- #1

- 1,868

- 0

## Main Question or Discussion Point

Hi guys

I wonder if you can explain this to me. In http://en.wikipedia.org/wiki/Ray_transfer_matrix_analysis#Table_of_ray_transfer_matrices it says under "Thin lens" that the ray matrix is given by

[tex]

\left( {\begin{array}{*{20}c}

{r_{final} } \\

{r_{final}^' } \\

\end{array}} \right) = \left( {\begin{array}{*{20}c}

1 & 0 \\

{ - 1/f} & 1 \\

\end{array}} \right)\left( {\begin{array}{*{20}c}

{r_{initial} } \\

{r_{initial}^' } \\

\end{array}} \right)

[/tex]

Here

The only argument my book uses is that "Immediately to the right of the lens the ray's displacement is the same as the initial displacement immediately to the left." But still, the displacements are not constant, so I cannot see why this argument works.

I wonder if you can explain this to me. In http://en.wikipedia.org/wiki/Ray_transfer_matrix_analysis#Table_of_ray_transfer_matrices it says under "Thin lens" that the ray matrix is given by

[tex]

\left( {\begin{array}{*{20}c}

{r_{final} } \\

{r_{final}^' } \\

\end{array}} \right) = \left( {\begin{array}{*{20}c}

1 & 0 \\

{ - 1/f} & 1 \\

\end{array}} \right)\left( {\begin{array}{*{20}c}

{r_{initial} } \\

{r_{initial}^' } \\

\end{array}} \right)

[/tex]

Here

*r*is the ray displacement and*r*' is the slope of the ray. Now the above ray matrix tells us that*r*_{final}=*r*_{initial}, but I cannot see why this is the case at all. I mean the displacement of the initial ray is constantly growing, while the displacement of the final ray is constantly decaying. Then how can we equal to the displacements?The only argument my book uses is that "Immediately to the right of the lens the ray's displacement is the same as the initial displacement immediately to the left." But still, the displacements are not constant, so I cannot see why this argument works.