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The arc Am B is the least-time path; the light from A does not focus on this arc. By contrast, the arc An B is of extremal, but not shortest time; this property goes together with the existence of a focusing point on this arc. Observing Fermat’s principle in action. 20. 16 This explains why the concave lens defocuses, while the convex lens focuses. 21 shows a ray from the bottom of a pool to the eye. The path shown in the figure is quicker than the straight path, since it “pays” to shorten the “expensive” underwater part where light is slower.

Thus the optimal point P is the tangency point between the line MN and a circle passing through A and B. There is precisely one such circle. The circle is easy to construct explicitly: its center is at the intersection of the perpendicular bisector of AB with vertical line through P. √ The best distance is the geometric mean of the heights, d = ab, where a and b are the heights of A and B. 17, we have OBD d = PQ = OD = O B 2 − B D2, where O B = O P = DQ = a+b 2 51 MINIMA AND MAXIMA and BD = b−a .

8) Using T1 = T2 we conclude: x = d; this solves the problem. A very compact answer! ♦ Just for comparison with a physical solution, here is a conventional one. A conventional solution. The combined perimeter is fixed: 4x + π d = L , so that d= L − 4x . 9) 43 MINIMA AND MAXIMA The combined area is therefore A(x) = x 2 + π d2 π = x2 + 4 4 L − 4x π 2 . Differentiating by x, we get A (x) = 2x + π ·2· 4 L − 4x π −4 π = 2x − 2 L − 4x . π From A (x) = 0 we obtain x− L − 4x = 0, π or x= L . 9) and simplifying we obtain d= In particular, x = d.

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Absurdity of the calculation of the intensity of gravitational radiation in the general theory of relativity


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