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What happens if the radii of curvature of the two spherical surfaces of a thin lens are equal?

According to the formula,

$$1/f = (n - 1)(1/R1 - 1/R2) $$

(Where $n$ is the refractive index of the material of glass with respect to air),

$f$ turns out to be infinity! Is it true? Or am I wrong?

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There is a sign convention for the radii $R_1$ and $R_2$ used in the equation. This convention is described in the Wikipedia article on the lensmaker's equation:

The signs of the lens' radii of curvature indicate whether the corresponding surfaces are convex or concave. The sign convention used to represent this varies, but in this article a positive $R$ indicates a surface's center of curvature is further along in the direction of the ray travel (right, in the accompanying diagrams), while negative $R$ means that rays reaching the surface have already passed the center of curvature. Consequently, for external lens surfaces as diagrammed above, $R_1 > 0$ and $R_2 < 0$ indicate convex surfaces (used to converge light in a positive lens), while $R_1 < 0$ and $R_2 > 0$ indicate concave surfaces.

So in your example of $|R_1| = |R_2| = R$ the equation for a biconvex lens would be:

$$ \frac{1}{f} = (n - 1)\left( \frac{1}{R} - \frac{1}{-R}\right) =\frac{2(n-1)}{R} $$

For a biconcave lens it would be:

$$ \frac{1}{f} = (n - 1)\left( \frac{1}{-R} - \frac{1}{R}\right) =\frac{-2(n-1)}{R} $$

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  • $\begingroup$ Oh, my folly. I didn't pay attention to the sign convention of the two radii. Thanks a lot! $\endgroup$ – dOWNSIDE uP Dec 30 '17 at 8:37

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