用中英文给我详细写出这个文档的答案

下列内容先给出英文解答，再给出相应的中文翻译与说明。为便于对照，两个版本按照同样的结构与编号给出。

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English Version

Question 1

We have a system of two coupled single-mode waveguides operating at a wavelength of 1550 nm. An optical signal is injected into waveguide $\#1$. We are given:

• The waveguides are “non-identical,” meaning each waveguide by itself would support a slightly different propagation constant (or effective index).
• After coupling, the new (super) modes of the combined system split into two distinct effective indices, with an effective index difference of $0.00065$.
• The original (uncoupled) difference in effective indices between the two single-mode waveguides is $0.0003$.

We define the following standard coupling notations in terms of effective indices:

• Let $\Delta$ be the difference in effective index between the two uncoupled waveguides (sometimes denoted $\Delta n$). Here, $\Delta = 0.0003$.
• Let $\kappa$ be the coupling coefficient (in effective-index form). We often see “$2\kappa$” referred to as the total coupling strength.
• Let $\gamma = \sqrt{\kappa^2 + (\Delta / 2)^2}$. Then $2\gamma$ is the total splitting of the new supermodes when both waveguides are present.

Given:

• The new supermodes split by $2\gamma = 0.00065$.
• The uncoupled difference between the waveguides is $\Delta = 0.0003$.

We can use:

$$2\gamma \;=\; 2\,\sqrt{\kappa^2 \;+\;\Bigl(\frac{\Delta}{2}\Bigr)^2}.$$

Hence:

$$\gamma \;=\; \sqrt{\kappa^2 + \Bigl(\frac{\Delta}{2}\Bigr)^2} \quad\Longrightarrow\quad \gamma = \frac{0.00065}{2} \;=\; 0.000325.$$

So

$$\gamma^2 =\; \kappa^2 \;+\;\Bigl(\tfrac{0.0003}{2}\Bigr)^2 =\; \kappa^2 \;+\; (0.00015)^2.$$

Numerically,

$$(0.000325)^2 = 1.05625\times 10^{-7}, \quad (0.00015)^2 = 2.25\times 10^{-8}.$$

Therefore,

$$\kappa^2 = 1.05625\times 10^{-7} \;-\; 2.25\times 10^{-8} = 8.3125\times 10^{-8},$$ $$\kappa = 9.115\times 10^{-5}, \quad 2\kappa \approx 1.82\times 10^{-4}.$$

Let us now address each part:

1(a)

Are the two waveguides identical or non-identical?

Because there is a nonzero difference in their individual effective indices ($\Delta = 0.0003 \neq 0$), the waveguides are non-identical.

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1(b)

Sketch (no detailed calculation needed) the new eigenmode(s) of the coupled system.

• There will be two supermodes: a (nearly) symmetric combination of each waveguide’s fundamental mode and an (approximately) antisymmetric combination.
• Qualitatively, you would draw one mode profile that has the fields in the same phase in waveguide $\#1$ and waveguide $\#2$, and another mode that has the fields out of phase.

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1(c)

Given the effective index difference between the two new supermodes is $0.00065$, determine the total mode splitting $2\gamma$.

They directly state that the new mode splitting in effective index is $0.00065$. Hence:

$$2\gamma = 0.00065.$$

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1(d)

Given that the “initial mode difference” (the uncoupled difference) between the two waveguides is $0.0003$, determine the coupling strength $2\kappa$.

From

$$\gamma^2 = \kappa^2 + \Bigl(\tfrac{\Delta}{2}\Bigr)^2,$$

with $\gamma = 0.000325$ and $\Delta = 0.0003$, we found

$$\kappa \approx 9.115\times 10^{-5}, \quad 2\kappa \approx 1.82\times 10^{-4}.$$

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1(e)

Determine the maximum power (in %) that is transferred to waveguide $\#2$, and the propagation distance at which this occurs.

