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Unformatted text preview: PHYS 302: Unit 13 Viewing Notes – Athabasca University PHYS 302: Vibrations and Waves – Unit 13 Viewing Notes
The lecture starts with a review of twoslit interference, between two neighbouring sources of interference among
thousands.
A review of high school geometry is needed to get the resultant in adding two general vectors of the same length A.
To see that the long side is 2 A cos 1 in length if the angle shown is δ (this would be the angle between the
2
vectors adding as the short sides), consider the large triangle as two smaller (identical) triangles with a common side
at the point where they join. Since δ plus the interior angle at the point where the triangles join add to 180º, and the
sum of all the angles in a triangle is also 180º, the two identical end angles must each be δ/2. These end angles are
the base angles in the small right angle triangles. The small triangles in turn have hypotenuse length A, so their
bases are A cos 1 long. Since there two of them, the total length is 2 A cos 1 (4:00). Although this description
2
2
is wordy, you can easily verify it with a sketch. Given that it is fundamental to what is done later, it is worth
visualizing clearly. Dr. Lewin refers to “Evectors” where E is the electric field, but the discussion could apply to
any wave variable. The sketch given in the lecture (completed at 5:40) shows the addition of three vectors, but the
results generalize to N vectors, and N could be very large. Each vector is at an angle δ to its neighbours, and these
angles are also the angles inside the circle, summing to Nδ as the total angle. The remaining angle to the vertical is
πNδ, and the halfangle in each wedge is δ/2. The radius of each wedge is R, and its halfwidth is A/2, so for a half
wedge we get sin 2 1
A
A
; that is, R 2
(8:00). The overall displacement from origin to end of the summed
sin R
2 1
2 vectors is OP, and we can use the identity above in triangle COP, which has two identical sides R and the external
angle πNδ. This gives OP 2 R cos 2N 2 R sin N2 . Substituting for R we get OP A sin
sin (9:30). 2 N
2 Even though some unknown quantities (such as R) were used in the derivation, the result features only known
quantities. The resulting length OP is the amplitude of the Efield and we need to square the Efield to get the
2 sin N2 , where I0 is the intensity that would arise from a single slit
intensity of light. So, we find I ( ) I 0 sin 2 (11:00). The functional form sin x
x is sometimes called a sinc function, and at 0 angle the 0/0 form reduces to N2, so
2 2 sin 2 2sin cos 2
2
2
2
Imax=N I0. For N=2, I ( ) I 0 I0 4 I 0 cos 2 , where a double angle formula sin 2 sin 2 2 was used. This result is already derived for twoslit interference (12:30). To plot this, it is preferable to use θ (the
d
angle in the lab) rather than the phase difference δ, but of course these are related by 2 sin . For this
reason, the plot uses sinθ and is labelled at λ/d intervals. The overall pattern is of principal maxima that are quite
high, with zeroes and secondary maxima in between. In addition, the spacing of the slits is inversely related to the
number of slits: the more slits, the better the spectral resolution (23:30). The increase in intensity at the maxima is
compensated by narrowing so that energy is conserved (24:30).
