Optical power of the lens depends on the radius of curvature R 1 and R 2 of its spherical surface and the refractive index n of the material from which the lens is made by expressed by the formula $(\frac{1}{r1} + \frac{1}{r2} ) (n-1) $
the question is what's the difference between the two radii, aren't they one?
[Physics] the power of a lens and radius of curvature relation
lensesoptics
Best Answer
According to the lens maker's formula,
$\frac{1}{f}=(\frac{n_2}{n_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$,
Here, $n_2$ is the refractive index of the lens and $n_1$ is the refractive index of the medium from which the light is incident on the lens. The medium surrounding the lens must be the same for this formula to work.
$R_1$ and $R_2$ are the radius of curvature of the curved surfaces of the lens from which the light enter the lens and emerges out of the lens respectively.
Sign Conventions play an important role here.
For a convex lens, $R_1$ is +ve (if light is incident from the left) and $R_2$ is -ve. In the above image, as you can see, the radius of the curvature of the convex lens' left curved surface $(R_1)$ lies in the direction of the incident ray, so it is +ve and the radius of curvature of the other surface $(R_2)$ points the direction opposite to the direction of incident ray, so its -ve.
So after applying the sign conventions for a convex lens, as explained above, we get the following formula,
$\frac{1}{f}=(\frac{n_2}{n_1}-1)(\frac{1}{R_1}+\frac{1}{R_2})$