• If we excite waveguide $\#1$ alone at $z=0$, the maximum fraction of power that can couple into waveguide $\#2$ (for non-identical waveguides) is $$P_{2,\text{max}} \;=\; \frac{\kappa^2}{\kappa^2 + \bigl(\tfrac{\Delta}{2}\bigr)^2} \,.$$ Numerically: $$\kappa^2 \;=\;(9.115\times 10^{-5})^2 \;\approx\;8.31\times 10^{-9}, \quad \Bigl(\frac{\Delta}{2}\Bigr)^2 = (0.00015)^2 = 2.25\times 10^{-8}.$$ $$P_{2,\text{max}} = \frac{8.31\times 10^{-9}}{8.31\times 10^{-9} + 2.25\times 10^{-8}} \;\approx\;0.27 \;=\; 27\%.$$
• The distance $z$ at which this maximum power transfer first occurs can be found by setting $\Omega z = \pi/2$, where $$\Omega = \sqrt{\kappa_\beta^2 + \bigl(\tfrac{\Delta\beta}{2}\bigr)^2} \;=\; k_0 \,\sqrt{\kappa_E^2 + \bigl(\tfrac{\Delta_E}{2}\bigr)^2} \;=\; k_0\,\gamma.$$ Here, $\gamma = 0.000325$ (dimensionless difference in index), and $k_0 = \tfrac{2\pi}{\lambda}$ at $\lambda=1550\,\text{nm}$. Numerically: $$k_0 = \frac{2\pi}{1.55\times 10^{-6}} \;\approx\;4.058\times 10^6\,\text{m}^{-1}, \quad \Omega \;=\; k_0\cdot 0.000325 \;\approx\;1320 \,\text{m}^{-1}.$$ Hence $$z_{\text{max}} \;=\; \frac{\pi/2}{\Omega} \;\approx\; \frac{1.5708}{1320} \;\approx\;0.00119\,\text{m} \;=\;1.19\,\text{mm}.$$

Therefore, about 27% of the power couples to waveguide $\#2$ at $z \approx 1.19$ mm.

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1(f)

Determine the shortest distance of light propagation when 10% of the power is transferred to waveguide $\#2$.

Still using

$$P_2(z) = P_{2,\max}\,\sin^2(\Omega z),$$

we want $P_2(z) = 0.10$. Since $P_{2,\max}\approx0.27$:

$$\sin^2(\Omega z) = \frac{0.10}{0.27} \;\approx\;0.37037.$$

Taking square roots and then inverse sine:

$$\Omega z = \arcsin\bigl(\sqrt{0.37037}\bigr) \;\approx\;0.642, \quad z = \frac{0.642}{1320} \;\approx\;0.000486\,\text{m} =\;0.486\,\text{mm}.$$

Thus, the shortest propagation distance to achieve 10% coupling is approximately 0.49–0.50 mm.

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Question 2

We have a Fabry–Pérot etalon with:

• Refractive index $n = 1.5$.
• Round-trip length $L = 1\,\text{cm}$. (Recall $L = 2\,d \cos\theta$.)
• Mirror reflectivity $R = 0.9$.
• Single-pass intensity loss $(1-G) = 0.01$ $\implies$ $G = 0.99$.

The power transmission formula is:

$$\frac{I_{\text{out}}}{I_{\text{in}}} =\; \frac{(1 - R)^2\,G}{\bigl(1 - G\,R\bigr)^2 + 4\,G\,R\;\sin^2\!\bigl(\tfrac{\delta}{2}\bigr)},$$

where $\delta$ is the round-trip phase. We answer each sub-question as follows:

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2(a)

Derive the maximum power transmission in terms of $G$ and $R$.

• The maximum occurs when $\sin^2(\delta/2) = 0$, i.e. $\delta = 2m\pi$.
• Then $$T_{\max} = \frac{I_{\text{out}}}{I_{\text{in}}}\bigg\rvert_{\sin^2(\delta/2)=0} = \frac{(1-R)^2\,G}{\bigl(1 - G\,R\bigr)^2}.$$

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2(b)

Calculate the maximum percentage of power transmitted.

