Consider a ray parallel to principal axis of a concave mirror of small aperture incident at M. The ray reflected along MF as shown. Let θ is the angle of incidences. As the aperture is small, angle θ is small and are PM is considered as straight segment.
From the Δle MCP
tan θ = \(\frac{PM}{PC}\)
tan θ = \(\frac{PM}{R}\)
since θ is small tan = θ
∴ θ = \(\frac{PM}{R}\)……. (1)
From the Δle PFM
tan 2θ = \(\frac{PM}{PF}\)
tan 2θ = \(\frac{PM}{F}\)
since θ is small tan 2θ = 2θ
∴ 2θ = \(\frac{PM}{F}\)
θ = \(\frac{PM}{2F}\)……..(2)
From (1) and (2)
\(\frac{PM}{F}\) = \(\frac{PM}{2F}\)
R = 2f
f = \(\frac{R}{2}\)
