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OPTI 507 - Solid-State Optics
HW 4 Solutions
Jason M. Auxier
Copyright ©c 2004
All rights reserved.
September 15, 2004
Homework 4 consists of problems 3.2 - 3.5 in the text: Introduction to Semi- conductor Optics by Peyghambarian et al [1].
1 Problem 3.2 - 10 Points
Prove Eq. (3.61).
1.1 Solution
Consider a prism with index n > 1. Looking at Fig. 1, we see that the input beam is headed to the normal of the output face. Let γ′^ = π/ 2 − γ, where γ is the incident angle. By Snell’s law,
Figure 1: The geometry of the prism along with the angle definitions.
Homework 4 Solutions 2
sin γ = n sin θ 1 n sin θ 2 = sin δ
Note that γ′^ + θ 1 + θ 2 = π/2, so
γ = θ 1 + θ 2 (2)
Thus, using a trig identity, we can write
sin γ = cos θ 1 sin θ 2 + sin θ 1 cos θ 2. (3)
Also,
cos θ 2 = sin(γ′^ + θ 1 ) = cos γ′^ sin θ 1 + sin γ′^ cos θ 1 = sin γ sin θ 1 + cos γ cos θ 1.
Now, using Eq. 4, sin θ 1 = sinn^ γ, sin θ 2 = sinn^ δ, and cos α =
1 − sin^2 α, we can write Eq. 3 as
sin γ =
n
sin δ n
(sin δ + sin γ cos γ) +
n^2
sin^3 γ. (5)
Multiplying both sides of this equation by n^2 and bringing sin^3 γ to the LHS, we have sin γ
n^2 − sin^2 γ
n^2 − sin^2 δ (sin δ + sin γ cos γ). (6)
Squaring both sides, bringing everything to the LHS, and factoring, we can write
( n^2 − sin^2 γ
n^2 sin^2 δ − sin^4 δ − sin^2 δ − 2 sin δ sin γ cos γ − sin^2 γ cos^2 γ
Now, n = ± sin γ is nonphysical, so the second term must vanish:
n^2 =
sin^2 δ
1 − cos^2 δ
- sin^2 δ + 2 sin δ sin γ cos γ + sin^2 γ cos^2 γ sin^2 γ
= 1 +
2 cos γ sin δ sin γ
sin^2 δ sin^2 γ
2 Problem 3.3 - 10 Points
Show that the reflectivity of a material with negative dielectric function ap- proaches unity.
Homework 4 Solutions 4
So, differentiating and dividing the common exponentials from both sides,
(+α 1 E, 0 , ikxE) = iξ c
(Hx, 0 , H 1 z ) for z > 0
(−α 2 E, 0 , ikxE) =
iξ c
(Hx, 0 , H 2 z ) for z < 0.
Matching these equations up at the boundary z = 0, from the x-component we find that α 1 = −α 2. (15)
Since absorption is positive or zero, this could only be satisfied if α 1 = α 2 = 0. This is not true for metals. Strictly speaking we are offering a proof by contradiction. Formally, we make the statement that for a metal, α 2 6 = 0 and assume that a TE mode exists. This assumption lead to a violation of α 2 6 = 0. So, by contradiction, metals cannot support a TE mode.
4 Problem 3.5 - 10 Points
When a very intense laser irradiates a metal, it ablates material near the surface, resulting in a gas of ionized particles (a plasma) expanding into free space. At which densities of the plasma will light with λ = 1μm be totally reflected?
4.1 Solution
Reflection occurs when the optical frequency is below the plasma frequency ωp.
ω^2 p =
4 πne^2 ≤∞me
Now, we want ω < ωp, so
ω^2 <
4 πne^2 ≤∞me
n >
ω^2 ≤∞me 4 πe^2
Also, ν = (^) λc , so ω = 2 πcλ. Therefore,
n >
2 πc λ
≤∞me 4 πe^2
πmec^2 e^2 λ^2
for ≤∞ ' 1.
REFERENCES 5
Remember that this is in Gaussian (cgs) units, so
1C = 3 × 109 esu e = 4. 81 × 10 −^10 esu me = 9. 11 × 10 −^28 g λ = 10−^4 cm c = 2. 9979 × 1010 cm/s.
Therefore,
n > 1 × 1021
e− cm^3
References
[1] Nasser N. Peyghambarian, Stephan W. Koch, and Andre Mysyrowicz. In- troduction to Semiconductor Optics. Prentice Hall, Englewood Cliffs, NJ,