Plugging in $R = 0.9$, $G = 0.99$:

$$(1 - R) = 0.1, \quad (1 - R)^2 = 0.01, \quad (1 - G\,R) = 1 - 0.99\times 0.9 = 1 - 0.891 = 0.109.$$

Hence

$$(1 - G\,R)^2 = 0.109^2 = 0.011881.$$

So

$$T_{\max} = \frac{0.01 \times 0.99}{0.011881} \;\approx\;0.8333 \;\equiv\;83.33\%.$$

Thus the maximum power transmission is about 83.3%.

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2(c)

Full width at quarter maximum (FWQM) definition and expression of $\Delta f_{\text{FWQM}}$.

• By definition, FWQM is the frequency range over which the transmission remains above 25% of its maximum level.

• Let $T(\delta)$ denote the transmission vs. $\delta$. We require $T(\delta) > 0.25\,T_{\max}$.

• In normalized form,

  $$\frac{T(\delta)}{T_{\max}} = \frac{1}{1 + \frac{4\,G\,R}{(1 - G\,R)^2}\,\sin^2(\delta/2)}.$$

  We want $\frac{T(\delta)}{T_{\max}} > 0.25 \;\Longrightarrow\; 1 + \frac{4\,G\,R}{(1 - G\,R)^2}\,\sin^2(\delta/2) < 4.$

• Let

  $$X = \frac{4\,G\,R}{(1 - G\,R)^2}.$$

  Then

  $$\sin^2\bigl(\tfrac{\delta}{2}\bigr) < \frac{3}{X}.$$

• Define $\delta_{1/4}$ to be the half-range phase shift (from the center resonance) for which the transmission just hits 25% of $T_{\max}$. We get

  $$\delta_{1/4} = 2\,\arcsin\!\Bigl(\sqrt{\tfrac{3}{X}}\Bigr).$$

• The full frequency width $\Delta f_{\text{FWQM}}$ is related to the phase width via

  $$\Delta f = \frac{c}{2\pi\,n\,L}\;\Delta \delta \quad\bigl(\text{since } \delta = \tfrac{2\pi\,n\,L}{c}\,f\bigr).$$

  Because the total width is $2\,\delta_{1/4}$ (from negative to positive side about resonance), we get

  $$\Delta f_{\text{FWQM}} = \frac{c}{2\pi\,n\,L}\,\Bigl( 2\,\delta_{1/4} \Bigr) = \frac{c}{\pi\,n\,L}\,\delta_{1/4} = \frac{c}{\pi\,n\,L}\,\;2\,\arcsin\!\Bigl(\sqrt{\tfrac{3}{X}}\Bigr).$$

  Substituting $X=4\,G\,R/(1-G\,R)^2$ yields the desired expression in terms of $G, R, n,$ and $L$.

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2(d)

Calculate the FWQM linewidth numerically.

Using:

• $R=0.9$, $G=0.99$,
• $n=1.5$, $L=0.01\,\text{m}$,
• $c=3\times10^8\,\text{m/s}$.

Steps:

1. Compute $$X = \frac{4\,G\,R}{(1 - G\,R)^2}.$$
2. Then $\sin^2(\delta_{1/4}/2)=\frac{3}{X}$.
3. Solve for $\delta_{1/4}$.
4. Finally $\Delta f_{\text{FWQM}} = \frac{c}{\pi\,n\,L}\,\delta_{1/4}.$

Plugging in the numbers typically yields a result around $\sim 1.28\,\text{GHz}$. A more precise numeric example:

• $(1 - G\,R) = 0.109,\quad (1 - G\,R)^2 = 0.011881.$
• Numerator $4\,GR = 4\times0.99\times0.9=3.564.$
• $X \approx 3.564 / 0.011881 \approx 300.$
• $\sin^2(\delta_{1/4}/2)=3/300=0.01 \implies \delta_{1/4}/2 = \arcsin(0.1)=0.100167,$ $\delta_{1/4} \approx 0.200334.$
• $\Delta f_{\text{FWQM}} = \dfrac{3\times10^8}{\pi \times1.5\times0.01}\times0.200334 \,\approx\,1.28\times10^9\;\text{Hz}.$

Hence the FWQM linewidth is about 1.28 GHz.

