Thin lens formula Can someone help me or guide me how the thin lens formula:
$$\frac{1}{s_1}+\frac{1}{s_0}=\frac{1}{f}$$
can be proven?
I was trying to prove it on my own using similar triangles, only to fail.
 A: I know this formula can be derived in 2 ways, and not sure of the best one for you, so I will show both of them. Summed up I would call them "Thin lens method" and "two surface method" or "two index method"
The thin lens method is most simple, so I will start with that...
Assume the thin lens (magically) turns the light at a changing slope that increases the further one enters away from the center of the lens, if you cannot do this, move on to the "two surface method".  In the real world the rate of changing slope as a ray enters further away from the center of the lens depends on a lot of things, but in this case let's just say this is linear. So let's use "$r$" to represent the distance from the center of the lens that the light ray enters and "$m$" to represent the slope of entry relative to the normal of the lens. I use "$P$" to represent the constant that "$r$" should be multiplied by to get the value of the slope change. So if we have an object at "$S_1$" from the lens and "$R$" from the center axis, we know one ray passing through the center of the lens will have no slope change, let's call this the "chief ray".  We can know the slope of a different ray with this
$$
m_{in} = { R - r \over S_1}
$$
Now this slope is changed with "$r$" and "$P$"...
$$
m_{out} = m_{in} + r * P = { R - r \over S_1} + r * P
$$
Remember 
$$
m_{chief}={R \over S_1}
$$
So given the new slope, we will observe an intersection with the "chief ray" at:
$$
S_0 = {r \over \left( m_{out}-m_{chief} \right)}
$$
$$
{1 \over S_0}={m_{out} \over r}-{m_{chief} \over r} ={{ R - r \over S_1} + r * P \over r}-{R \over r * S_1}={{ R - r \over r * S_1} + P }-{R \over r * S_1}
$$
Now we can know the relation between "$P$", "$S_1$", and "$S_2$" is:
$$
{1 \over S_0}+{R \over r * S_1}-{{ R - r \over r * S_1} } = P ={1 \over S_1} + {1 \over S_0}
$$
Thus:
$$
P={1 \over f}
$$
$$
\space
$$
The "two surface method" uses the index of refraction properties, so unless Snell's Law has been covered, this method probably cannot be. Also keep in mind an ideal lens is made with parabolic curves, though most real lenses use spherical surfaces. Mostly for ease of manufacturing and compatibility with other lenses(also spherical), this is why we get spherical aberrations. I am not going to try and explain the math behind spherical lenses, as they do not work like ideal lenses. So we start with Snell's  Law and move toward defining one over the focal length. 
If we are using a parabolic lens than we know the surface has a slope relative to the line perpendicular to the lens of:
$$
m_{s1} = {1 \over r} \space \space \space \space \space \space and \space on \space the \space other \space side  \space \space \space \space \space \space  m_{s0} = {- 1 \over r} 
$$
Using the same variables as before:
$$
m_{in} = { R - r \over S_1}
$$
$$
m_{chief}={R \over S_1}
$$
Only this time a difference between the distance from the object to the actual lens surface and the distance "$S_1$" exists. Just to note that fact. We will say that the plane, of which "$S_1$" is the distance to from the object, is the plane an ideal lens would be placed at to simulate the same result as this real system with an index "$n$". We will also assume air has an index of one. For the "chief ray" we know the ray should be directly perpendicular to the surface. Thus this ray will exit at a point on the lens that also has a surface perpendicular to the ray.  
