![]() The heat generated at that point caused a fire to ignite and destroy the entire Roman fleet. Supposedly, he tilted the mirror toward the sun in such a way that all the sun's rays when reflected off the mirror went through the focal point. There is a legend about Archimedes, in which he is said to have used a parabolic mirror to defeat the Roman General Marcellus. Now the ray reflects from OP1 at angle r (from laws of reflection i r) and reaches. This will be true for any point \(A\) on the parabola. Ray S will reflect on mirror OP1 at point O and with an angle i. Like the parallelogram, it also has rotational. Suppose the parabola has focus \(F\) and directrix \(\mathscr\). A rectangle has reflectional symmetry when reflected over the line through the midpoints of its opposite sides. A parabolic mirror is one whose surface is generated by rotating a parabola about its axis. In this case, the angles are measured with respect to the tangent line, and the same argument used by Heron shows that if \(C\) is such that the angle of incidence equals the angle of reflection, then \(|AC| |BC|\) is minimized.Īn interesting application of the Law of Reflection arises in the case of a light beam sent toward a parabolic mirror, where the light beam is parallel to the axis of the parabola. We may replace the line in Heron's problem by any concave curve (a curve is concave if it lies entirely on one side of any tangent line). In fact, the mirror in the Equal Angle Law of Reflection need not be flat. This as we have found, is true when the value of \(|AC| |BC|\) is at its lowest: so light always takes the shortest path! It was Euclid who, over three hundred years earlier, had noted the now well-known Reflection Law for light:Įuclid's Law of Reflection states that if a beam of light is sent toward a mirror, then the angle of incidence equals the angle of reflection. The situation at almost-glancing angle is that the incoming and reflected light waves almost perfectly cancel each other out (opposite phase, almost-equal magnitude), leaving almost no fields on one side of the boundary and since theres almost no transmitted light, theres almost no fields on the other side of the boundary too. ![]() This is the equal angle law of reflection. In addition \(m∠2=m∠3\) because \(C\) is at an equal angle to \(B\) and \(B'\), and \(m∠1=m∠3\) since these are vertically opposite angles. Any other position of \(C\) will increase \(|AC| |BC|\). The shortest path from \(A\) to \(B′\) is a straight line, so the point \(C\) that minimizes \(|AC| |CB′|\) should be on the line \(AB'\). He noticed that if \(B\) is reflected across the line, to some point\(B′\), then for any point \(C\) on the line, \(|CB|=|CB′|\), and hence minimizing \(|AC| |CB|\) is equivalent to minimizing \(|AC| |CB′|\). However, here is the mathematical approach used by Heron. You can explore and try to find the minimum distance yourself. Given two points A and B on one side of a line, find C a point on the straight line, that minimizes AC BC.Ĭlick here for a visual of the problem. A classical problem in mathematics is Heron's Shortest Distance Problem: Symmetry is frequently used in solving problems involving shortest paths.
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