Gaussian Optics / Paraxial Approximation

[DivGradCurl]

Derive

\frac{n_1}{s_o} + \frac{n_2}{s_i} = \frac{n_2-n_1}{R}

for Gaussian optics from the following equation

\frac{n_1}{l_o} + \frac{n_2}{l_i} = \frac{1}{R} \left( \frac{n_2s_i}{l_i} - \frac{n_1 s_o}{l_o} \right)

by approximating

l_o = \sqrt{R^2 + \left( s_o + R \right) ^2 - 2R\left( s_o + R \right) \cos \phi}

and

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

with the aid of

\cos \phi \approx 1

\hline

Here is what I have:

l_o \approx \sqrt{R^2 + \left( s_o + R \right) ^2 - 2R\left( s_o + R \right) } = \sqrt{R^2 + s_o ^2 + R^2 + 2s_o R - 2s_o R - 2R^2 } = s_o

and

l_i \approx \sqrt{R^2 + \left( s_i - R \right) ^2 - 2R\left( s_i - R \right) } = \sqrt{R^2 + s_i ^2 + R^2 -2s_i R - 2Rs_i + 2R^2} = 2R-s_i

But, I expected to find l_i \approx s_i .... what I have doesn't seem to work out. I can't find where I made a mistake, though.

Thanks

[marlon]

Well keep in mind that the li and l0 are in the denominator...

You need to use the fact that \frac{1}{2R(1-\frac{s_i}{2R})} = \frac{1}{2R}(1 + \frac{s_i}{2R} +...)

Because of the paraxial approximation you have that si >>> R


regards
marlon : ps are you sure about these formulas ??? Shouldn't there be a minus sign in the left hand side of your first given formula...

[DivGradCurl]

I've double checked my equations. Unfortunately, they're exactly equal to those in my calc book (Calculus: Concepts and Contexts 2nd Ed. by James Stewart - page 632). Well, I did some more work, but couldn't get the desired result. Anyway, here you go:

If

\frac{n_1}{s_o} = \frac{n_1}{l_o}

and

\frac{n_2}{s_i} = \frac{n_2}{2R-s_i} = \frac{n_2}{2R\left( 1 - \frac{s_i}{2R} \right)} = \frac{n_2}{2R} \sum _{n=0} ^{\infty} \left( \frac{s_i}{2R} \right)^n \approx \frac{n_2}{2R}

Then

\frac{n_1}{s_o} + \frac{n_2}{s_i} = \frac{n_1}{l_o} + \frac{n_2}{2R} = \frac{2Rn_1 + s_o n_2}{2Rs_o} = \frac{1}{R}\left( \frac{Rn_1}{s_o} + \frac{n_2}{2} \right) \neq \frac{n_2 - n_1}{R}

In other words, I'm stuck! :smile: