Lens surrounded by two different materials? [closed]

If have a material refractive index $n_1$ then a lens refractive index $n_2$ then another material on the other side of the lens (but in contract with the lens) refractive index $n_3$. How do I find the focal length of this system (/lens makers formula)? I have tried doing this the standard way with a prism, but the algebra gets very hard, very quickly. I then thought to use the thin lens in contact formula, this however would mean that there would need to be a boundary between the two lenses where we change from the medium refractive index $n_1$ to $n_2$. Can you give me any hints of how to do it, thanks? closed as off-topic by AccidentalFourierTransform, ZeroTheHero, Kyle Kanos, Bill N, Cosmas ZachosJun 24 '18 at 18:11

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• What if you look at this system as a system of two effective lenses $n_1$-$n_2$-$n_1$ and $n_3$-$n_2$-$n_3$, where $n_2$ lenses have twice the focal length of the original lens, and the distance between them is zero, so you effectively would have $n_1$-$n_2n_2$-$n_3$? – Ruslan Dec 3 '14 at 8:39
• A lens is actually two curved surfaces. The lensmaker's formula tells you how to deal with each curved surface. Try using that to reproduce the thin lense formula for air-glass-air. Then apply it to your own situation. – Bill N Jun 20 '18 at 21:19

You can use the formula for thin lenses in contact, but you have to add an extra diopter. For example, if you want to calculate the focal length on the right side, then you consider a lens of index n1 and a lens of index n2, both living in a medium of index n3, like the following:

n3  | n1 ( n2 )  n3

The interface on the left is flat, so it does not alter a beam of light parallel to the optical axis, and thus does not affect the right focus. If you are instead interested in the left-side focal length, then you would add the extra interface on the right side, like this:

n1  ( n2 ) n3 |  n1

You will end up with two different focal lengths, unless n1 = n3. This is expected.

The formula is a little complicated:

$$\frac{1}{f'}=\dfrac{n-n_b}{n_p\ r_1}-\dfrac{n-n_p}{n_p\ r_2}-\dfrac{(n-n_b)(n-n_p)}{n\cdot n_p}\cdot\frac{x}{r_1r_2}$$

and

$$f=-\frac{n_b}{n_p}f'$$

With $n_b$ = index before. $n_p$ = index after. $x$=thickness of the lens.

You can derive this using the dioptres' usual formulas.

• It's very rude to downvote withot explaining why. We all can learn and improve our answers. – FGSUZ Jun 18 '18 at 12:53