# Combining lenses

1. Nov 4, 2009

### E92M3

1. The problem statement, all variables and given/known data
2 thin lenses separated by 10cm. First has a focal length of 20cm, the second has a focal length of 30cm.

2. Relevant equations
Lens transfer matrix for each lens:
$$\begin{bmatrix} 1-\frac{D_2d_{21}}{n_{t1}} &-D_1-D_2+\frac{D_1D_2d_{21}}{n_{t1}} \\ \frac{d_{21}}{n_{t1}} & 1-\frac{D_2d_{21}}{n_{t1}} \end{bmatrix}$$

3. The attempt at a solution
Since the thickness of the 2 lens is assumed to be zero we have:
$$A_1=\begin{bmatrix} 1 & \frac{-1}{f_1}\\ 0 & 1 \end{bmatrix}$$
For the first lens and
$$A_2=\begin{bmatrix} 1 & \frac{-1}{f_2}\\ 0 & 1 \end{bmatrix}$$
for the second. Therefore we have:
$$A=A_2TA_1=A_2=\begin{bmatrix} 1 & \frac{-1}{f_2}\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0\\ d & 1 \end{bmatrix}\begin{bmatrix} 1 & \frac{-1}{f_1}\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1-\frac{d}{f_2} & \frac{-1}{f_1}-\frac{1}{f_2}+\frac{d}{f_1f_2}\\ d & 1-\frac{d}{f_1} \end{bmatrix}=\begin{bmatrix} \frac{2}{3} & \frac{-1}{15}cm^{-1}\\ 10cm & \frac{1}{2} \end{bmatrix}=\begin{bmatrix} A_{11} & A_{12}\\ A_{21} & A_{22} \end{bmatrix}$$

Since the system is assumed to be in air, the focal length is the same on both sides of the system. SO we have:

$$f=\frac{1}{-A_{12}}=15cm$$
$$\overline{H_1O_1}=(1-A_{11})f=5cm$$
$$\overline{H_2O_2}=(1-A_{22})f=7.5cm$$

Where H denotes the principle planes and O denotes the lenses.
So here's the problem, recall that the lenses are separated by only 10cm, this forces principle plane #2 to be in front (left of) principle plane #1. How can this be true? What did i do wrong?

Last edited: Nov 4, 2009