There are several ways to get minima (27:30). As a demonstration, Dr. Lewin shines a 3 mm laser beam on a grating
with spacing 1.9×106 m, so that about 1600 grating slits are within the beam. The maxima of order n are at
sin n nd , so in this case the angles should be 0º, 16.3º, 34º, and 57º for the 0 through 3rd (highest possible)
orders for green laser light of wavelength 532 nm (31:00). Dr. Lewin follows this with some grating demos,
including a reflection grating in which the slits are ruled on a reflective surface (37:00). The many orders are visible,
each corresponding to a peak in the diffraction pattern we saw before, but each shows many colours that correspond
to the colours in white light. When a monochromatic (red) source is used, one gets only those peaks expected for red
(38:50). Gratings are used to see the spectra of atoms, such as He (helium). The various wavelengths present spread
out in angles (41:30) by the grating (note: the example in the video lecture does not line up the way one would see it
visually). Ne (neon) has many lines in its spectrum (42:20), due to its complex structure. 1 PHYS 302: Unit 13 Viewing Notes – Athabasca University Break
Using Huygens’ principle, a single opening should interfere with itself. An artificial distinction is drawn between
“interference” from multiple slits and “diffraction” from a single slit, but these are really the same phenomenon. In
the lecture, Dr. Lewin refers to interference of one edge of the slit with the centre of the slit. In this case, there is
destructive interference if ½ Dsinθ = ½ λ. This approach is clever, because if the upper edge interferes with the
centre, then similar pairs of points can be found as one moves down, so the argument applies to the whole slit. The
angle β is defined as the phase difference corresponding generally to the halfwidth of the slit, 2 1 D sin D sin (47:15). Bekefi and Barrett 8.7 (47:35) is like Eq. 827 in the French text, which is
2
based on Eq. 825 (p. 291). The intensity is I ( ) I 0 sin 2 . Drafting a table will facilitate making a sketch. When sinθ=0, then β=0, sinβ=0, and I is I0. The last result can come from l’Hopital’s rule, but another way to get it
is simply to put in the expression for the expansion of sinβ as 1
1 3! 3 5! 5 ... and make the high order terms disappear in the limit as β→0.
It is expected that the middle of the slit should reveal the most light. If sinθ=λ/D, then β=π, sinβ=0, and I is 0 (as
expected from the beginning of the discussion). A similar thing happens when sinθ=2λ/D, and this happens at
infinitely many other points. Making the sketch as a function of sinθ, note that the width of the central maximum is
twice that of the other maxima, and its halfwidth in angle is about λ/D. For 600 nm (red) light, with D=0.1 mm and
viewed at L=3 m, λ/D=6×103 radians, the width at the screen will be 2 cm, which is 200 times wider than the slit. In
fact, the smaller the slit, the wider the pattern. The first minimaximum, at 1.5 λ/D, has β=1.5π, sinβ=1, and
I=0.045I0. The slide before the demonstration shows the wavelength dependence, with the width proportional to the
wavelength, and with overlap for white light. The demo, with a variablesize slit, clearly shows these effects, even
when starting with a 1 mmwide slit (1:00:00). The whole image becomes fainter as the slit width is reduced, since
less light can get through, but the widening of the central maximum is evident. If you recall the grating demo, the
width of the individual grating slits showed that all the maxima were not, in fact, the same height. The real grating
equation, taking the effects both of number of slits and of slit width into account, is I I0 sin 2 sin 2 sin 2 N 2 (1:03:50). The slit function modulates the strength of the grating maxima (which, compared to the slit diffraction, is
close together due to their larger spacing). This effect (which was also clear in the original grating demo) is clear,
with the 5thorder grating maximum being eliminated by the slit function (1:08:00).
Similar diffraction effects take place for circular apertures, although the math is slightly more complex. With
circular apertures, the angle to minima is multiplied by 1.22 (1:08:20). In the demo, Dr. Lewin uses a laser of
λ=594 nm and a screen distance of L=4 m. The pattern, arising from a 1/8 mm opening (pinhole), should be about
5 cm across. The pinhole allows little light through, but the rings are clear, though faint (1:11:15). In addition to
laser light, a bright LED has monochromatic light, which serves as a good light source for personal observation of
these effects, although it could not be captured on video. Diffraction ultimately limits the angular sizes we can see.
In the case of double stars, two circular patterns arise, which overlap if they are close together. According to the socalled Rayleigh criterion, things are perceived as double if they are separated by the distance to the first minimum,
that is, 1.2λ/D. This only works if the atmosphere does not contribute additional blurring. For the Hubble Space
Telescope (with a diameter of 2.4 m) for 500 nm green light, 1.2λ/D is about 1/20 of an arc second. Telescopes on
Earth can be much bigger than the Hubble Space Telescope, but due to atmospheric turbulence (astronomers call
this seeing) they typically cannot resolve angles smaller than about 0.5 arc seconds. The eye, with its small aperture,
can resolve about 5 arc minutes, in theory. In practice, this number is really about 1 to 2 arc minutes (demo). 2 ...
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This note was uploaded on 02/18/2011 for the course PHYS 320 taught by Professor Martinconner during the Spring '10 term at Open Uni..
 Spring '10
 martinconner

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