Chapter 8.8, Problem 28E

a.

To determine

To derive: equation 3 for Gaussian optics from equation1 by approximating $\cos \varphi$ in equation 2 by its first-degree Taylor Polynomial.

Answer to Problem 28E

  $\frac{n_1}{s_0}+\frac{n_2}{s_i}=\frac{n_2-n_1}{R}.$

Explanation of Solution

Given information: First-degree Taylor Polynomial.

Calculation:

First degree Taylor Polynomial for $\cos \varphi$ at a =0.

  $\begin{array}{c}f(\varphi )=\cos \varphi \Rightarrow f(0)=\cos 0=1.\\ f\text{'}(\varphi )=-\sin \varphi \Rightarrow f\text{'}(0)=-\sin 0=0.\\ T_1(\varphi )=1+0.\varphi =1.\\ \text{Substitute}{T}_1\text{in}\text{equation}\text{of}{\mathcal{l}}_0\text{and}{\mathcal{l}}_1,\text{and}\text{simplify};\end{array}$

  $\begin{array}{c}{}_0=\sqrt{R^2+{(s_0+R)}^2-2R(s_0+R)\cos \varphi }\\ \approx \sqrt{R^2+{(s_0+R)}^2-2R(s_0+R)}\\ \approx \sqrt{R^2+s^2{}_0+R^2+2R{s}_0-2R{s}_0-2R^2}\\ \approx \sqrt{s_0{}^2}\\ \approx s_0\end{array}$

  $\begin{array}{c}{\mathcal{l}}_i=\sqrt{R^2+{(s_i-R)}^2+2R(s_i-R)\cos \varphi }\\ \approx \sqrt{R^2+{(s_i-R)}^2+2R(s_i-R)}\\ \approx \sqrt{R^2+s_i{}^2+R^2-2R{s}_i+2R{s}_i-2R^2}\\ \approx s_i\end{array}$

  $\text{Substitute}\text{into}\text{Hecht's}\text{equation}$

  $\begin{array}{c}\frac{n_1}{{\mathcal{l}}_0}+\frac{n_2}{{\mathcal{l}}_i}=\frac{1}{R}(\frac{n_2{s}_i}{{\mathcal{l}}_i}-\frac{n_1{s}_0}{{\mathcal{l}}_0})\\ \frac{n_1}{s_0}+\frac{n_2}{s_i}=\frac{1}{R}(\frac{n_2{s}_i}{s_i}-\frac{n_1{s}_0}{s_0})\end{array}$

  $\begin{array}{c}\frac{n_1}{s_0}+\frac{n_2}{s_i}=\frac{1}{R}(n_2-n_1)\\ \frac{n_1}{s_0}+\frac{n_2}{s_i}=\frac{n_2-n_1}{R}.\end{array}$

b.

To determine

To show: that if $\cos \varphi$ is replaced by its third −degree Taylor polynomial in equation 2, then equation1 becomes equation 4 for third-order optics.

Explanation of Solution

Given information:

Calculation:

  $\begin{array}{l}f(\varphi )=\cos \varphi .\\ \text{The}\text{third}\text{dergee}\text{Taylor}\text{polynomial}\text{is}\text{found}\text{from:}\\ $ T_3=f(a)+\frac{f\text{'}(a)}{1}(\varphi -a)+\frac{f\text{'}\text{'}(a)}{2}{(\varphi -a)}^2+\frac{f^{(3)}(a)}{6}{(\varphi -a)}^3.$\end{array}$

  $\begin{array}{c}f(\varphi )=\cos \varphi \Rightarrow f(0)=\cos 0=1.\\ f\text{'}(\varphi )=-\sin \varphi \Rightarrow f\text{'}(0)=-\sin 0=0.\\ f\text{'}\text{'}(\varphi )=-\cos \varphi \Rightarrow f\text{'}\text{'}(0)=-\cos 0=-1.\\ f^{(3)}(\varphi )=\sin \varphi \Rightarrow f^{(3)}(0)=\sin 0=0.\end{array}$

