Case of Concave Mirror
Consider a ray AB parallel to principal axis incident on a concave mirror at point B and is reflected along BF. The line CB is a normal to the mirror as shown in Fig.
Let θ be angle of incidence, ∠ABC.
∴ ∠ABC = ∠BCP = ∠θ (Alt. ∠S)
Draw BD ⊥ CP.
In right angled ∆ BCD,
tan θ = \(\frac{BD}{CD}\) ............(i)
In right angled ∆ BFD,
tan 2θ = \(\frac{BD}{FD}\) ............(ii)
Dividing (i) and (ii), we get
\(\frac{tan\ 2\ \theta}{tan\ \theta} = \frac{CD}{FD}\) ..........(iii)
If θ is very small, then
tan θ = θ and tan 2θ = 2θ
Since the aperture of the mirror is small, therefore, point B lies very close to P.
∴ CD ≈ CP and FD ≈ FP.
From (iii), \(\frac{2\ \theta}{\theta} = \frac{PC}{PF} = \frac{R}{f}\)
R = 2f
