2}\simeq1-1/(2n^2).$$$$\boxed{f_{\mathrm{esc}}\simeq1/(4n^2)=.0400}\quad(4.17\%\text{ below exact}).$$

Physical check

The denominator includes all directions, not only the upward hemisphere. This is a first-pass geometric fraction; coatings, texturing and photon recycling alter extraction.