# How to find focal length of a single convex lens?

I need help finding the focal length of a single convex lens. The radius of curvature is 200mm. left side is air and the glass has a index of 1.5. I search on google but there was only formulas for think lens, convex and converge lends. I need the formula for a single convex lens. Any idea? I know the ray transfer matrix.

$$\left(\begin{matrix} 1 & 0 \\ \frac{n-n'}{nR} & \frac{n'}{n} \end{matrix}\right)$$

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I formatted the matrix with standard TeX. Are you sure that's an $n'$ in the denominator of the lower left entry? I don't know too much about this formalism, but the Wikipedia table has something very similar but with an $n$ there. – Chris White Sep 12 '12 at 23:18
Thanks a lot for formatting it. You are also correct, it should be a n in the lower left. Thank you. – user101699 Sep 12 '12 at 23:25
If this is homework, please add the homework tag. – Ben Crowell Sep 13 '12 at 1:34
Doesn't en.wikipedia.org/wiki/Lens_(optics)#Lensmaker.27s_equation give you the equation you need? – John Rennie Sep 13 '12 at 9:14

A lens has two surfaces. One of these surfaces still has refracting power, due to the difference in refractive index and angle of incidence.

## Refraction on one surface

Let parallel rays start in glas. Radius sign convention results $R=-200\,mm$ since incident light is on same side as center of curvature.

Radius of curvature and the two media influence the focal point. For your rays propagate from right to left (air) the FFL (front focal length, as depicted) is $$f=R\cdot \frac{n_1}{n_1-n_2}=R\cdot \frac{1}{1-n_2}$$ This is the reciprocal element of your transfer matrix. Focal length is proportional to radius of curvature. Increasing refractive index $n_2$ of glas shortens it. $$f=-200\,mm \cdot \frac{1}{-0.5}=400\,mm$$

## Remark

The cited above lensmaker equation with $R_1=\infty$ will return the same result. Since on-axis parallel rays are not refracted on plane surface, as can be seen from Snell's law.

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