The relationship between focallenght and radius of curvature is
Answers
Answered by
0
The focal length of a spherical mirror is equal to half of its radius of curvature
{f = 1/2 R}.Proof: 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
{f = 1/2 R}.Proof: 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
Anonymous:
mark it as the brainliest
Similar questions