光电子学与光子学的原理及应用s.o.kasap 课后答案

(1)
The amplitude of the reflected beam is Areflected i.e. = A1 + A2 + A3 + ... (2) Areflected/A0 = r1 + t1t2r2e−j2φ − t1t2r1r22e−j4φ + t1t2r12r23e−j6φ + ...
Chapter 1/2/3/7
1 − t1t 2 = 1 −
4n1 n2 n + n + 2n1n2 − 4 n1n2 = 2 (n1 + n2 ) (n1 + n2 )2
2 1 2 1 2 2 + n1 − 2n1 n2 n1 n −n = = 1 2 = r12 2 (n1 + n2 ) n1 + n2 2
r12 =
n1 − n2 n1 − n1n3 = = n1 + n2 n1 + n1n3
1− 1+
n3 n1 n3 n1
For light traveling in medium 2 incident on the 2-3 interface at normal incidence,
r23 =
n2 − n3 = n2 + n3
For TiO2 d= d=
λ
4 n2
= =
900 × 10−9 m = 0.15 µm 4(1.5) 900 × 10−9 m = 0.10 µm 4(2.3)
λ
4n2
1.8
Thin film coating and multiple reflections:
Assume that n1 < n2 < n3 and that the thickness of the coating is d. For simplicity, we will assume normal incidence. The phase change in traversing the coating thickness d is φ = (2π/λ)n2d where λ is the free space wavelength. The wave has to be multiplied by exp(−jφ) to account for this phase difference. The coefficients are given by, n −n r1 = r12 = 1 2 = −r21 , n1 + n2 and
so that the k-th term for k > 1 is
Areflected k tt = − 1 2 (− r1 r2 e− j 2φ ) A0 k r1
so that the reflection coefficient is A tt r = reflected = r1 − 1 2 A0 r1
t 1 = t 12 = 2n1 , n1 + n2 n2 − n3 n2 + n3 2 n2 , n1 + n2
r2 = r23 = t 2 = t 21 =
t 23 =
2n3 , n2 + n 3
Consider 1 − t1t2,
Solutions for Optoelectronics and Photonics: Principles and Practices
n1n3 − n3 = n1n3 + n3
n1 n −1 1 − 3 n3 n1 = n1 n +1 1 + 3 n3 n1
thus,
r23 = r12
Significance? For an efficient antireflection effect, waves A (reflected at 1-2) and B (reflected at 2-3) in Figure 1Q4 below should interfere with near “total destruction”. That means they should have the same magnitude and that requires that the reflection coefficient between 1 and 2 should be the same as that between 2 and 3; r12 = r23. Thus, the layer 2 can act as an antireflection coating if its index n2 = (n1n3)1/2. This can be achieved by r12 = r23. The best antireflection coating has to have a refractive index n2 such that n2 = (n1n3)1/2 = [(1)(3.5)]1/2 = 1.87. Given a choice of two possible antireflection coatings, SiO2 with a refractive index of 1.5 and TiO2 with a refractive index of 2.3, both are close . The phase change for wave B going through the coating of thickness d is 2k2d where k2 = n2ko and ko = wavevector in free space = 2π/λ. This should be 180° or π. Thus we need 2n2(2π/λ)d = π or For SiO2
Solutions for Optoelectronics and Photonics: Principles and Practices
CLeabharlann Baiduapter 1/2/3/7
1.4
Antireflection coating
For light traveling in medium 1 incident on the 1-2 interface at normal incidence,