So using Snell's Law we can find the new slope of the ray as it is inside the lens:
$$
{\sin\left(\theta_{going-in}\right) \over \sin\left(\theta_{inside}\right)} = {1 \over n} \space \space \space \space \space \space \rightarrow \space \space \space \space \space \space {{\sin\left(\arctan\left(m_{s1}\right)-\arctan\left(m_{in}\right)\right) \over \sin\left(\arctan\left(m_{inside}\right)-\arctan\left({-1 \over m_{s1}}\right)\right)}} = {1 \over n} 
$$
OK, so let's make that a bit more solvable, by assuming this ray is directly perpendicular to what would be the thin lens as it enters this lens (basically "$R=r$" and "$m_{in}=0$"):
$$
\sin\left(\arctan\left(m_{inside}\right) - \arctan\left({-1 \over m_{s1}}\right)\right) = \sin\left(\arctan\left(m_{s1}\right)\right) * n 
$$
Now to find the exit angle with the same Snell's Law:
$$
{\sin\left(\theta_{inside}\right) \over \sin\left(\theta_{going-out}\right)} = {n \over 1} \space \space \space \space \space \space \rightarrow \space \space \space \space \space \space n={{\sin\left(\arctan\left(m_{inside}\right)\right) \over \sin\left(\arctan\left({{r+m_{chief}*S_0} \over S_0}\right)\right)}} 
$$
Now we can start getting something from:
$$
n*\sin\left(\arctan\left({{r+m_{chief}*S_0} \over S_0}\right)\right)=\sin\left(\arctan\left(m_{inside}\right)\right)
$$
using 
$$
\arctan\left(m_{inside}\right) = \arcsin\left( \sin\left(\arctan\left(m_{s1}\right) * n\right) + \arctan\left({-1 \over m_{s1}}\right)\right) 
$$
$$
n*\sin\left(\arctan\left({{r+m_{chief}*S_0} \over S_0}\right)\right)=\sin\left(\arcsin\left( \sin\left(\arctan\left(m_{s1}\right)\right) * n\right) + \arctan\left({-1 \over m_{s1}}\right) \right)
$$
$$
n*\sin\left(\arctan\left({R \over S_0}+{R \over S_1}\right)\right)=\sin\left(\arcsin\left( \sin\left(\arctan\left({1 \over R}\right)\right) * n\right) + \arctan\left({-R}\right) \right)
$$
$$
\arcsin\left(n*\sin\left(\arctan\left({R \over S_0}+{R \over S_1}\right)\right)\right)=\arcsin\left( \sin\left(\arctan\left({1 \over R}\right)\right) * n\right) + \arctan\left({-R}\right)
$$
$$
\sin\left(\arctan\left({R \over S_0}+{R \over S_1}\right)\right)= \sin\left(\arctan\left({1 \over R}\right)\right) + {\sin\left(\arctan\left(-R\right) \right) \over n}
$$
OK so that is a pain, but I think you can now see the "${1 \over S_0}+{1 \over S_1}$" forming with a trigonometric relation to "$n$". I would recommend using Jone's Vectors, but I suppose this is meant to be a proof,  Good Luck.
A: Well I was hoping somebody else would present this high school solution, since I don't have any way of posting drawings or graphs etc.    But you should be able to follow this.
In my sign convention (here) all quantities are positive.
......oh|<----x---->|<---f--->|<---f'-->|<--y-->|ih'.....
.............<---------l---------> <------l'------->|........
So my object (o) of height (h) is distant (x) from the front focal point, which is distant (f) from the thin lens. And we have  x + f = l
From the thin lens to the back focal point is (f') where f and f' are equal.  My image (i) of height h' is distant (y) from the back focal point, and  y + f' = l' 
Now I need another line which you will have to imagine.   It goes from the top of the object (oh) at height (h) , through the front focal point, and continues to the lens, at a height (down)  h.f/x = h.f/(l-f) 
Now this ray passed through the focal point, and therefore it refracts parallel to the axis, and must pass through the top of the image, at height h' (down)
Ergo  h' = h.f/x = h.f/(l-f)
By symmetry, if I reverse everything, I can write:-
 h = h'.f'/y = h'f'/(l'-f')..... = h.f/x .f'/y = h.f/(l-f) .f'/(l'-f')  

By substituting  for h and h' in the two sets of equations.
Then one can see that:    h = h.f/x.f'/y or x.y = ff' = f^2
This is Newton's thin lens formula.  Then from the other equations I have:
 h = h.f/(l-f).f'/(l'-f') , so f.f' = (l-f).(l'-f') = l.l'- f(l + l') + f.f' ;(f'=f)

So finally l.l' = f(l + l')  or 1/f = l/l.l' + l'/l.l' = 1/l + 1/l'
QED
Sorry to the mathematicians for dragging out what you can do in your head.    OP should take note of Newton's formula,    xy = f^2    x and y measured from the focus.
