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