# Equivalent focal length of two lenses

1. Oct 24, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
Two thin lenses have focal lengths of -5cm and +20cm. Determine their equivalent focal lengths when (a) cemented together and (b) separated by 10cm.

2. Relevant equations
$\frac{1}{f_{eq}}=\sum_{n=1}^k \frac{1}{f_n}$

3. The attempt at a solution
So the first part was easy since its just with $f_1=-5cm$ and $f_2=20cm$, $f_{eq}=(\frac{1}{f_1}+\frac{1}{f_2})^{-1}=-6.667cm$ but for the second part I'm not sure how to find it since there is separation in-between the two lenses. I thought it was given by the equation $\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1f_2}$ where $d$ is the distance between the lenses but this doesn't give the correct answer.

Edit: Solved it

Last edited: Oct 24, 2015
2. Oct 24, 2015

### Simon Bridge

Well done. How did you solve it? (Helps others with the same problem.)

3. Oct 25, 2015

### Potatochip911

We will use the formula $\frac{1}{f}=\frac{1}{s}+\frac{1}{s^{'}}$, now from the problem we want to know the focal length of the system, if the object is coming in from infinity then the final image will be at the focal point of infinity. So $\frac{1}{f_1}=\frac{1}{s_1}+\frac{1}{s_1^{'}}$ and $s_1\to\infty$ so $f_1=s_1^{'}$. Now our image position will become our object for the next lens thus we want to find an equation for the distance from the second lens that will agree with our sign convention (+ if left, - if right of the lens). This is given by $s_2=d-f_1$ where $d$ is the separation in-between the lenses. Applying the thin lens equation again we have $\frac{1}{f_2}=\frac{1}{s_2}+\frac{1}{s_2^{'}}$, substitute our expression for $s_2$ and note that the image position must be the focal point of the system since the rays of light exited it parallel. So $\frac{1}{f_2}=\frac{1}{d-f_1}+\frac{1}{f_{system}}\Rightarrow f_{system}=(\frac{1}{f_2}-\frac{1}{d-f_1})^{-1}$ with $f_1=-5cm$, $f_2=20cm$ and $d=10cm$. There is also another focal point of the system going the other way but it's pretty much the same concept so I'll leave it out.