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inÑnitesimal vectors with the requirement that the angle they make shall increase,
not linearly, but as the sợuøre of the length of the curve. To construect that
curve involves slightly advanced mathematics, but we can always construct it by
actually drawing the arrows and measuring the angles. In any case, we get the
marvelous curve (called Cornu's spiral) shown in Eig. 30-8. Now how do we use
this curve?
Tf we want the intensity, let us say, at point , we add a lot of contributions
of diferent phases from point Ï2 on up to infnity, and from D down only to
point Ởp. So we start at p ¡n Fig. 30-8, and draw a series Of arrows OÝ ©Ver-
increasing angle. 'Pherefore the total contribution above point p all goes along
the spiraling curve. If we were to stop integrating at some place, then the total
amplitude would be a vector from 7? to that point; in this particular problem we
are going to infñnity, so the total answer is the vector p¿. Now the position
on the curve which corresponds to point p on the object depends upon where
point ? ¡s located, since point D, the inflection point, always corresponds to
the position of point . 'Thus, depending upon where ? is located above , the
beginning point will fall at various positions on the lower left part of the curve,
and the resultant vector p.., will have many maxima and minima (Fig. 30-9).
--- Trang 528 ---
S2
Fig. 30-8. The addition of amplitudes for many In-phase oscillators
whose phase delays vary as the square of the distance from point D of
the previous figure.
1.0 R
0.25F----=--—~z
Fig. 30-9. The Iintensity near the edge of a shadow. The geometrical
shadow edge Is at xo.
--- Trang 529 ---
Ôn the other hand, if we are at Q, on the other side of , then we are using
only one end of the spiral curve, and not the other end. In other words, we do not
even start at JD, but at Hạ, so on this side we get an intensity which continuously
falls of as Q goes farther into the shadow.
One point that we can immediately calculate with ease, to show that we really
understand it, is the intensity exactly opposite the edge. 'Phe intensity here 1s
1/4 that of the incident light. Reason: Pxactly at the edge (so the endpoint ÖØ of
the arrow is at D in Fig. 30-8) we have half the curve that we would have had iŸ
we were far into the bright region. If our point †## is far into the light we go from
one end of the curve to the other, that is, one full unit vector; but if we are at
the edge of the shadow, we have only half the amplitude——1/4 the intensity.
In this chapter we have been finding the intensity produced in various direc-
tỉons from various distributions of sources. As a fñnal example we shall derive
a formula which we shall need for the next chapter on the theory of the index
of refraction. p to this point relative intensities have been sufficient for our
purpose, but this time we shall fnd the complete formula for the field in the
following situation.
30-7 The field of a plane of oscillating charges
Suppose that we have a plane full of sources, all oscillating together, with
their motion in the plane and all having the same amplitude and phase. What is
the fñeld at a finite, but very large, distance away from the plane? (We cannot
get very close, of course, because we do not have the right formulas for the field
close to the sources.) IÝ we let the plane of the charges be the zz-plane, then
we want to fñnd the field at the point ? far out on the z-axis (Eig. 30-10). We
Oscillating charge
€' :
Sheet of oscillating charges
Fig. 30-10. Radiation field of a sheet of oscillating charges.
--- Trang 530 ---
suppose that there are ? charges per unit area of the plane, and that each one of
them has a charge g. All of the charges move with simple harmonic motion, with
the same direction, amplitude, and phase. We let the motion of each charge, ¿th
respect to is on querage postfion, be #øọ cos (‡. Ôr, using the complex notation
and remembering that the real part represents the actual motion, the motion
can be described by #zoe“t,
Now we fnd the fñield at the point from all of the charges by fñnding the
ñeld there rom each charge g, and then adding the contributions from all the
charges. We know that the radiation field is proportional to the acceleration of
the charge, which is —œ2zoe”“ (and is the same for every charge). The electric
fñeld that we want at the point due to a charge at the point @ is proportional
to the acceleration of the charge g, but we have to remember that the feld at the
point ?P at the instant £ is given by the acceleration of the charge at the earlier
time f“ = £— r/c, where r/e is the time i% takes the waves to travel the distance r
from @ to P. Therefore the field at is proportional to
— 02zgefe—r/6), (30.10)
Using this value for the acceleration as seen from in our formula for the electric
fñeld at large distances from a radiating charge, we get
Electric feld at ? q 2zge⁄2ữ—r/e)
lim charge at Q ) ¬. . Ặ—— (30.11)
Now this formula is not quite right, because we should have used øø‡ the
acceleration of the charge but ?s cormmponen‡ perpendicular to the line Q?P. We
shall suppose, however, that the point ? is so far away, compared with the
distance of the point Q from the axis (the distance ø in Eig. 30-10), for those
changes that we need to take into account, that we can leave out the cosine facbor
(which would be nearly equal to 1 anyway).
To get the total fñeld at , we now add the efects of all the charges in the
plane. We should, of course, make a øecfor sum. But since the direction of the
electric feld is nearly the same for all the charges, we may, in keeping with the
approximation we have already made, just add the magnitudes of the fñelds. 'lo
our approximation the field at depends only on the distance z, so all charges
at the same z produce equal fñelds. So we add, frst, the felds of those charges In
a ring of width đø and radius ø. hen, by taking the integral over all ø, we will
obtain the total fñeld.
--- Trang 531 ---
The number of charges in the ring is the product of the surface area of the
ring, 2ø đo, and ?, the number of charges per unit area. We have, then,
2 iœ(t—r/c)
Total ñeld at = J _—1 “I0 n.2mpdp, (30.12)
47coc2 T
We wish to evaluate this integral from ø = 0 to ø = œ. The variable ứ, of
course, is to be held fñxed while we do the integral, so the only varying quantities
are ø and r. Leaving out all the constant factors, ineluding the ƒactor e”*°t, for