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中文版本

下面以与英文解答相同的结构进行说明与计算。

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问题1

给定两个单模光波导，工作波长为1550 nm，光信号从波导$\#1$注入。由于两个波导之间存在耦合作用，形成两个新的模式（超模），从而导致模的有效折射率出现劈裂。题目给出：

• 两个原始波导（未耦合时）的有效折射率差$\Delta=0.0003$，说明它们并非完全相同。
• 耦合后形成的新超模的有效折射率差为$0.00065$。
• 定义$\kappa$为耦合系数（以有效折射率差的方式表示），$\gamma = \sqrt{\kappa^2 + (\Delta/2)^2}$。则两个新超模之间的总劈裂为$2\gamma$。

根据以上信息，可列方程

$$2\gamma = 2\,\sqrt{\kappa^2 + \Bigl(\tfrac{\Delta}{2}\Bigr)^2},$$

并由题意给出：

$$2\gamma = 0.00065 \;\Longrightarrow\; \gamma = 0.000325,$$ $$\Delta = 0.0003.$$

于是

$$(0.000325)^2 = \kappa^2 + (0.00015)^2,$$

数值计算：

$$\kappa \approx 9.115\times 10^{-5}, \quad 2\kappa \approx 1.82\times 10^{-4}.$$

以下分别解答各小问。

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1(a)

判断这两个波导是“相同”还是“不同”？

由于波导间原本就存在$0.0003$的有效折射率差（$\Delta\neq0$），故可知波导并不完全相同，即它们是非相同波导。

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1(b)

画出耦合后新模式的场分布（无需具体数值计算）。

在耦合系统中会出现两个超模：一个是（近似）对称叠加，另一个是（近似）反对称叠加。简单示意时，可画出：

• 对称模：在两个波导中场“同相”分布。
• 反对称模：在两个波导中场“反相”分布。

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1(c)

若新形成的两个超模的有效折射率差为$0.00065$，求总劈裂$2\gamma$。

题目已直接给出：新超模之间的有效折射率差即为$2\gamma$，因此

$$2\gamma = 0.00065.$$

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1(d)

已知原本两个波导的有效折射率差$\Delta=0.0003$，求波导间的耦合强度$2\kappa$。

使用

$$\gamma^2 = \kappa^2 + \Bigl(\frac{\Delta}{2}\Bigr)^2, \quad \gamma = 0.000325,\; \Delta=0.0003.$$

数值计算得$\kappa\approx 9.115\times 10^{-5}$，故

$$2\kappa \approx 1.82\times 10^{-4}.$$

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1(e)

求能耦合到波导$\#2$的最大功率百分比，以及在哪个传播距离达到此最大值。

• 对于非相同波导，若$\#1$在$z=0$处单独激励，波导$\#2$中最大可耦合的功率比例为

  $$P_{2,\max} = \frac{\kappa^2}{\kappa^2 + \bigl(\frac{\Delta}{2}\bigr)^2}.$$

  代入$\kappa=9.115\times 10^{-5}$、$\Delta=0.0003$，可得

  $$P_{2,\max} \approx 0.27,$$

  即约27%。

• 达到此最大值的传播距离$z$可由$\Omega z = \pi/2$得到，其中 $\Omega = k_0 \cdot \gamma$，$\gamma=0.000325$，$k_0=2\pi/\lambda\approx4.058\times10^6\,\text{m}^{-1}$（$\lambda=1550$ nm）。计算：

  $$\Omega \approx 4.058\times10^6 \times 0.000325 \approx 1320\,\text{m}^{-1}, \quad z_{\max} = \frac{\pi/2}{\Omega} \approx \frac{1.5708}{1320} \approx 0.00119\,\text{m} = 1.19\,\text{mm}.$$

  即约1.19 mm时，波导$\#2$所获功率达到最大27%。

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1(f)

