Prove that R=2f

Asked by cool_mohit981994 | 14th Apr, 2009, 08:01: PM

Expert Answer:

You can't actually prove it. And it never is (practically). It's just close enough for physics.

But , an approximate derivation could be done.

A ray of light passing parallel to the principal axis and close to the principal axis after incident at a point Q on the mirror is reflected according to the law of reflection ( i = r ) and passes through the focus F.

If O is the center of curvature of the mirror, consider the triangle OQF.

Angle OQf = r the angle of reflection.

Angle QOf = i the angle of incidence (the line OQ passes through two parallel lines). And therefore the Angle Angle QOf = i

As i = r the triangle is a isosceles tri angle and hence QF = OF

If P is the pole of the mirror, considering large radius of curvature we can take PF =QF and hence OF = PF or half the radius of curvature.

Answered by  | 23rd Apr, 2009, 02:35: PM

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