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

Mary , 6 Years ago
Grade 6
anser 1 Answers
Arun

Last Activity: 6 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.

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