Derive the relationship between the focal length $(f)$ and the radius of curvature $(R)$ for a spherical mirror.

(N/A) Let $C$ be the centre of curvature and $P$ be the pole of the spherical mirror.
Consider a light ray parallel to the principal axis striking the mirror at point $M$. The line $CM$ is normal to the mirror at $M$.
Let $\theta$ be the angle of incidence. According to the law of reflection,the angle of reflection is also $\theta$.
From the geometry,$\angle MCP = \theta$ (alternate interior angles) and $\angle MFP = 2\theta$ (exterior angle of $\triangle MCF$ is equal to the sum of opposite interior angles).
Let $MD$ be the perpendicular from $M$ to the principal axis. For paraxial rays,$D$ is very close to $P$,so $CD \approx CP = R$ and $FD \approx FP = f$.
In $\triangle MDC$,$\tan \theta = \frac{MD}{CD} \approx \frac{MD}{R}$. For small $\theta$,$\tan \theta \approx \theta$,so $\theta \approx \frac{MD}{R}$.
In $\triangle MDF$,$\tan 2\theta = \frac{MD}{FD} \approx \frac{MD}{f}$. For small $\theta$,$\tan 2\theta \approx 2\theta$,so $2\theta \approx \frac{MD}{f}$.
Substituting $\theta$ from the first equation into the second: $2(\frac{MD}{R}) = \frac{MD}{f}$.
Therefore,$R = 2f$ or $f = \frac{R}{2}$.