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Multiple Slit Diffraction

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Multiple Slit Diffraction
CHAPTER 27 | WAVE OPTICS
wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive
interference increases with λ , so that spectra (measurements of intensity versus wavelength) can be obtained.
Example 27.2 Calculating Highest Order Possible
Interference patterns do not have an infinite number of lines, since there is a limit to how big
interference possible with the system described in the preceding example?
m can be. What is the highest-order constructive
Strategy and Concept
d sin θ = mλ (for m = 0, 1, −1, 2, −2, … ⎞⎠ describes constructive interference. For fixed values of d and λ , the larger
m is, the larger sin θ is. However, the maximum value that sin θ can have is 1, for an angle of 90º . (Larger angles imply that light goes
backward and does not reach the screen at all.) Let us find which m corresponds to this maximum diffraction angle.
The equation
Solution
Solving the equation
d sin θ = mλ for m gives
m = d sin θ .
λ
Taking
(27.8)
sin θ = 1 and substituting the values of d and λ from the preceding example gives
m=
Therefore, the largest integer
(0.0100 mm)(1)
≈ 15.8.
633 nm
(27.9)
m can be is 15, or
m = 15.
(27.10)
Discussion
The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations.
However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen
has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the
center. Consequently, not all 15 fringes may be observable.
27.4 Multiple Slit Diffraction
An interesting thing happens if you pass light through a large number of evenly spaced parallel slits, called a diffraction grating. An interference
pattern is created that is very similar to the one formed by a double slit (see Figure 27.16). A diffraction grating can be manufactured by scratching
glass with a sharp tool in a number of precisely positioned parallel lines, with the untouched regions acting like slits. These can be photographically
mass produced rather cheaply. Diffraction gratings work both for transmission of light, as in Figure 27.16, and for reflection of light, as on butterfly
wings and the Australian opal in Figure 27.17 or the CD pictured in the opening photograph of this chapter, Figure 27.1. In addition to their use as
novelty items, diffraction gratings are commonly used for spectroscopic dispersion and analysis of light. What makes them particularly useful is the
fact that they form a sharper pattern than double slits do. That is, their bright regions are narrower and brighter, while their dark regions are darker.
Figure 27.18 shows idealized graphs demonstrating the sharper pattern. Natural diffraction gratings occur in the feathers of certain birds. Tiny, fingerlike structures in regular patterns act as reflection gratings, producing constructive interference that gives the feathers colors not solely due to their
pigmentation. This is called iridescence.
Figure 27.16 A diffraction grating is a large number of evenly spaced parallel slits. (a) Light passing through is diffracted in a pattern similar to a double slit, with bright regions
at various angles. (b) The pattern obtained for white light incident on a grating. The central maximum is white, and the higher-order maxima disperse white light into a rainbow
of colors.
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CHAPTER 27 | WAVE OPTICS
Figure 27.17 (a) This Australian opal and (b) the butterfly wings have rows of reflectors that act like reflection gratings, reflecting different colors at different angles. (credits: (a)
Opals-On-Black.com, via Flickr (b) whologwhy, Flickr)
Figure 27.18 Idealized graphs of the intensity of light passing through a double slit (a) and a diffraction grating (b) for monochromatic light. Maxima can be produced at the
same angles, but those for the diffraction grating are narrower and hence sharper. The maxima become narrower and the regions between darker as the number of slits is
increased.
The analysis of a diffraction grating is very similar to that for a double slit (see Figure 27.19). As we know from our discussion of double slits in
Young's Double Slit Experiment, light is diffracted by each slit and spreads out after passing through. Rays traveling in the same direction (at an
angle θ relative to the incident direction) are shown in the figure. Each of these rays travels a different distance to a common point on a screen far
away. The rays start in phase, and they can be in or out of phase when they reach a screen, depending on the difference in the path lengths traveled.
