Compound lens separated by a distance

[Toby_phys]

Homework Statement

Homework 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:
$$\frac{1}{s} +\frac{1}{s'}=\frac{1}{f_1}$$

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

 

[BvU]

Hello Toby,

Note - I feel my mistake is here, I think my algebra is correct after this.

Nope. Just fine.

The total focal length is given by

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).