# Compound lens separated by a distance

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1. Feb 17, 2017

### Toby_phys

1. The problem statement, all variables and given/known data

2. Relevant equations

We will call $s$ the distance of the object from the first lens, $s'$ the distance of the image from the first lens and $s''$ the distance of the image from the second lens.

The Gauss's lens equation:
$$\frac{1}{s} +\frac{1}{s'}=\frac{1}{f_1}$$

3. The attempt at a solution

Using Gauss's lens equation 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.

Thank you in advance

2. Feb 17, 2017

### BvU

Hello Toby,
Nope. Just fine.
Are $s$ and $s''$ wrt the same zero point ? No.
And: there is another unknown: the position $x$ of the equivalent lens wrt e.g. lens 1.... So that$${1\over f} = {1\over s + x} + {1\over d + s - x} \ ...$$
Meaning you'll need another equation.

I made a drawing with f1 = f2 = 5 cm, s= 10 cm and d = 3 cm. It comes out pretty neatly (but the ${3\over 100}$ is a rather small term).

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