# Focal length of lens at border of two liquids

Tags:
1. May 1, 2015

### matej1408

1. The problem statement, all variables and given/known data
I need to find optical power (reciprocal focal length) of this system with thin lens

2. Relevant equations
I tried to solve this using spherical diopter equation
n1/a+n2/b=(n2-n1)/R
where a is object distance and b is image distance

3. The attempt at a solution
equation for first diopter
n1/a+n/b'=(n-n1)/R
equation for second diopter
-n/b'+n2/b=-(n2-n)/R
adding these two equation i have
n1/a+n2/b=(2*n-n1-n2)/R
putting a→∞, b should be focal lenght then
so:
1/f=1/b=(2*n-n1-n2)/(n2*R)
but textbook solution is:

where -x=a and x'=b
i figured out this solution doesn't have two different focal length solutions (one from each side) and my solution have that. They solved problem using ray path and geometry.
I'm wondering what is wrong with my solution

2. May 2, 2015

### andrevdh

What might work is treating the two surfaces of the lens separately. That is the first produces an image which becomes the object for the second. Start out with an object far away - sorry, that is what your solution does in effect.

Last edited: May 2, 2015
3. May 3, 2015

### rude man

Look at the derivation of the lensmaker's formula and modify it to the case where n1 ≠ 1.

4. May 3, 2015

### matej1408

in my textbook is solved by derivation of lensmaker's formula, but i want to know why my solution isn't correct

5. May 5, 2015

### andrevdh

The power of the lens seems to be the sum of the powers of the two surfaces, which would then be
P = Pleft + Pright
= (n - n1)/R + (n2 - n)/-R
= 1/R (2n - n1 - n2)
I am referencing Jenkins and White - Fundamentals of Optics.
Each of the surfaces of the lens has two associated focal lengths.
So who knows what the f in the formula in the book refers to!

Last edited: May 7, 2015