物理光学习题参考答案

Er ∴r = E0
r12 = − r21
2 t 12 t 21 =1− r12
E0
E1
E2
E3
E4
1
∞
2
d
=
r12 + r23e 1 + r12r23e iΔϕ
iΔ ϕ
3
2 2 2 2 r12 + r23 + 2r12r23 cos Δϕ (1 − r12 )(1 − r23 ) = 1− ∴R = 2 2 2 2 1 + r12r23 + 2r12r23 cos Δϕ 1 + r12r23 + 2r12r23 cos Δϕ
第一章 小结
§1.1 电磁波谱及电磁场基本方程 §1.2 光波在各向同性介质中的传播 §1.3 光波在介质界面上的反射和折射 §1.4 光波场的频率谱
1
⎧ ⎧ ⎪ ⎪ E 0 e i ( kz − ω t + ϕ 0 ) ⎪ v ⎪ 2 ? ? v ∂ E ? ⎪ E 0 i ( kz − ω t + ϕ 0 ) ⎪ 2 ⎪ ⇒ ∇ E = με ∂ t 2 ⇒ E = f ( z − υ t ) ⇒ E = ⎨ r e ⎪ ⎪ r ⎪ E 0 i ( kz − ω t + ϕ 0 ) r ∂B ⎫ ⎪ e ⎪ ∇×E = − ⎪ ⎪ ⎩ r ∂t ⎪ ⎪ r sin( θ i − θ t ) ⎧ ⎪ ⎪ ∇⋅D = ρ ⎪ rs = − r ⎪ ⎬⇒⎨ sin( θ i + θ t ) ∇ ⋅B = 0 v v ⎪ ⎪ ⎪ v r ⎪ ⎪ ⎧u × (E 1 − E 2 ) = 0 ⎪ 2 sin θ t cos θ i r r ∂D v ts = ⎪v v ⎪ ⎪ ? u ⋅ (B − B ) = 0 ? ⎪ ∇×H = j+ sin( θ i + θ t ) ⎪ ∂t ⎪ ⎪⇒ ⎪ v v 1 v 2 ⎭ ⇒⎨ ⎨ ⎪ ⎪u ⋅ ( D 1 − D 2 ) = 0 ⎪ r = tan( θ i − θ t ) v v ⎪ ⎪v ⎪ p tan( θ i + θ t ) ⎪ ⎩u × ( H 1 − H 2 ) = 0 ⎪ ⎪ 2 sin θ t cos θ i ⎪t = ⎪ ⎪ p sin( θ i + θ t ) cos( θ i − θ t ) ⎩ ⎩ E ( f x , f y , f z ,ν ) = F [E ( x , y , z , t ) ] =
13
P33(1-9)
sin(θ i − θ t ) θi − θt n 2 − n1 r1s = − ≈− ≈− sin(θ i + θ t ) n1 + n 2 θi + θt tan(θ i − θ t ) r1p = ≈ tan(θ i + θ t )
2
θi − θt ≈ θi + θt
n 2 − n1 n1 + n 2
⎛ n 2 − n1 ⎞ R1 = ⎜ ⎟ ⎜ n + n ⎟ = 4.3% 2 ⎠ ⎝ 1 n2 sin θ c = < sin 45o n1 ∴ R2 = 1 R 3 = R1 I out = I 0 (1 − R 1 )R 2 (1 − R 3 ) = I 0 (1 − R 1 ) 2 = 92%I 0
=
ω
= 3 × 108 (m / s )
ˆ = − 1 i + 3 j λ = υT = υ = π = 3.14(m ) 振动方向 ： E 0 2 2 ν k = − 3i − j k = 2(1 / m ) ˆ = − 3 i− 1 j 传播方向 ： k 2 2 ω = 6 × 108 (rad / s )
2 i
9
P33(1-7)
sin(θ i − θ t ) rs = − sin(θ i + θ t ) 2 sin θ t cos θ i ts = sin(θ i + θ t ) tan(θ i − θ t ) rp = tan(θ i + θ t ) 2 sin θ t cos θ i tp = sin(θ i + θ t ) cos(θ i − θ t )
5
反射率：
2 ⎧ ⎧ 2 2 ⎤ ⎡ cos θ − n − sin θ i i 2 ⎪ ⎪R = r = ⎢ ⎥ s s 2 2 ⎪ ⎪ ⎢ cos θ i + n − sin θ i ⎥ ⎪ ⎣ ⎦ ⎪θ < θ c⎨ 2 ⎪ i 2 2 2 ⎤ ⎪ ⎡ n cos θ − n − sin θ ⎪ i i 2 ⎨ ⎥ ⎪ R p = rp = ⎢ 2 2 2 ⎪ ⎢ n cos θ i + n − sin θ i ⎥ ⎪ ⎣ ⎦ ⎩ ⎪ ⎪ ⎧R s = 1 ⎪θ i ≥ θ c ⎨ ⎪ ⎩R p = 1 ⎩
8
P33(1-4)
θ i = 30o , n = 1.7
