Theory_2_Solution - 39th International Physics Olympiad...

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Solution 1. A θ C B D E D’ Figure 1 Let us consider a plane containing the particle trajectory. At , the particle position is at point A. It reaches point B at 0 t = 1 tt = . According to the Huygens principle, at moment , the radiation emitted at A reaches the circle with a radius equal to AD and the one emitted at C reaches the circle of radius CE. The radii of the spheres are proportional to the distance of their centre to B: 1 0 << ( ) () 1 1 CE 1 const CB / ct t n n β == = v The spheres are therefore transformed into each other by homothety of vertex B and their envelope is the cone of summit B and half aperture 1 2 Arcsin n π ϕ = =− , where is the angle made by the light ray CE with the particle trajectory. 1.1. The intersection of the wave front with the plane is two straight lines, BD and BD'. 1.2. They make an angle 1 Arcsin n = with the particle trajectory.
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2 . The construction for finding the ring image of the particles beam is taken in the plane containing the trajectory of the particle and the optical axis of the mirror. We adopt the notations: S – the point where the beam crosses the spherical mirror F – the focus of the spherical mirror C – the center of the spherical mirror IS – the straight-line trajectory of the charged particle making a small angle α with the optical axis of the mirror. I θ C F O M N S A P Q Figure 2 CF = FS = f CO//IS CM//AP CN//AQ n FCO = FO f n n MCO OCN == MO f = × We draw a straight line parallel to IS passing through the center C. The line intersects
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This note was uploaded on 11/08/2011 for the course PHYS 0000 taught by Professor Na during the Spring '11 term at Rensselaer Polytechnic Institute.

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Theory_2_Solution - 39th International Physics Olympiad...

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