Prove any ray parallel to the principal axis after refraction will pass through the focus
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C is the centre of curvature,F is the principal focus,P is the pole,focal length f=CP,radius of curvature R=2f
Now angle r = angle i(angle of incidence = angle of reflection) = angle BCF (alternate interior angles formed by parallel lines and a transversal)
So in triangle BCF, BF=CF
But CF=FP,so BF = FP.
But this is a contradiction.
Because the radius is the one and only point inside a circle which is equidistant from 2 distinct points on the circle(B and P)
But here the radius is CP not FP.
Now angle r = angle i(angle of incidence = angle of reflection) = angle BCF (alternate interior angles formed by parallel lines and a transversal)
So in triangle BCF, BF=CF
But CF=FP,so BF = FP.
But this is a contradiction.
Because the radius is the one and only point inside a circle which is equidistant from 2 distinct points on the circle(B and P)
But here the radius is CP not FP.
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