求使波导$\#2$获得10%功率时的最短传播距离。

令$P_2(z)=0.1$。若$P_2(z)=P_{2,\max}\sin^2(\Omega z)$，且$P_{2,\max}\approx0.27$，则

$$0.1 = 0.27 \times \sin^2(\Omega z) \;\Longrightarrow\; \sin^2(\Omega z)=\frac{0.1}{0.27}\approx0.37037.$$ $$\sin(\Omega z)=\sqrt{0.37037}\approx0.6087, \quad \Omega z=\arcsin(0.6087)\approx0.642.$$ $$z = \frac{0.642}{1320}\approx0.000486\,\text{m}=0.486\,\text{mm}.$$

因此，波导$\#2$首次获得10%功率的最短传播距离约为0.49 mm左右。

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问题2

一个法布里–珀罗(Fabry–Pérot)标准具满足：

• 折射率$n=1.5$。
• 回程光程$L=1\,\text{cm}$（即$L=2\,d \cos\theta$）。
• 镜面反射率$R=0.9$。
• 单程损耗$(1-G)=0.01\Rightarrow G=0.99$。

其透射率公式为

$$\frac{I_{\text{out}}}{I_{\text{in}}} = \frac{(1-R)^2\,G}{(1-G\,R)^2 + 4\,G\,R\,\sin^2(\delta/2)},$$

下列分别求解：

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2(a)

推导该标准具的最大透射率（用$G$和$R$表示）。

• 当$\sin^2(\delta/2)=0$（$\delta=2m\pi$）时，分母里$\sin^2(\delta/2)$那一项为零，透射率达到最大： $$T_{\max} = \frac{(1-R)^2\,G}{(1-G\,R)^2}.$$

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2(b)

计算其最大透射率（百分比）。

令$R=0.9$，$G=0.99$，则

$$(1-R)=0.1,\quad (1-R)^2=0.01,\quad (1-G\,R)=1-0.99\times0.9=1-0.891=0.109,$$ $$(1-G\,R)^2=0.109^2=0.011881.$$

所以

$$T_{\max} =\frac{0.01\times0.99}{0.011881}\approx0.8333\quad(\text{即}83.33\%).$$

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2(c)

四分之一最大值(FWQM)带宽定义及其线宽公式$\Delta f_{\text{FWQM}}$。

• 当透射率超过最大值的25%时，对应的相位区间$\delta$即定义了四分之一最大值的带宽。
• 在归一化形式下，要求 $$\frac{T(\delta)}{T_{\max}}>0.25 \;\Longrightarrow\; 1+\frac{4\,G\,R}{(1-G\,R)^2}\,\sin^2\Bigl(\frac{\delta}{2}\Bigr)<4.$$
• 令$X=\frac{4\,G\,R}{(1-G\,R)^2}$，则$\sin^2(\delta/2)<3/X$。可求出边界相位$\delta_{1/4}$，并把这个相位区间转换到频率区间得到$\Delta f_{\text{FWQM}}$。因为 $$\delta = \frac{2\pi\,n\,L}{c}\,f \;\Longrightarrow\; d\delta = \frac{2\pi\,n\,L}{c}\,df \;\Longrightarrow\; df = \frac{c}{2\pi\,n\,L}d\delta.$$
• 对应完整相位展宽$2\,\delta_{1/4}$，则 $$\Delta f_{\text{FWQM}} = \frac{c}{2\pi\,n\,L}\,\bigl(2\,\delta_{1/4}\bigr) = \frac{c}{\pi\,n\,L}\,\delta_{1/4}.$$

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2(d)

数值计算该标准具的FWQM线宽。

• 仍取$R=0.9,\;G=0.99,\;n=1.5,\;L=0.01\,\text{m},\;c=3\times10^8\,\text{m/s}$。
• 计算过程示例：
  1. $(1-G\,R)=0.109$，所以$(1-G\,R)^2=0.011881$。
  2. $4\,G\,R=4\times0.99\times0.9=3.564$。
  3. $X=3.564/0.011881\approx300$。则$\sin^2(\delta_{1/4}/2)=3/300=0.01$，$\delta_{1/4}/2=\arcsin(0.1)=0.100167$，$\delta_{1/4}=0.200334$。
  4. 最终 $$\Delta f_{\text{FWQM}} = \frac{c}{\pi\,n\,L}\,\delta_{1/4} \approx \frac{3\times10^8}{3.14159\times1.5\times0.01}\times0.200334 \approx 1.28\times10^9\,\text{Hz}.$$
  即约1.28 GHz。