As seen in the figure, each ray travels a distance d sin θ different from that of its neighbor, where d is the distance between slits. If this distance
equals an integral number of wavelengths, the rays all arrive in phase, and constructive interference (a maximum) is obtained. Thus, the condition
necessary to obtain constructive interference for a diffraction grating is
d sin θ = mλ, for m = 0, 1, –1, 2, –2, … (constructive),
where
(27.11)
d is the distance between slits in the grating, λ is the wavelength of light, and m is the order of the maximum. Note that this is exactly the
d . However, the slits are usually closer in diffraction gratings than in double slits, producing fewer
same equation as for double slits separated by
maxima at larger angles.
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CHAPTER 27 | WAVE OPTICS
Figure 27.19 Diffraction grating showing light rays from each slit traveling in the same direction. Each ray travels a different distance to reach a common point on a screen (not
shown). Each ray travels a distance
d sin θ
different from that of its neighbor.
Where are diffraction gratings used? Diffraction gratings are key components of monochromators used, for example, in optical imaging of particular
wavelengths from biological or medical samples. A diffraction grating can be chosen to specifically analyze a wavelength emitted by molecules in
diseased cells in a biopsy sample or to help excite strategic molecules in the sample with a selected frequency of light. Another vital use is in optical
fiber technologies where fibers are designed to provide optimum performance at specific wavelengths. A range of diffraction gratings are available for
selecting specific wavelengths for such use.
Take-Home Experiment: Rainbows on a CD
d of the grooves in a CD or DVD can be well determined by using a laser and the equation
d sin θ = mλ, for m = 0, 1, –1, 2, –2, … . However, we can still make a good estimate of this spacing by using white light and the
The spacing
rainbow of colors that comes from the interference. Reflect sunlight from a CD onto a wall and use your best judgment of the location of a
strongly diffracted color to find the separation d .
Example 27.3 Calculating Typical Diffraction Grating Effects
Diffraction gratings with 10,000 lines per centimeter are readily available. Suppose you have one, and you send a beam of white light through it
to a screen 2.00 m away. (a) Find the angles for the first-order diffraction of the shortest and longest wavelengths of visible light (380 and 760
nm). (b) What is the distance between the ends of the rainbow of visible light produced on the screen for first-order interference? (See Figure
27.20.)
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CHAPTER 27 | WAVE OPTICS
x = 2.00 m
x -direction. In other words, the rainbow pattern extends out of the page.
Figure 27.20 The diffraction grating considered in this example produces a rainbow of colors on a screen a distance
along the screen are measured perpendicular to the
from the grating. The distances
Strategy
The angles can be found using the equation
d sin θ = mλ (for m = 0, 1, –1, 2, –2, …)
once a value for the slit spacing
(27.12)
d has been determined. Since there are 10,000 lines per centimeter, each line is separated by 1/10,000 of a
centimeter. Once the angles are found, the distances along the screen can be found using simple trigonometry.
Solution for (a)
The distance between slits is
and
d = (1 cm) / 10,000 = 1.00×10 −4 cm or 1.00×10 −6 m . Let us call the two angles θ V for violet (380 nm)
θ R for red (760 nm). Solving the equation d sin θ V = mλ for sin θ V ,
sin θ V =
where
mλ V
,
d
(27.13)
m = 1 for first order and λ V = 380 nm = 3.80×10 −7 m . Substituting these values gives
Thus the angle
−7
sin θ V = 3.80×10 −6 m = 0.380.
1.00×10 m
(27.14)
θ V = sin −1 0.380 = 22.33º.
(27.15)
−7
sin θ R = 7.60×10 −6 m .
1.00×10 m
(27.16)
θ R = sin −1 0.760 = 49.46º.
(27.17)
θ V is
Similarly,
Thus the angle
θ R is
Notice that in both equations, we reported the results of these intermediate calculations to four significant figures to use with the calculation in
part (b).
Solution for (b)
The distances on the screen are labeled
y V and y R in Figure 27.20. Noting that tan θ = y / x , we can solve for y V and y R . That is,
y V = x tan θ V = (2.00 m)(tan 22.33º) = 0.815 m
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(27.18)
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