rs = rp = cos θ i − n 2 − sin θ i2 cos θ i + n − sin θ
2 2 2 2 i
= −0.305 = 0.213
n 2 cos θ i − n 2 − sin θ i2 n cos θ i + n − sin θ
n t cos θ t 2 sin 2θ i sin 2θ t = t s t′ Ts = ts = s 2 sin (θ i + θ t ) n i cos θ i n t cos θ t 2 sin 2θ t cos 2θ i Tp = tp = = t p t′ p 2 2 n i cos θ i sin (θ i + θ t ) cos (θ i − θ t )
其中 : Δϕ =
2πn 2d
λ0
n 0 − n1 n1 − n 2 , r12 = , r23 = n 0 + n1 n1 + n 2
3
附加题1
2 2 2 2 r12 + r23 + 2r12r23 cos Δϕ (1 − r12 )(1 − r23 ) = 1− R= 2 2 2 2 1 + r12r23 + 2r12r23 cos Δϕ 1 + r12r23 + 2r12r23 cos Δϕ
A
Ar
A′t′
A′r′
A′
At
10
P33(1-7)
sin(θ t − θ i ) rs′ = − sin(θ t + θ i ) 2 sin θ i cos θ t t′ = s sin(θ t + θ i ) ′ rp = t′ = p tan(θ t − θ i ) tan(θ t + θ i ) 2 sin θ i cos θ t sin(θ t + θ i ) cos(θ t − θ i )
附加题1
6
第一章习题
作业： 1 4 7 9 13
7
P33(1-1)
E = E 0 e i ( k ⋅r − ωt + ϕ 0 ) E 0 = −2i + 2 3 j 振幅： E 0 = 4( v / m )
k ω 3 × 108 ν= = = 9.55 × 107 ( Hz ) π 2π
υ=
λ
T
14
P34(1-13)
c 299792 × 10 3 = 1.96526 × 108 (m / s ) υp = = n 1.52546 λ dλ ⎞ ⎛ 8 υg = υp ⎜ 1 + ⎟ = 1.90052 × 10 (m / s ) n dn ⎠ ⎝
15
第二章 小结
§2.1 光干涉的条件 §2.2 双光束干涉 §2.3 多光束干涉 §2.4 光学薄膜 §2.5 典型的干涉仪及其应用 §2.6 光的相干性
3
反射比：
附加题1
n2 = n 21 = n ˆ ˆ n 1 = 1 . 5 , n 2 = 1, n1
⎧ cos θ i − n 2 − sin θ i2 ⎪rs = ⎪ cos θ i + n 2 − sin θ i2 ⎪ θi < θc ⎨ n 2 cos θ i − n 2 − sin θ i2 ⎪ ⎪rp = 2 n cos θ i + n 2 − sin θ i2 ⎪ ⎩ ⎧ cos θ i − i sin θ i2 − n 2 ⎪rs = = e iϕ rs ⎪ cos θ i + i sin θ i2 − n 2 ⎪ θi ≥ θc ⎨ n 2 cos θ i − i sin θ i2 − n 2 ⎪ iϕ rp = = e rp ⎪ n 2 cos θ i + i sin θ i2 − n 2 ⎪ ⎩
若 : Δϕ = 2Nπ， 即 : n 2d = Nλ0 ⎛ r12 + r23 ⎞ ⎛ n 0 − n 2 ⎞ 2 ⎟ =⎜ ⎟ = r13 , 与不镀膜相同 则： R = ⎜ ⎜ 1+ r r ⎟ ⎜ n + n ⎟ 12 23 ⎠ 2 ⎠ ⎝ ⎝ 0
A
Ar
A′t′
A′r′
A′
At
11
P33(1-7) 由：
sin(θ i − θ t ) rs = − sin(θ i + θ t ) 2 sin θ t cos θ i ts = sin(θ i + θ t ) tan(θ i − θ t ) rp = tan(θ i + θ t ) 2 sin θ t cos θ i tp = sin(θ i + θ t ) cos(θ i − θ t ) rs′ = − t′ = s sin(θ t − θ i ) sin(θ t + θ i )
4
位相差：
附加题1
sin(θ i − θ t ) ⎧ rs = − ⎪ sin(θ i + θ t ) ⎪ θi < θc ⎨ ⎪r = tan(θ i − θ t ) ⎪ p tan(θ i + θ t ) ⎩