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总结

1. 耦合波导问题中，首先通过给定的新模有效折射率差及单个波导的差值，求耦合系数$\kappa$，再根据耦合方程计算最大耦合功率以及特定转移功率所需的传播距离。
2. 法布里–珀罗标准具的透射率问题，需分别求最大透射率、数值大小以及四分之一最大值(FWQM)的带宽与具体数值。通过代入给定的镜面反射率、单程损耗系数等，可获得约83.3%的最大透射，以及约1.28 GHz的FWQM线宽。

以上即为本作业各部分的详细解答(中英文对照)。祝学习顺利!

Below is a step-by-step outline of how to solve each part:

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1. Original System: Two Identical Waveguides

Given:

• Each waveguide has unperturbed effective index $n_{0} = 2.0$.
• Because they are identical, the uncoupled index difference is $\Delta = 0$.
• The observed normal-mode splitting of the coupled system (in terms of effective index) is $$\Delta n_{\text{split}} \;=\; 0.005.$$ By convention, we often denote the splitting of the two supermodes as $$2\gamma \;=\; \Delta n_{\text{split}}.$$ Thus, $$2\gamma = 0.005 \quad\Longrightarrow\quad \gamma = 0.0025.$$
• For identical waveguides (i.e., $\Delta = 0$), the formula for the supermode splitting becomes $$2\gamma \;=\; 2\kappa \quad\Longrightarrow\quad \kappa \;=\;\gamma \;=\;0.0025.$$ Hence the “coupling strength” (in index units) is $2\kappa = 0.005$.

Interpretation:

• In the usual coupled-mode theory, the difference in the supermode propagation constants (in $\text{rad/m}$) is $\Delta \beta = k_0 \times 2\kappa$ where $k_0 = \frac{2\pi}{\lambda}$.
• Here, $\kappa = 0.0025$ is measured in index units, so to convert to a wavenumber domain we multiply by $k_0$.

(a) Distance at which 50% of the optical power is in waveguide #2

1. Power transfer formula (identical waveguides, excited solely in waveguide #1 at $z=0$):

   $$P_2(z) \;=\; \sin^2\!\bigl( \kappa_\beta\,z \bigr),$$

   where $\kappa_\beta = k_0\,\kappa$ is the coupling constant in units of $\mathrm{m}^{-1}$. Equivalently,

   $$P_2(z) \;=\; \sin^2\!\bigl( \tfrac{\Delta \beta}{2}\,z \bigr) \quad\text{with}\quad \Delta\beta = k_0\,2\kappa.$$
   • Since these are identical waveguides, $\Delta=0$, so there is no phase mismatch term.

2. We want $P_2(z) = 0.5$. Hence,

   $$\sin^2(\kappa_\beta z) = 0.5 \quad\Longrightarrow\quad \kappa_\beta z = \frac{\pi}{4} \quad\bigl(\text{or }3\pi/4, \ldots\bigr).$$

   The shortest such distance is

   $$z_{50\%} = \frac{\pi}{4\,\kappa_\beta} = \frac{\pi}{4\,k_0\,\kappa}.$$

3. Numerical evaluation at $\lambda=1310\,\mathrm{nm}$:

   • $k_0 = \dfrac{2\pi}{\lambda} \approx \dfrac{2\pi}{1.31\times10^{-6}} \approx 4.79\times10^{6}\,\mathrm{m}^{-1}.$
   • $\kappa = 0.0025.$
   • Thus, $\kappa_\beta = k_0 \times 0.0025 \approx 4.79\times10^{6} \times 0.0025 = 1.20\times10^{4}\,\mathrm{m}^{-1}.$

   Hence,

   $$z_{50\%} = \frac{\pi}{4 \times 1.20\times10^{4}} \;\approx\; \frac{3.14159}{4.8\times10^{4}} \;\approx\; 6.5\times10^{-5}\,\mathrm{m} = 0.065\,\mathrm{mm} = 65\,\mu\mathrm{m}.$$

So about 65 µm of propagation is needed for half the power to appear in waveguide #2 for the original, identical-waveguide system.

