- #1

Pendleton

- 20

- 3

- Homework Statement
- Prove the reflection and transmission coefficients of light crossing a sharp boundary between uniform volumes of different refractive indices sum to one.

- Relevant Equations
- $$R \equiv {I_R \over I_I} = \left( {E_{0_R} \over E_{0_I}} \right)^2, \quad\quad T \equiv {I_T \over I_I} = {\epsilon_2 v_2 \over \epsilon_1 v_1} \left( {E_{0_T} \over E_{0_I}} \right)^2, \quad\quad \tilde E_{0_R} = {1-\beta \over 1+\beta}E_{0_I}, \quad\quad \tilde E_{0_T} = {2 \over 1+\beta}E_{0_I}, \quad\quad \beta = {\mu_1n_2\over\mu_2n_1}$$

Consider polarized light crossing a sharp boundary between two volumes, each of a different but uniform refraction index ##n_1## or ##n_2##.

Prove that the sum of the transmission and reflection coefficients of this light ##R+T=1##, where

$$R \equiv {I_R \over I_I} = \left( {E_{0_R} \over E_{0_I}} \right)^2, \quad\quad T \equiv {I_T \over I_I} = {\epsilon_2 v_2 \over \epsilon_1 v_1} \left( {E_{0_T} \over E_{0_I}} \right)^2, \quad\quad \tilde E_{0_R} = {1-\beta \over 1+\beta}E_{0_I}, \quad\quad \tilde E_{0_T} = {2 \over 1+\beta}E_{0_I}, \quad\quad \beta = {\mu_1n_2\over\mu_2n_1}$$

First, we note that the real amplitudes are

$$E_{0_R} = \left|{1-\beta \over 1+\beta}E_{0_I}\right|, \quad\quad E_{0_T} = {2 \over 1+\beta}E_{0_I}$$

Next, we substitute them into the coefficient formulae

$$R={\left(\left|{1-\beta \over 1+\beta}E_{0_I}\right|\right)^2 \over E_{0_I}^2} = \left({1-\beta \over 1+\beta}\right)^2,

\quad\quad

T= {\epsilon_2 v_2 \over \epsilon_1 v_1}{\left({2 \over 1+\beta}E_{0_I}\right)^2 \over E_{0_I}^2} = {\epsilon_2 v_2 \over \epsilon_1 v_1}\left(2 \over 1+\beta\right)^2$$

Therefore,

$$R+T = \left({1-\beta \over 1+\beta}\right)^2 + {\epsilon_2 v_2 \over \epsilon_1 v_1}\left(2 \over 1+\beta\right)^2 = {(1-\beta)^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over (1+\beta)^2} $$

Expanding the binomials,

$${1-2\beta +\beta^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over 1+2\beta+\beta^2}$$

Substituting for beta,

$${1-2\left({\mu_1n_2\over\mu_2n_1}\right) +\left({\mu_1n_2\over\mu_2n_1}\right)^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over 1+2\left({\mu_1n_2\over\mu_2n_1}\right)+\left({\mu_1n_2\over\mu_2n_1}\right)^2}$$

Multiplying the top and bottom by the squared denominator of beta,

$${(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2+4{\epsilon_2 v_2 \over \epsilon_1 v_1}(\mu_2n_1)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

Refactoring and restating the ##v## as ##n##,

$${(4{\epsilon_2 n_1 \over \epsilon_1 n_2}+1)(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

By definition,

$$n \equiv \sqrt{\epsilon\mu\over\epsilon_0\mu_0}$$

Therefore,

$$\epsilon = n^2\epsilon_0{\mu_0\over\mu}$$

Therefore,

$$4{\epsilon_2 n_1 \over \epsilon_1 n_2} = 4{n_2^2\epsilon_0{\mu_0\over\mu_2} n_1 \over n_1^2\epsilon_0{\mu_0\over\mu_1} n_2}$$

$$= 4{n_2\mu_1 \over n_1\mu_2}$$

Substituting back,

$${(4{n_2\mu_1 \over n_1\mu_2}+1)(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

Multiplying the top and bottom by the denominator of the fraction in the numerator,

$${(4n_2\mu_1+n_1\mu_2)(\mu_2n_1)^3-2\mu_1n_2(\mu_2n_1)^2 +n_1\mu_2(\mu_1n_2)^2 \over (\mu_2n_1)^3+2\mu_1n_2(\mu_2n_1)^2+n_1\mu_2(\mu_1n_2)^2}$$

Putting the Greek letters first,

$${(4\mu_1 n_2+\mu_2 n_1)(\mu_2n_1)^3-2\mu_1n_2(\mu_2n_1)^2 +\mu_2 n_1(\mu_1 n_2)^2 \over (\mu_2n_1)^3+2\mu_1n_2(\mu_2n_1)^2+\mu_2n_1(\mu_1n_2)^2}$$

I am out of ideas.

