Blog: How can focal length be equal to radius of curvature?

Thursday, July 12, 2012

How can focal length be equal to radius of curvature?

Hello!

It is not clear whether you speak about a (spherical) mirror or about a lens. Let's consider both options.

1. For a spherical mirror with the radius of curvature R the focal length is equal (up to a sign) to the half of $\displaystyle{R},$ i.e. $\displaystyle\frac{{R}}{{2}}.$ Therefore the radius of curvature and the focal length cannot be equal.

2. There are more degrees of freedom for a lens with spherical surfaces: the radius of curvature of the second lens' side and the index of refraction (which depends on the lens' material).

Lensmaker's equation says that for a thin double convex lens in air

$\displaystyle\frac{{1}}{{{f}}}\approx{\left({n}-{1}\right)}\cdot{\left(\frac{{1}}{{R}_{{1}}}+\frac{{1}}{{R}_{{2}}}\right)},$

where $\displaystyle{n}\gt{1}$ is the refractive index of a material.

(here $\displaystyle{R}_{{1}}$ and $\displaystyle{R}_{{2}}$ are considered positive to not think about the sign convention).

If $\displaystyle{f{=}}{R}_{{1}}$ as you ask then $\displaystyle\frac{{1}}{{{R}_{{1}}}}\cdot{\left({2}-{n}\right)}=\frac{{1}}{{R}_{{2}}},$

so for $\displaystyle{1}\lt{n}\lt{2}$ it is possible for $\displaystyle{R}_{{2}}=\frac{{R}_{{1}}}{{{2}-{n}}}.$

One can also consider non-thin lens.

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Thursday, July 12, 2012

How can focal length be equal to radius of curvature?

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