Conflicting lensmaker's equation representation

Question

I keep finding these two conflicting forms of the lensmaker's equation. I understand that (n2-n1)/n1 is the same as (n-1) if referring to a lens in air, but I can not make sense of why the reciprical of R2 is subtracted in one and added in the other. The second equation makes more sense to me, since if the radii of curvature were equal in both sides than the top one would evaluate to 0 which isnt true about lenses. So is the top one just wrong?

$\frac{1}{f} = \left( \frac{n_2 - n_1}{n_1} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$

I did ask generative AI and it hinted at sign conventions for concave or convex interfaces but I didn't fully understand.

Thanks!

Answer 1

Yes, the second one assumes the lens is in air or vacuum (i.e. $n_1 = 1$), but that has nothing to do with the fact that $1/R_2$ has opposite signs in these equations. Neither equation is wrong, they just use different conventions in defining the radii of curvature. First of all, the usual assumption is that the radius of curvature is positive if the lens surface is convex, and negative if it is concave. The question is, are we looking at the surface from the outside, or the inside?

Imagine a biconvex lens, light is incident on it from the left. The left surface of the lens, as seen by this light, has $R_1 > 0$. The right surface is concave according to the light indident on it from the left (i.e. from inside the lens), so $R_2 < 0$. The first equation uses this convention, so $1/R_1 - 1/R_2$ is the sum of the positive numbers.

Now take the same biconvex lens. Look at the left surface from the outside: it is convex, so $R_1 > 0$. Look at the right surface from the outside. It is also convex, so $R_2 > 0$. This is the convention your second equation uses, so again $1/R_1 + 1/R_2$ is the sum of two positive numbers.

Answer 2

$R > 0$ if the surface curves to the left and $< 0$ if it curves to the right.

For a biconvex lens, $R_1 > 0$ and $R_2 < 0$. You find that $1/f > 0$ as expected.

For a biconcave lens, $R_1 < 0$, $R_2 > 0$, and $1/f > 0$.

If both surfaces curve to the left with the same radius, $R_1 = R_2$, and $1/f =0$. To a first approximation, such a lens is a bent window, and like a window it does not focus light. In better approximations, the thickness makes a difference and there are aberrations. So such a lens has some small effect.