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frequency, or slightly diferent wavelength, how close together in wavelength could |
they be such that the grating would be unable to tell that there were really |
two diferent wavelengths there? 'The red and the blue were clearly separated. |
But when one wave ¡is red and the other is slightly redder, very close, how close |
can they be? Thịis is called the resolưing pouer of the grating, and one way of |
analyzing the problem is as follows. Suppose that for light of a certain color |
we happen to have the maximum of the difracted beam occurring at a certain |
angle. TÝ we vary the wavelength the phase 2zđdsin Ø/À is different, so oÝ course |
the maximum occurs at a different angle. 'Phat is why the red and blue are |
spread out. How different in angle must i% be in order for us to be able to see |
it? Tƒ the two maxima are exactly on top of each other, of course we cannot see |
--- Trang 522 --- |
them. Tf the maximum oŸ one is far enough away from the other, then we can see |
that there ¡is a double bump in the distribution of light. In order to be able to |
Jjust make out the double bump, the following simple criterion, called iayleigh s |
criterion, is usually used (Eig. 30-6). It is that the first minimum from one bump |
should sit at the maximum of the other. Now it is very easy to calculate, when |
one minimum sits on the other maximum, how much the diference in wavelength |
is. The best way to do it is geometrically. |
"~. —— h ` —— - |
Fig. 30-6. lllustration of the Rayleigh criterion. The maximum of one |
pattern falls on the first minimum of the other. |
In order to have a maximum for wavelength ÀJ, the distance A (Eig. 30-3) |
must be A7, and if we are looking at the znth-order beam, it is nA/. In other |
words, 2zdsinØ/ÀX' = 2m, so ndsin 9, which is A, is rmwÀ! tìmes m, or wwÀ/. For |
the other beam, of wavelength À, we want to have a mứnữmum at thĩs angle. |
That is, we want A to be exactly one wavelength À more than mnA. That is, |
A = mnÀ + À = mmnÀ'. Thus iŸ Ä' = À + A^A, we find |
AA/A= 1/mn. (30.9) |
The ratio À/AA is called the resolung pouer oŸ a grating; we see that it is equal |
to the total number of lines in the grating, times the order. lt is not hard to |
prove that this formula is equivalent to the formula that the error in Ífrequency 1s |
cqual to the reciprocal time diference between extreme paths that are allowed |
to interfere: |
Aw = 1/T. |
In fact, that is the best way to remember i%, because the general formula works |
not only for gratings, but for any other instrument whatsoever, while the special |
formula (30.9) depends on the fact that we are using a grating. |
* In our case 7'= A/c = mnÀ/c, where c is the speed of light. The frequency = c/À, so |
Au=eAA/A2. |
--- Trang 523 --- |
30-4 The parabolic antenna |
Now let us consider another problem in resolving power. 'Phis has to do |
with the antenna of a radio telescope, used for determining the position of radio |
sources in the sky, i.e., how large they are in angle. Of course if we use any old |
antenna and fnd signals, we would not know tom what direction they came. VWWe |
are very interested to know whether the source is in one place or another. Ône |
way we can find out is to lay out a whole series of equally spaced dipole wires on |
the Australian landscape. Then we take all the wires from these antennas and |
feed them into the same receiver, in such a way that all the delays in the feed |
lines are equal. Thus the receiver receives signals from all of the dipoles in phase. |
That is, it adds all the waves from every one of the dipoles in the same phase. |
Now what happens? If the source is directly above the array, at inÑnity or nearly |
so, then its radiowaves will excite all the antennas in the same phase, so they all |
feed the receiver together. |
Now suppose that the radio source is at a slight angle Ø from the vertical. |
Then the various antennas are receiving signals a little out of phase. The receiver |
adds all these out-of-phase signals together, and so we get nothing, if the angle Ø |
is too big. How bịg may the angle be? Ansuer: we get zero If the angle A/Ù = 9 |
(Fig. 30-3) corresponds to a 360° phase shift, that is, if A is the wavelength À. |
'This is because the vector contributions form together a complete polygon with |
zero resultant. The smallest angle that can be resolved by an antenna array of |
length Ù is Ø = À/L. Notice that the receiving pattern oŸ an antenna such as |
this is exactly the same as the intensity distribution we would get if we turned |
the receiver around and made it into a transmitter. 'This is an example of |
what is called a reciprocitU principle. It turns out, in fact, to be generally true |
for any arrangement of antennas, angles, and so on, that if we frst work out |
what the relative intensities would be in various directions if the receiver were a |
transmitter instead, then the relative directional sensitivity of a receiver with |
the same external wiring, the same array of antennas, is the same as the relative |
intensity of emission would be if it were a transmitter. |
Some radio antennas are made in a diferent way. Instead of having a whole |
lot of dipoles in a long line, with a lot of feed wires, we may arrange them not in |
a line but in a curve, and put the receiver at a certain point where it can detect |
the scattered waves. This curve is cleverly designed so that if the radiowaves |
are coming down from above, and the wires scatter, making a new wave, the |
wires are so arranged that the scattered waves reach the receiver all at the same |
--- Trang 524 --- |
time (Eig. 26-12). In other words, the curve is a øaraboïa, and when the source |
is exactly on is axis, we get a very strong intensity at the focus. In this case we |
understand very clearly what the resolving power of such an instrument is. The |
arranging of the antennas on a parabolie curve is not an essential point. It is only |
a convenient way to get all the signals to the same point with no relative delay |
and without feed wires. The angle such an instrument can resolve is still Ø = À/1, |
where Ù is the separation of the first and last antennas. I$ does not depend on |
the spacing of the antennas and they may be very close together or in fact be |
all one piece of metal. NÑow we are describing a telescope mirror, of course. We |
have found the resolving power of a telescopel (Sometimes the resolving power is |
written Ø = 1.22À/L, where Ƒ is the diameter of the telescope. The reason that it |
is not exactly ÀA/ is this: when we worked out that Ø = À/Ù, we assumed that all |
the lines of dipoles were equal in strength, but when we have a circular telescope, |
which is the way we usually arrange a telescope, not as much signal comes from the |
outside edges, because it is not like a square, where we get the same intensity all |
along a side. We get somewhat less because we are using only part of the telescope |
there; thus we can appreciate that the efective diameter ¡s a little shorter than |
the true diameter, and that is what the 1.22 factor tells us. In any case, it seems |
a little pedantic to put such precision into the resolving power formula.*) |
30-5 Colored fÌms; crystals |
The above, then, are some of the efects of interference obtained by adding |
the various waves. But there are a number of other examples, and even though |
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