Question about Fabry interferometer

[Clara Chung]

Homework Statement

Homework Equations

The Attempt at a Solution

I need help on part b and c.
Here is my attempt:
Ai=Ai,0 exp(ikz-wt) (1 + √R1√R2 exp(iδ) + [√R1√R2 exp(iδ)]^2 +......)
=Ai,0 exp(ikz-wt) / (1- √R1√R2 exp(iδ))
=> |Ai| = |Ai,0| |(1 + √R1√R2 exp(iδ))| / ( (1 - √R1√R2 cos(δ))² + sin²(δ))
= |Ai,0| |(1 + √R1√R2 exp(iδ))| / ( 1 - 2√R1√R2 cos(δ) +R1R2)
Then I don't know what to do

For part c I don't know the right way to start, can you give me some ideas?
Thank you

 

[Charles Link]

One thing that is helpful is to use Fresnel coefficients in doing these sums of the electric field amplitude from multiple reflections. If energy reflectivity coefficient is $R$, the $\rho$ for the Fresnel reflection coefficient satisfies $|\rho|^2=R$, so that $|\rho|=\sqrt{R}$. $\\$ In a discussion that appeared earlier today on Physics Forums, there was some question whether this Fresnel reflection coefficient also contains a phase factor of $e^{i \pi/2}$. Here is a "link" to that discussion : https://www.physicsforums.com/threads/interference-puzzle.942715/#post-5963889 The phase change probably won't affect the computed result in any tremendously significant way, so setting the phase change to zero (making phase factor equal to $1$) should work for computational purposes. $\\$ Once you compute the electric field amplitude, by summing the results of multiple reflections, you then need to compute intensity $I$ which is proportional to $|E_T|^2$, where $|E_T|$ is the total electric field amplitude.(Intensity $I=nE^2$ where $n$ is the index of refraction. Here $n =1$. This equation is in units that simplify the computation). $\\$ And it can also be helpful to use a Fresnel transmission coefficient $\tau=\sqrt{1-R }$. $\\$The Optics textbook by Hecht and Zajac does a good job of showing how to sum these multiple reflections for the Fabry-Perot interference. $\\$ Editing: Scratch this part: What you have for part b in the OP works quite well...Scratch: $\big{[}$There is actually an easier way to work this problem than summing multiple reflections. I don't have a "link" for you, but can describe the method which uses the Fresnel coefficients and involves the complex $E$ amplitudes. Just outside and to the left of the cavity there is $E_{A \, right (incident)}$ and an $E_{A \, left \, (reflected)}$. Just inside the cavity on the left side there is $E_{B \, left}$ and $E_{B \, right}$. On the right side of the interior of the cavity there is $E_{C \, left}$ and $E_{C \, right}$. Finally to the very right, just to the right of the cavity there is $E_{D \, right \, (transmitted)}$ which is right going. Where the $E$ amplitude has a subscript "left" it means wave "traveling in the left direction" and a subscript "right" means wave traveling to the right. The various amplitudes are related by phase factors and/or Fresnel coefficients. Once you write down all of the equations connecting them, the solution simply involves solving for $E_{D \, right \, (transmitted)}$ in terms of $E_{A \, right \, (incident) }$. $\\$ e.g. one equation with the Fresnel coefficients is $E_{A \, left \, (reflected)}=\rho_1 E_{A \, right \, (incident)}+\tau_1 E_{B \, left}$. $\\$ Note: I did get this other method to work and also give the same answer, but for a single layer, the method of summing the infinite geometric series of multiple reflections is simpler. This other method can be very useful when there are multiple layers (e.g. multiple layers of thin films as in thin film optical filters). For even a double layer, counting multiple reflections is extremely difficult, and this other method is necessary, but for a single layer, summing the infinite geometric series is the easiest way to compute the transmitted electric field amplitude. $\big{]}$ $\\$ Editing: For part b, when you add up the contributions from all of the multiple reflections, there will be a factor $\tau_1=\sqrt{1-R_1}$ to get the wave into the cavity, and a factor $\tau_2=\sqrt{1-R_2}$ to get each wave contribution out of the cavity. When you get the result for $A_t=E_t$, you then need to take the absolute value squared of the result for $E_t$ by multiplying it by its complex conjugate to get the intensity, i.e. $I_t=|E_t|^2=(E_t)(E_t^*)$. A factor of $1-\sqrt{R_1} \sqrt{R_2} e^{-i \delta}$ gives you the necessary complex conjugate in the denominator term. (No need to multiply numerator and denominator by $(1+... )$ as you show in the OP. That step is not the correct one to get the denominator free of the imaginary term).

