Cerenkov Radiation

Cerenkov radiation is an electromagnetic shock wave caused by a charged particle moving with a velocity v which is faster than the velocity of light c/n in a medium with index of refraction n.

a) Show that the shock wave is emitted at an angle θ_c relative to the particle direction, where

b) Show that a spherical mirror with radius of curvature R will focus this shock wave onto a ring in the focal plane of the mirror.

c) Find the radius of the ring.

SOLUTION

a) At each point in the passage of the charged particle through the medium, a spherical wave is produced whose rate of travel is c/n, while the particle is travelling at a velocity v (see Figure 1.1a). The lines perpendicular to the wave front give the direction of propagation of the radiation and the angle θ_c, as required:

b) The spherical mirror and the cone of radiation produced by the charged particle are azimuthally symmetric, so we may solve the problem in two dimensions. We now must show that the parallel rays striking the mirror

Figure 1.1a

Figure 1.1b

will be focussed to a point on the “focal line” of the mirror. The focal length of a spherical mirror with radius of curvature R is R/2. Consider two rays incident on the mirror at an angle θ_c to the horizontal, one which passes through the center of the circle and another on the opposite side of the focus, which strikes the mirror a a distance x away as measured along the axis (see Figure 1.1b). In the paraxial approximation, we may use the standard relation between the image distance i, the object distance o, and the focal length f = R/2:

(1)

Checking for the ray leaving the center of the circle (object at 2f), we have

Obviously, a ray along a radius of the circle will be reflected back on itself. It passes through the point P along the focal line at a distance (R/2) tan θ_c from the axis. Now, a ray originating at the point E will strike the mirror at the point B and form a virtual image at point A on the other side of the mirror. Using (1) to find the image distance and thereby

(2)

(3)

where the length of the segment  is made positive (although the image distance is negative). To find the point where the reflected ray crosses the focal line, we will use similar triangles, ∆APF ~ ∆ABD. So

(4)

First, we find that and then in the same paraxial approximation (see Figure 1.1b)

(5)

so

(6)

c) Restoring the cylindrical symmetry to the problem, the point in the focal line becomes a circle in the focal plane of radius

مواضيع ذات صلة

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