  $\text{Now}\text{substitute}\text{in}\text{the}\text{Taylor}\text{polynomial}\text{formula}\text{with}a=0:$

  $T_3=\cos 0+\frac{0}{1}(\varphi )+\frac{1}{2}{(\varphi )}^2+\frac{0}{6}{(\varphi )}^3$

  $T_3=1+\frac{1}{2}{(\varphi )}^2$

  $\begin{array}{c}\text{Since}\varphi =\sin \varphi \\ T_3=1-{\sin }^2\varphi \\ {\mathcal{l}}_0=\sqrt{R^2+{(s_0+R)}^2-2R(s_0+R)T_3}\end{array}$

  ${\mathcal{l}}_0=\sqrt{s_0{}^2+R{\sin }^2\varphi (s_0+R)}$

  ${\mathcal{l}}_0{}^{-1}=(s_0{}^2+R{\sin }^2\varphi {(s_0+R)}^{-\frac{1}{2}}$

  $\begin{array}{c}{\mathcal{l}}_0{}^{-1}=\frac{1}{s_0}-\frac{R{\sin }^2\varphi (s_0+R)}{2s_0{}^3}\\ \text{Similarly;}\\ {\mathcal{l}}_i{}^{-1}=\frac{1}{s_i}-\frac{R{\sin }^2\varphi (s_i+R)}{2s_i{}^3}\end{array}$

  $\begin{array}{c}\text{Finally}\text{}\text{,have}\text{to}\text{prove}\text{it}\text{so}:\\ \frac{n_1}{{\mathcal{l}}_0}+\frac{n_2}{{\mathcal{l}}_i}=\frac{n_1{s}_0}{R{\mathcal{l}}_0}+\frac{n_1{s}_i}{R{\mathcal{l}}_i}\end{array}$

  $\text{Because}\text{have}\text{many}\text{terms}\text{will}\text{deal}\text{half}\text{of}\text{them}\text{at}\text{a}\text{time:}$

  $\begin{array}{c}\frac{n_1}{{\mathcal{l}}_0}=-\frac{n_1{s}_0}{R{\mathcal{l}}_0}\\ \frac{n_1}{s_1}-\frac{n_{1}R{\sin }^2\varphi (s_0+R)}{2s_0{}^3}=-\frac{n_1}{R}+\frac{n_1{\sin }^2\varphi (s_0+R)}{2s_0{}^3}\end{array}$

  $\frac{n_1}{s_1}=-\frac{n_1}{R}+\frac{n_1{\sin }^2\varphi }{2s_0}[\frac{(R^2+R{s}_0)}{s_0{}^2}+\frac{(R+s_0)}{s_0}]$

  $\begin{array}{c}Let\text{}{\sin }^2\theta =\frac{h^2}{R^2}\\ \frac{n_1}{s_1}=-\frac{n_1}{R}+\frac{n_1{h}^2}{2s_0}[\frac{1}{s_0{}^2}+\frac{1}{R{s}_0}+\frac{1}{R{s}_0}+\frac{1}{R^2}].\end{array}$

  $\frac{n_1}{s_1}=-\frac{n_1}{R}+\frac{n_1{h}^2}{2s_0}{[\frac{1}{s_0}+\frac{1}{R}]}^2.$

  $\text{Similarly}.$

  $\frac{n_2}{s_i}=-\frac{n_2}{R}+\frac{n_2{h}^2}{2s_i}{[\frac{1}{s_i}+\frac{1}{R}]}^2.$

  $\text{Finally}\text{add}\text{the}\text{two}\text{halves}\text{and}\text{get}:$

  $\frac{n_1}{s_0}+\frac{n_2}{s_i}=\frac{n_2-n_1}{R}+h^2[\frac{n_1}{2s_0}{(\frac{1}{s_0}+\frac{1}{R})}^2+\frac{n_1}{2s_i}{(\frac{1}{s_i}+\frac{1}{R})}^2]$