# Geoemtrical Optics: Principal Plane question

1. Oct 25, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
An equiconvex lens having spherical surfaces of radius 10cm, a central thickness of 2cm, and a refractive index of 1.61 is situated between air and water (n=1.33). An object 5cm high is placed 60cm in front of the lens surface. Find the cardinal points for the lens and the position and size of the image formed.

2. Relevant equations
$\frac{f^{'}}{s^{'}}+\frac{f}{s}=1$
f' is the second focal point, f is the first focal point, s is the object distance, and s' is the image distance.
$\frac{n_i}{s}+\frac{n_f}{s^{'}}=\frac{n_f}{f^{'}}$
3. The attempt at a solution
I made the ray transfer matrix to solve this question, I won't go through the steps because I know that it is 100% correct.
$$\begin{bmatrix} \frac{744}{805} & \frac{200}{161}\\ -0.065322 & \frac{2220}{3059} \end{bmatrix}=\begin{bmatrix} A & B \\ C & D \end{bmatrix}$$ For my sign convention I have the distance from the first vertex to the first principal plane as $h=\frac{D-1}{C}$, the distance from the second vertex to the second principal plane as $h^{'}=\frac{A-1}{C}$ and the second focal point as $f^{'}=-\frac{1}{C}$. In my sign convention if h>0 then it is to the right of the first vertex and if h'>0 then it is to the left of the second vertex. So I get the values (I know these are also correct).
$h=4.12cm \hspace{3mm} h^{'}=1.16cm \hspace{3mm} f^{'}=15.309cm$ and then from $f=\frac{n}{n^{'}}f^{'}\Rightarrow f=11.511cm$. So in order to use the equation I must get the object distance relative to the first plane which is clearly $s=60cm+4.12cm==64.12cm$ but then why I plug that and the other values into the formula $\frac{n_i}{s}+\frac{n_f}{s^{'}}=\frac{n_f}{f^{'}}$ I get 18.65cm. The correct answer is given in my book as 18.94cm which occurs when they use 60cm as the object distance. This makes no sense to me since this is the distance from the first lens not the first principal plane.

2. Oct 30, 2015