Notice that x = y = f gives l + l' = 4f which is the minimum object to image distance.
Sorry about the bush algebra; I can write it with a #2 pencil in 1/10th the time.
A: Another high school proof of the general lens equation:
Let's suppose you have a lens of refractive index $n_2$ seperating two media of refractive index $n_1$ and $n_3$, then by the equation of curved surface refraction, we can write image as refracted by first surface of lens:
$$ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R_1} \tag{1}$$
And, by the second surface of lens as:
$$ \frac{n_3}{v'} - \frac{n_2}{u'} = \frac{n_3-n_2}{R_2} \tag{2}$$
Variables used:
$v$ image formed by first surface
$u$ object of the first surface
$v'$ image formed by second surface
$R_1$ radius of curvature of first surface
$R_2$ radius of curvature of second surface
Fact: $u'=v$ because the image of first surface is object of second and hence (2) becomes:
$$  \frac{n_3}{v'} - \frac{n_2}{v} = \frac{n_3 - n_2}{ R_2} \tag{3}$$
Adding (3) and (1):
$$\frac{n_3}{v'}- \frac{n_1}{u}= \frac{n_2-  n_1}{R_1} + \frac{n_3- n_2}{R_2}$$
If, $n_3=n_1=1$ that is the ambient mediums on both side are air then the previous equation becomes:
$$ \frac{1}{v'} - \frac{1}{u} = (n_2-n_1) \left[  \frac{1}{R_1} + \frac{1}{R_2}\right]$$
To find focal length, just send $u\to \infty$ which makes:
$$ \frac{1}{f} = (n_2-n_1) \left[  \frac{1}{R_1} + \frac{1}{R_2}\right]$$
Hence,
$$ \frac{1}{v'} - \frac{1}{u} =  \frac{1}{f}$$

Generalization, let's say we have $j$ lens 'pieces' in series to each other and each has refractive index $n_j$ and radius of curvature $R_1$, then:
$$ \frac{n_2}{v_1} - \frac{n_1}{u_1} = \frac{n_2 - n_1}{R_1}$$
$$ \frac{n_3}{v_2} - \frac{n_2}{u_2} = \frac{n_3 - n_2}{R_1}$$
$$ \vdots$$
$$\frac{n_j}{v_j} - \frac{n_{j-1} }{u_{j-1} } = \frac{n_j - n_{j-1}}{R_j}$$
Noticing that $v_k = u_{k+1}$ and adding all the equatoins:
$$ \frac{n_j}{v_j} - \frac{n_1}{u_1}= \sum_{k=0}^n \frac{n_k -n_{k-1}}{R_k} \tag{4}$$
A: A lot depends on how much you want to learn, just to teach elementary (geometrical) optics.   There are many good books, but often they cover too much that you aren't interested in.
By far the best textbook on actual geometrical optics, relating to lens design, including aberrations etc, is one that as a mathematician, you would enjoy for yourself.
Applied Optics & Optical Design, By A. E. Conrady written in 1926 , taught all of the world's good optical designers of WW-II, on all sides of the conflict. It is fully mathematical, but high school level.  It's a two volume Dover Press paper back, that you can buy at Amazon, or B&N for peanuts.  If you get a free account for the Internet Archive, you can also borrow it hourly.
High school, algebra, geometry, and trigonometry required.  NO calculus needed, and it will show you lens formulae that are much more useful, than the one you give above.
I should add, that all of the computations are done with logarithm tables, which was the best in those days, but now you simply use a calculator.
I managed to convince a good number of employers for over 50 years, that I knew a little bit of optics; solely because of Conrady.
His daughter, Hilda Conrady Kingslake, was the wife of chief Kodak optics guru, Rudolph Kingslake, and she was for years, historian for the Optical Society of America.   Sadly, both are now doing optics in the clouds.
