InterviewSolution
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InterviewSolution
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To prove focal length is half of radius of curvature. |
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Answer» Focal length and radius of curvature:\tLet F be focus and C be center of curvature of a concave mirror.\tConsider an incident ray travelling parallel to principle axis and meet at point N.\tIt passes through focus after reflection.\tCN is the normal drawn to the mirror(normals drawn to mirror meet at C).\tNow, PC = R PF = f ∠FNC = ∠FCN =\xa0 ∠MFN = 2\xa0\tIn ΔMNC [For paraxial rays, M and P coincide]\tIf\xa0\xa0is small,\xa0 ------------------------(1)\xa0\tIn ΔMNF [For paraxial rays, M and P coincide]\tIf\xa0\xa0is small,\xa0 ------------------------(2)\tSubstituting (1) in (2), we get 2PF = PC R = 2f f = R/2\tTherefore, focal length is half the radius of curvature.\xa0\xa0In fig, theta is taken instead of alpha.\xa0\xa0 Taking a concave mirror, the\xa0curved mirror will have a principal axis near which a ray of light is incident\xa0on the mirror parellel to it.Now the angle between\xa0the radius of curvature and principal axis will be equal to the angle at which ray is incident\xa0and due to reflections law incident angle would be equal to angle cbf. L abc = L cbf (law of reflection) L abc = L fcb ( alternate)Hence,\xa0cbf = fcb ⇒ bf = fcSince b very close to\xa0p; pf = fc ⇒ pc = 2 pfNow considering convex mirror, similar to concave mirror\xa0L abn = L nbd\xa0{reflection law}L abn = Lbcf {corresponding}Hence,\xa0bf = fcSince, b is very close to\xa0p, pf = fc ⇒ pc = 2pfHence, in both cases\xa0Radius is double the focal length. |
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