# Show that the principal focus of a curve mirror is related to the center of curvature by this relation  f=r/2

Arun
25757 Points
5 years ago
 In below figure1 and 2, a ray of light BP' travelling parallel to the principal axis PC is incident on a spherical mirror PP'. After reflection, it goes along P'R, through the focus F P is the pole and F is the focus of the mirror. The distance PF is equal to the focal length f. C is the centre of curvature. The distance PC is equal to the radius of curvature R of the mirror. P'C is the normal to the mirror at the point of incidence P'. For a concave mirror:In above figure,
 ∠BP'C = ∠P'CF (alternate angles) and ∠BP'C = ∠P'F (law of reflection,∠i=∠r) Hence ∠P'CF = ∠CP'F ∴ FP'C is isosceles. Hence, P'F = FC
 If the aperture of the mirror is small, the point P' is very close to the point P,
 then P'F = PF ∴ PF = FC = 1/2 PC or f = 1/2 R
 For a convex mirror: In above figure,
 ∠BP'N = FC∠P' (corresponding angles) ∠>BP'N = ∠NP'R (law of reflection, ∠i=∠r) and ∠NP'R = ∠CP'F (vertically opposite angles) Hence ∠FCP' = ∠CP'F ∴  FP'C is isosceles. Hence, P'F = FC
 If the aperture of the mirror is small, the point P' is very close to the point P.
 Then P'F = PF ∴PF = FC = 1/2 PC or f = 1/2 R
 Thus, for a spherical mirror {both concave and convex), the focal length is half of its radius of curvature.