Prove that the sum of the transmission and reflection coefficients of this light ##R+T=1##, where

$$R \equiv {I_R \over I_I} = \left( {E_{0_R} \over E_{0_I}} \right)^2, \quad\quad T \equiv {I_T \over I_I} = {\epsilon_2 v_2 \over \epsilon_1 v_1} \left( {E_{0_T} \over E_{0_I}} \right)^2, \quad\quad \tilde E_{0_R} = {1-\beta \over 1+\beta}E_{0_I}, \quad\quad \tilde E_{0_T} = {2 \over 1+\beta}E_{0_I}, \quad\quad \beta = {\mu_1n_2\over\mu_2n_1}$$

First, we note that the real amplitudes are

$$E_{0_R} = \left|{1-\beta \over 1+\beta}E_{0_I}\right|, \quad\quad E_{0_T} = {2 \over 1+\beta}E_{0_I}$$

Next, we substitute them into the coefficient formulae

$$R={\left(\left|{1-\beta \over 1+\beta}E_{0_I}\right|\right)^2 \over E_{0_I}^2} = \left({1-\beta \over 1+\beta}\right)^2,

\quad\quad

T= {\epsilon_2 v_2 \over \epsilon_1 v_1}{\left({2 \over 1+\beta}E_{0_I}\right)^2 \over E_{0_I}^2} = {\epsilon_2 v_2 \over \epsilon_1 v_1}\left(2 \over 1+\beta\right)^2$$

Therefore,

$$R+T = \left({1-\beta \over 1+\beta}\right)^2 + {\epsilon_2 v_2 \over \epsilon_1 v_1}\left(2 \over 1+\beta\right)^2 = {(1-\beta)^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over (1+\beta)^2} $$

Expanding the binomials,

$${1-2\beta +\beta^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over 1+2\beta+\beta^2}$$

Substituting for beta,

$${1-2\left({\mu_1n_2\over\mu_2n_1}\right) +\left({\mu_1n_2\over\mu_2n_1}\right)^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over 1+2\left({\mu_1n_2\over\mu_2n_1}\right)+\left({\mu_1n_2\over\mu_2n_1}\right)^2}$$

Multiplying the top and bottom by the squared denominator of beta,

$${(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2+4{\epsilon_2 v_2 \over \epsilon_1 v_1}(\mu_2n_1)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

Refactoring and restating the ##v## as ##n##,

$${(4{\epsilon_2 n_1 \over \epsilon_1 n_2}+1)(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

By definition,

$$n \equiv \sqrt{\epsilon\mu\over\epsilon_0\mu_0}$$

Therefore,

$$\epsilon = n^2\epsilon_0{\mu_0\over\mu}$$

Therefore,

$$4{\epsilon_2 n_1 \over \epsilon_1 n_2} = 4{n_2^2\epsilon_0{\mu_0\over\mu_2} n_1 \over n_1^2\epsilon_0{\mu_0\over\mu_1} n_2}$$

$$= 4{n_2\mu_1 \over n_1\mu_2}$$

Substituting back,

$${(4{n_2\mu_1 \over n_1\mu_2}+1)(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

Multiplying the top and bottom by the denominator of the fraction in the numerator,

$${(4n_2\mu_1+n_1\mu_2)(\mu_2n_1)^3-2\mu_1n_2(\mu_2n_1)^2 +n_1\mu_2(\mu_1n_2)^2 \over (\mu_2n_1)^3+2\mu_1n_2(\mu_2n_1)^2+n_1\mu_2(\mu_1n_2)^2}$$

Putting the Greek letters first,

$${(4\mu_1 n_2+\mu_2 n_1)(\mu_2n_1)^3-2\mu_1n_2(\mu_2n_1)^2 +\mu_2 n_1(\mu_1 n_2)^2 \over (\mu_2n_1)^3+2\mu_1n_2(\mu_2n_1)^2+\mu_2n_1(\mu_1n_2)^2}$$

I am out of ideas.