# Compound lenses separated by a distance

Tags:
1. Feb 16, 2017

### Toby_phys

Let $s$ be the object distance for the first lens #s'# the distance to the image for the first lens and $s''$ the final image distance:

Using Gauss's lens equation $\frac{1}{s}+\frac{1}{s'}=\frac{1}{f_1}$ from this we get:
$$s'=\frac{sf_1}{s-f_1}$$

we can use the image from the first lens as the object for the second lens.

$$\frac{-1}{s'-d}+\frac{1}{s''}=\frac{1}{f_2}$$
Note - I feel my mistake is here, I think my algebra is correct after this.

This gets us:
$$\frac{1}{s''}=\frac{1}{f_2}+\frac{1}{s'-d}=\frac{s-f_1}{sf_1-sd+f_1d}+\frac{1}{f_2}$$

The total focal length is given by:

$$\frac{1}{f}=\frac{1}{s} +\frac{1}{s''}=\frac{1}{s}+\frac{1}{f_2}+\frac{s-f_1}{sf_1-sd+f_1d}$$

Which doesn't get the desired result. The correct result is/should be:

$$frac{1}{s}+\frac{1}{s'}=\frac{1}{f_1}+\frac{1}{f_2} + \frac{d}{f_1f_2}$$