Gaussian optics

Question

We are given $\frac{n_1}{l_0}+\frac{n_2}{l_i}=\frac{1}{R}(\frac{n_2s_i}{l_i}-\frac{n_1s_0}{l_0})$

$l_0=\sqrt{R^2+(s_0+R)^2-2R(s_0+R)cos(\phi)}$

$l_i=\sqrt{R^2+(s_i-R)^2+2R(s_i-R)cos(\phi)}$

$h= R\;sin(\phi)$

The equation we want to derive is $\frac{n_1}{s_0}+\frac{n_2}{s_i}=\frac{n_2-n_1}{R}+h^2(\frac{n_1}{2s_0}(\frac{1}{s_0}+\frac{1}{R})^2+\frac{n_2}{2s_i}(\frac{1}{R}-\frac{1}{s_i})^2)$

We are told to proceed by replacing cos(ϕ) with its third degree taylor polynomial as well as approximating ϕ by sin(ϕ).I can't seem to get this solution even using these hints.Please let me know how to prove this identity.

Answer 1

You want to compute $$ \frac{n_1}{s_0} + \frac{n_2}{s_i} $$ to the lowest non-vanishing power of $h/R$, given that

$$ \frac{n_1}{l_0} + \frac{n_2}{l_i} = \frac{1}{R}\left( \frac{n_2\,s_i}{l_i} - \frac{n_1\,s_0}{l_0} \right) \qquad (1) $$

where $$ l_0^2 = R^2 + (s_0 + R)^2 - 2R\,(s_0 + R)\cos\phi $$ $$ l_i^2 = R^2 + (s_i - R)^2 + 2R\,(s_i - R)\cos\phi $$ and $$ h = R\sin\phi $$

---

Start by writing $\cos\phi = (1 - 1 + \cos\phi)$ so you can complete the squares in $l_0^2$ and $l_i^2$ and write $$ l_0^2 = [\,R - (s_0 + R)\,]^2 + 2R\,(s_0 + R)(1 - \cos\phi) $$ $$ l_i^2 = [\,R + (s_i - R)\,]^2 - 2R\,(s_i - R)(1 - \cos\phi) $$ or $$ l_0^2 = s_0^2 + 2R\,(s_0 + R)(1 - \cos\phi) $$ $$ l_i^2 = s_i^2 - 2R\,(s_i - R)(1 - \cos\phi) $$

Now use the trigonometric identity $\cos(2\theta) = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta$ to write $$ 1 - \cos\phi = 2\sin^2(\frac{\phi}{2}) $$ so $$ l_0^2 = s_0^2 + 4R\,(s_0 + R)\sin^2(\frac{\phi}{2}) $$ $$ l_i^2 = s_i^2 - 4R\,(s_i - R)\sin^2(\frac{\phi}{2}) $$

For $|\,h/R\,| \ll 1$, $\sin\phi \approx \phi$ so, in that approximation, $$ l_0^2 \approx s_0^2 + R\,(s_0 + R)\frac{h^2}{R^2} = s_0^2 \left(1 + \frac{s_0 + R}{R\,s_0^2}\,h^2 \right) $$ $$ l_i^2 \approx s_i^2 - R\,(s_i - R)\frac{h^2}{R^2} = s_i^2 \left(1 - \frac{s_i - R}{R\,s_i^2}\,h^2 \right) $$

Now take the inverse square root, allowing for $(1+x)^{-1/2} \approx 1 - x/2$ when $|\,x\,| \ll 1$: $$ \frac{1}{l_0} \approx \frac{1}{s_0}\left(1 - \frac{1}{2}\,\frac{s_0 + R}{R\,s_0^2}\,h^2 \right) $$ $$ \frac{1}{l_i} \approx \frac{1}{s_i}\left(1 + \frac{1}{2}\,\frac{s_i - R}{R\,s_i^2}\,h^2 \right) $$

Therefore, $$ \frac{1}{R}\left( \frac{n_2\,s_i}{l_i} - \frac{n_1\,s_0}{l_0} \right) \approx \frac{1}{R}\left\{ n_2\left(1 + \frac{1}{2}\,\frac{s_i - R}{R\,s_i^2}\,h^2 \right) - n_1\left(1 - \frac{1}{2}\,\frac{s_0 + R}{R\,s_0^2}\,h^2 \right) \right\} $$ or

$$ \frac{1}{R}\left( \frac{n_2\,s_i}{l_i} - \frac{n_1\,s_0}{l_0} \right) \approx \frac{n_2 - n_1}{R} + \frac{h^2}{R}\left\{ \frac{n_1}{2s_0}\left( \frac{1}{R} + \frac{1}{s_0} \right) + \frac{n_2}{2s_i}\left( \frac{1}{R} - \frac{1}{s_i} \right) \right\} \qquad (2) $$

On the other hand, we also have $$ \frac{n_1}{l_0} + \frac{n_2}{l_i} \approx \frac{n_1}{s_0}\left(1 - \frac{1}{2}\,\frac{s_0 + R}{R\,s_0^2}\,h^2 \right) + \frac{n_2}{s_i}\left(1 + \frac{1}{2}\,\frac{s_i - R}{R\,s_i^2}\,h^2 \right) $$ or

$$ \frac{n_1}{l_0} + \frac{n_2}{l_i} \approx \frac{n_1}{s_0} + \frac{n_2}{s_i} - n_1\,\frac{h^2}{2s_0^2}\left(\frac{1}{R} + \frac{1}{s_0}\right) + n_2\,\frac{h^2}{2s_i^2}\left(\frac{1}{R} - \frac{1}{s_i} \right) \qquad (3) $$

Since we're given (1), we can now equate (3) and (2) to obtain $$ \frac{n_1}{s_0} + \frac{n_2}{s_i} - n_1\,\frac{h^2}{2s_0^2}\left(\frac{1}{R} + \frac{1}{s_0}\right) + n_2\,\frac{h^2}{2s_i^2}\left(\frac{1}{R} - \frac{1}{s_i} \right) \approx \frac{n_2 - n_1}{R} + \frac{h^2}{R}\left\{ \frac{n_1}{2s_0}\left( \frac{1}{R} + \frac{1}{s_0} \right) + \frac{n_2}{2s_i}\left( \frac{1}{R} - \frac{1}{s_i} \right) \right\} $$ or

$$ \frac{n_1}{s_0} + \frac{n_2}{s_i} \approx \frac{n_2 - n_1}{R} + \frac{n_1\,h^2}{2s_0}\left(\frac{1}{s_0} + \frac{1}{R} \right)^2 + \frac{n_2\,h^2}{2s_i}\left( \frac{1}{R} - \frac{1}{s_i}\right)^2 $$

which is the expression to be derived.