Thin lens

The thin lens approximation ignores optical effects due to the thickness of lenses and simplifies ray tracing calculations. It is often combined with the paraxial approximation in techniques such as ray transfer matrix analysis.

Focal length

The focal length, f, of a lens in air is given by the lensmaker's equation:

${\displaystyle \frac{1}{f}=(n-1)\left[\frac{1}{R_1}-\frac{1}{R_2}+\frac{(n-1)d}{n{R}_1{R}_2}\right],}$

where n is the index of refraction of the lens material, R₁ and R₂ are the radii of curvature of the two surfaces, and d is the thickness of the lens. Here R₁ is taken to be positive if the first surface is convex, and negative if the surface is concave. The signs are reversed for the back surface of the lens: R₂ is positive if the surface is concave, and negative if it is convex. This is an arbitrary sign convention; some authors choose different signs for the radii, which changes the equation for the focal length.

For a thin lens, d is much smaller than one of the radii of curvature (either R₁ or R₂). In these conditions, the last term of the Lensmaker's equation becomes negligible, and the focal length of a thin lens in air can be approximated by

${\displaystyle \frac{1}{f}\approx \left(n-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right].}$

Derivation using Snell's law

Consider a thin lens with a first surface of radius $R$ and a flat rear surface, made of material with index of refraction $n$.

Applying Snell's law, light entering the first surface is refracted according to ${\displaystyle \sin i=n\sin r_1}$, where ${\displaystyle i}$ is the angle of incidence on the interface and ${\displaystyle r_1}$ is the angle of refraction.

For the second surface, ${\displaystyle n\sin r_2=\sin e}$, where ${\displaystyle r_2}$ is the angle of incidence and ${\displaystyle e}$ is the angle of refraction.

For small angles, $\sin x\approx x$. The geometry of the problem then gives:

$${\displaystyle \begin{array}{cc}e& \approx n{r}_2\\ & =n(i-r_1)\\ & \approx n(i-\frac{i}{n})\end{array}}$$

If the incoming ray is parallel to the optical axis and distance $h$ from it, then $${\displaystyle \sin i=\frac{h}{R}⟹i\approx \frac{h}{R}.}$$

Substituting into the expression above, one gets $${\displaystyle e\approx \frac{h}{R}(n-1).}$$

This ray crosses the optical axis at distance ${\displaystyle f}$, given by $${\displaystyle \tan e=\frac{h}{f}⟹e\approx \frac{h}{f}}$$

Combining the two expressions gives $\frac{1}{f}=\frac{1}{R}(n-1)$.

It can be shown that if two such lenses of radii $R_1$ and $-R_2$ are placed close together, the inverses of the focal lengths can be added up giving the thin lens formula:

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