 

[Clara Chung]

One thing that is helpful is to use Fresnel coefficients in doing these sums of the electric field amplitude from multiple reflections. If energy reflectivity coefficient is $R$, the $\rho$ for the Fresnel reflection coefficient satisfies $|\rho|^2=R$, so that $|\rho|=\sqrt{R}$. $\\$ In a discussion that appeared earlier today on Physics Forums, there was some question whether this Fresnel reflection coefficient also contains a phase factor of $e^{i \pi/2}$. Here is a "link" to that discussion : https://www.physicsforums.com/threads/interference-puzzle.942715/#post-5963889 The phase change probably won't affect the computed result in any tremendously significant way, so setting the phase change to zero (making phase factor equal to $1$) should work for computational purposes. $\\$ Once you compute the electric field amplitude, by summing the results of multiple reflections, you then need to compute intensity $I$ which is proportional to $|E_T|^2$, where $|E_T|$ is the total electric field amplitude.(Intensity $I=nE^2$ where $n$ is the index of refraction. Here $n =1$. This equation is in units that simplify the computation). $\\$ And it can also be helpful to use a Fresnel transmission coefficient $\tau=\sqrt{1-R }$. $\\$The Optics textbook by Hecht and Zajac does a good job of showing how to sum these multiple reflections for the Fabry-Perot interference. $\\$ Editing: Scratch this part: What you have for part b in the OP works quite well...Scratch: $\big{[}$There is actually an easier way to work this problem than summing multiple reflections. I don't have a "link" for you, but can describe the method which uses the Fresnel coefficients and involves the complex $E$ amplitudes. Just outside and to the left of the cavity there is $E_{A \, right (incident)}$ and an $E_{A \, left \, (reflected)}$. Just inside the cavity on the left side there is $E_{B \, left}$ and $E_{B \, right}$. On the right side of the interior of the cavity there is $E_{C \, left}$ and $E_{C \, right}$. Finally to the very right, just to the right of the cavity there is $E_{D \, right \, (transmitted)}$ which is right going. Where the $E$ amplitude has a subscript "left" it means wave "traveling in the left direction" and a subscript "right" means wave traveling to the right. The various amplitudes are related by phase factors and/or Fresnel coefficients. Once you write down all of the equations connecting them, the solution simply involves solving for $E_{D \, right \, (transmitted)}$ in terms of $E_{A \, right \, (incident) }$. $\\$ e.g. one equation with the Fresnel coefficients is $E_{A \, left \, (reflected)}=\rho_1 E_{A \, right \, (incident)}+\tau_1 E_{B \, left}$. $\\$ Note: I did get this other method to work and also give the same answer, but for a single layer, the method of summing the infinite geometric series of multiple reflections is simpler. This other method can be very useful when there are multiple layers (e.g. multiple layers of thin films as in thin film optical filters). For even a double layer, counting multiple reflections is extremely difficult, and this other method is necessary, but for a single layer, summing the infinite geometric series is the easiest way to compute the transmitted electric field amplitude. $\big{]}$ $\\$ Editing: For part b, when you add up the contributions from all of the multiple reflections, there will be a factor $\tau_1=\sqrt{1-R_1}$ to get the wave into the cavity, and a factor $\tau_2=\sqrt{1-R_2}$ to get each wave contribution out of the cavity. When you get the result for $A_t=E_t$, you then need to take the absolute value squared of the result for $E_t$ by multiplying it by its complex conjugate to get the intensity, i.e. $I_t=|E_t|^2=(E_t)(E_t^*)$. A factor of $1-\sqrt{R_1} \sqrt{R_2} e^{-i \delta}$ gives you the necessary complex conjugate in the denominator term. (No need to multiply numerator and denominator by $(1+... )$ as you show in the OP. That step is not the correct one to get the denominator free of the imaginary term).

Very detailed!