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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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View Full Document2
. 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 straightline 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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 Spring '11
 NA
 Physics

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