Lasers (Homework 1, Optics)

1. Starting from the ray equation, prove that if $n$ does not depend on $z$, that is $n=n(x,y)$ we have $d^2x/dz^2=\frac{1}{2n^2(x_0,y_0)\cos^2(\gamma_0)}*\frac{d n^2}{d x}$. here the initial ray starts from $(x_0,y_0)$ and makes an angle $\gamma_0$ with the $z$ axis.

Ray equation is

$(d/d s) (n*d x/d s)=∂n/∂x$

$d n/d s*d x/d s+n*(d^2 x)/(d s^2 )=∂n/∂x$

with $d s=\sqrt{(1+(d x/dz)^2)} dz$    expressed in $z$ coordinate
$1/\sqrt{(1+(d x/dz)^2)}*(d n/dz)*(d x/d s)+n*(d^2 x)/(d s^2 )=∂n/∂x$

but $n=n(x,y)$  so that  $d n/dz=0$

it remains:   $n*(d^2 x)/(d s^2 )=∂n/∂x$
or $2n^2*(d^2 x)/(d s^2 )=2n*∂n/∂x$

or $(d^2 x)/(d s^2 )=1/(2n^2 )*(∂(n^2))/∂x$

since $d s^2=(1+(d x/dz)^2 )dz^2$    we can write
$[1/(1+(d x/dz)^2 )]*[(d^2 x)/(dz^2 )]=[1/(2n^2 )]*(∂(n^2))/∂x$    
with $d x/( dz)=\tan⁡ γ$

$(d^2 x)/(dz^2 )=(1+\tan^2⁡γ)/(2n^2 )*(∂(n^2))/∂x$


$(d^2 x)/(dz^2 )=1/(2n^2*cos^2⁡γ )*(∂(n^2))/∂x$     because $1+ \tan^2⁡x=sec^2⁡x$

2. Using the ray matrices prove the Gaussian thin lens formula $(1/s_0)+(1/s_i)=(1/f)$. Suppose the object is located on the optical axis of the lens.

We have the initial point  $\begin{pmatrix} r_1\\  \theta_1 \end{pmatrix}$
and the final point $\begin{pmatrix} r_2\\  \theta_2 \end{pmatrix}$ .

The transfer matrix space-lens-space is

$T=\begin{pmatrix}1 & d_2\\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0\\ -1/f & 1\end{pmatrix}\begin{pmatrix}1 & d_1\\ 1 & 1\end{pmatrix}=\begin{pmatrix}1-(d_2/f) & d_1+d_2-(d_1 d_2/f))\\ -1/f & 1-(d_1/f))\end{pmatrix}$

 So that the total is:

$\begin{pmatrix}r_1\\\theta_1\end{pmatrix} =\begin{pmatrix}1-(d_2/f) & d_1+d_2-(d_1 d_2/f))\\ -1/f & 1-(d_1/f))\end{pmatrix}*\begin{pmatrix}r_2\\ \theta_2\end{pmatrix}$


$r_2=(1-d_2/f) r_1+(d_1+d_2-(d_1 d_2)/f) θ_1$

But because all incident rays on the object at $r_1$ must converge on the image at $r_2$ and this does not not depend on their incident angle (this is called the imaging condition) it follows that

$(d_1+d_2-(d_1 d_2)/f)=0$

or $1/d_1 +1/d_2 =1/f$