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2. Modified System: Waveguide #2 Becomes Slightly Wider

Now waveguide #2 is changed so that its unperturbed effective index becomes $n_{0}''=2.005$, while waveguide #1 remains $n_{0}'=2.000$. The difference in the uncoupled waveguide indices is thus

$$\Delta = n_{0}'' - n_{0}' = 2.005 \;-\;2.000 = 0.005.$$

Meanwhile, the coupling strength remains the same as before:

$$\kappa = 0.0025.$$

(b) New Supermode Splitting $\Delta n_{\text{new}}$

When the waveguides are non-identical, the two new supermodes split by

$$2\gamma \;=\; 2\,\sqrt{\kappa^2 + \Bigl(\tfrac{\Delta}{2}\Bigr)^2}.$$

Here:

• $\kappa=0.0025$
• $\Delta=0.005$

Thus,

$$2\gamma = 2\;\sqrt{(0.0025)^2 \;+\; (0.005/2)^2}.$$

Numerically:

1. $(0.0025)^2 = 6.25 \times 10^{-6}.$
2. $\Delta/2 = 0.0025,\;\;(0.0025)^2 = 6.25 \times 10^{-6}.$
3. Sum inside the square root: $6.25\times10^{-6} + 6.25\times10^{-6} = 1.25\times10^{-5}.$
4. $\sqrt{1.25\times10^{-5}} \approx 3.5355\times10^{-3}.$
5. Multiply by 2: $$2\gamma \;\approx\; 2 \times 3.5355\times10^{-3} \;=\; 7.07\times10^{-3}.$$

Hence the new normal-mode splitting is

$$\Delta n_{\text{new}} \;=\; 2\gamma \;\approx\; 0.00707.$$

That is the difference between the two new supermode effective indices in the modified system.

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Final Answers

1. (a) For the original system (two identical waveguides with splitting 0.005 at 1310 nm):

   • The coupling coefficient is $\kappa=0.0025$.
   • The distance for 50% power in waveguide #2 is $$z_{50\%} \;\approx\; 65\,\mu\mathrm{m}.$$

2. (b) When waveguide #2’s index is changed to 2.005 (while waveguide #1 remains 2.000), and the coupling strength $\kappa$ remains the same:

   • The new waveguide mismatch is $\Delta=0.005$.
   • The two supermodes now split by $$2\gamma \;=\; \boxed{0.00707\;(\text{approximately}).}$$

   Thus the new eigenmode effective indices differ by about 0.0071 instead of the original 0.005.

下图及文字所展示的，是在两条波导彼此耦合时，所形成的“上、下”两个新超模(也叫正常模、超级模)在有效折射率上的分裂情况。它分为两种典型情形：

1. Identical（相同）波导：两条波导原本(未耦合时)的模式有效折射率完全相同，没有初始差异。
2. Non‐identical（不相同）波导：两条波导的模式有效折射率存在一定差异（即“失配”），因此在耦合后出现更为复杂的分裂。

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1. 左图：Identical 波导时的模式分裂

• 水平方向(横坐标)通常是“波导间距”或者“波导中心间隔”在改变时，观测到两个新模式的有效折射率(纵坐标)如何变化。波导越靠近(间隔越小)，模式间的相互耦合越强，导致两个新模式在有效折射率上分得更开。
• 纵坐标是耦合后形成的两个模式的有效折射率值，可以看到上下两支曲线分别对应“对称超模”和“反对称超模”。
• 若两波导“完全相同”，那么在耦合后，这两个新模式相对于“原先单个波导模式的折射率”向上、向下各偏离一些，上下两条曲线正好对称分布，其间的“分裂”称为 $$\Delta n \;=\;\frac{2\kappa}{k_0}$$ 其中
  • $\kappa$是“耦合常数”，常用来度量波导间能量互相耦合的强弱；
  • $k_0 = \tfrac{2\pi}{\lambda}$是自由空间的波数。
    换句话说，若直接以“有效折射率差”($\Delta n$)来度量，在相同波导的情况下总的模分裂就是$2\kappa/k_0$。图中实线是更精确的数值解，虚线是简化的耦合模理论解，两者走势相当接近。

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2. 右图：Non‐identical 波导时的模式分裂

• 若两条波导原本(未耦合时)就存在一定的模式有效折射率差，通常可记为

  $$\Delta \;=\;\tfrac{1}{2}\bigl(n_{\mathrm{eff},1}-n_{\mathrm{eff},2}\bigr),$$

  或者在很多教材中直接记为$\Delta = \bigl(\beta_1 - \beta_2\bigr)/2$（视使用的符号而定）。总之，它代表“失配”或“初始不一致”。

• 这时候，两波导之间的总的模式分裂（图中标注为$\Delta n_t$）不再只是由耦合系数$\kappa$单独决定，而要综合耦合强度$\kappa$和失配量$\Delta$。耦合模理论给出的结果是

  $$2\gamma \;=\; 2\,\sqrt{\kappa^{2} \;+\;\Delta^{2}}.$$

  如果用“折射率差”来衡量，则

  $$\Delta n_t \;=\;\frac{2\gamma}{k_0} \;=\;\frac{2}{k_0}\,\sqrt{\kappa^{2} + \Delta^{2}}.$$

  其中$\Delta$便是“非相同”导致的额外项。对比“相同波导”时的$\Delta n=2\kappa/k_0$，可以看出现在多了$\Delta^{2}$这个“失配”贡献，所以non-identical时的分裂量一般会更大，并且上下两条新模式不再对称地分布在“平均折射率”的上下，而会出现一定的偏移。

• 图中右边的示意就是：原先两个波导的有效折射率就不同，因此两个新模式的折射率曲线彼此都被往上或往下拉开，不再对称围绕某个中心线；它们之间的“总分裂”就是

  $$2\gamma\,(\text{或 }2\sqrt{\kappa^2 + \Delta^2}),$$

  而两条模曲线的中心“位置”与那一开始的失配$\Delta$也有关系（也就是说，在右图上，两条曲线看似整体向上或向下偏移了）。

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3. 文字小结

• 耦合强度 (Coupling Strength)：
  记为$2\kappa$（或$\kappa$本身）。在只考虑波导相同又紧密并排放置时，耦合越强，新模式分裂就越大。若用折射率差量度，则$\Delta n=2\kappa/k_0$。

• 失配 (Mismatch) $\Delta$：
  当两个波导本身就不一样，它们的模式有效折射率存在差值$\Delta$，则耦合后出现的总模式分裂为$2\gamma=2\sqrt{\kappa^2 + \Delta^2}$。也可以写成折射率差形式$\Delta n_t = 2\gamma/k_0$。

• 图示差别：

  • 左图(Identical)中，分裂只由$\kappa$决定：$\Delta n = 2\kappa/k_0$，上下两条曲线相对对称。
  • 右图(Non‐identical)中，除了$\kappa$还多了一个$\Delta$使分裂变为$2\sqrt{\kappa^2+\Delta^2}$，上下两支也不再严格对称地围绕同一个中心。

• 关于$\Delta=1/2$“两个波导初始有效折射率之差”：
  某些文献里定义$\Delta$时就是把两波导各自的折射率差除以2，使得耦合方程里有$\Delta$出现在$\Delta^2$处，最终$2\gamma=2\sqrt{\kappa^2+\Delta^2}$的形式更简洁。这是常见的记号习惯之一。

因此，总的来说：

1. 相同波导(identical)时，两个超模分裂量就是$2\kappa$（或$\Delta n=2\kappa/k_0$）。
2. 不相同波导(non‐identical)时，则要在$\sqrt{\kappa^2 + \Delta^2}$里同时考虑耦合和失配，分裂变为$2\gamma = 2\sqrt{\kappa^2 + \Delta^2}$。

上面这段文字以及图示就是在说明：波导耦合后，超模的有效折射率会向上下两个极端偏移，其间的差值就是“模式分裂”；若波导不一样，还要加上原始失配$\Delta$的影响，最终让分裂量和偏移位置都发生变化。