Ординатура / Офтальмология / Английские материалы / Optics Learning by Computing with Examples using MATLAB_Dieter Moller_2007
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3.4. FAR FIELD AND FRAUNHOFER DIFFRACTION |
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2.Use FileFig 3.9 for the amplitude grating with d 0.0018 mm, λ 0.0005 mm, a 0.0036 mm, and θ −1.401 to 1.4.
a.How is the width of the main maxima dependent on the diffraction angle
θ?
b.What happens to the main maxima if we change d to 0.0009?
c.For blue light use λ 0.0004 mm and for red light λ 0.0007 mm. Plot for both the interference and diffraction factor on one graph and compare the two intensity patterns for different values of N.
d.Use close values of the two wavelengths, (e.g., 0.0005 and 0.00055) and determine when two peaks of different wavelengths and at different orders may be resolved depending on choices of N. (For changes in the angle, we have that 1 rad is about 5 to 6 degrees.)
3.From the interference factor find the numerical values of sin θ for which the interference factor is equal to N2 (or 1 in the normalized case). Express the numerical factor in terms of a, m, and λ; this is the grating equation.
FileFig 3.10 (D10FAGRDSLS)
Graph of the intensity P (Y ) of a double slit for d 0.02, λ 0.0005, a 0.2, X 4000, N 2, and Y −800, −799.9 to 800. A photograph is shown in Figure 3.18.
D10FAGRDSLS is only on the CD.
Application 3.10.
1.When d is changed from d a/2 to d a, what happens to the diffraction factor D(Y ) and how is the intensity P (Y ) D(Y )I (Y ) affected, where I (Y ) is the interference factor?
2.From Figure 3.16 one can obtain (a sin θ) mλ with an elementary derivation. This is the grating equation. This equation may also be obtained from the formula of the intensity for the main maxima. Using this formula make a
FIGURE 3.18 Diffraction pattern of a double slit where the separation a of the slits is much larger than the width d. The central part of the diffraction pattern shows an interference pattern similar to that observed in Young’s experiment (from M. Cagnet, M. Francon, and J. C. Thrieer, Atlas of Optical Phenomena, Springer-Verlag, Heidelberg, 1962).
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FIGURE 3.19 Coordinates for the amplitude grating. The incident light has angle ψ with the X direction.
list of the maxima depending on Y , for d a/10, and compare with P (Y ) of the graph.
3.Missing order: Use 4d a and start counting the maxima of P (Y ) from the center at 0. The fourth maximum of I (Y ) is suppressed by the factor D(Y ). Use other ratios of a/d.
4.Continue (3) on missing order. Change the wavelength to twice its value and to 1/2 and see what happens to the pattern.
3.4.5.2 Amplitude Grating with Incidence Light at an Angle
An experimental setup may require that the incident light enter the grating not in the direction of the normal but under an angle ψ with respect to the grating surface (Figure 3.19). The only difference now is that the optical path difference (i.e., a(sin θ) of the two parallel beams; see Eq. (3.54) is enlarged by the term a(sin θ). One obtains for the intensity
I(θ) {[sin(πd(sin θ + sin ψ)/λ)]/[πd(sin θ + sin ψ)/λ)]}2
·{sin[πNa(sin θ + sin ψ)/λ]/N sin[πa(sin θ + sin ψ)/λ]}2. (3.61)
The interference diffraction pattern is similar to the pattern of the amplitude grating used at normal incidence. The difference is that the main maximum is not at θ 0. The normalized interference factor is equal to 1 for the main maximum, that is, for (sin θ + sin ψ) 0, or at sin θ −sinψ. The zeroth order is shifted from θ 0 to θ −ψ. This is shown in the graph of FileFig 3.11.
FileFig 3.11 (D11FAGRANGS)
A graph is shown of the diffraction pattern of an amplitude grating with the incident light under an angle ψ to the normal.
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D11FAGRANG is only on the CD.
Application 3.11. Find the numerical values of (sin θ + sinψ) for which the normalized interference factor is equal to N1. Express the numerical factor in terms of a, m, and λ and compare with the grating equation.
3.4.5.3 Echelette Grating (Phase Grating)
A transmission grating with periodic differences in thickness is called a phase grating. When the grating has steps with angle ε, as shown in Figure 3.20, it is called an echelette grating. It may be produced from materials of refractive index n or with a metal surface such as a reflection grating. To calculate the interference diffraction pattern of the echelette grating, we have to determine the optical path difference between equivalent wavelets from neighboring steps and integrate over all the steps. The result is an interference factor, depending on the periodicity constant a of the steps and a diffraction factor, depending on
−θ
ε
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−ε
− θ − ε) =
θ + ε)
FIGURE 3.20 Coordinates for an echelette grating: (a) refracted and diffracted light and interference; (b) coordinates for the diffraction on one facet. Integration of the diffraction factor over the facet uses the optical path difference yi sin(θ + ε). The maximum is at θ −ε. We set d d.
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the shape of the step. To make the calculation of the diffraction factor easy, we assume that the periodicity constant is equal to the width of the steps d and we choose the angle of incidence such that the transmitted light is refracted in a direction perpendicular to the plane of the step (Figure 3.20a). The angle ε is then the angle between the normal of the facet and the normal of the plane of periodicity. The intensity is again given as the product of the interference and diffraction factor. The interference factor is the same as the one obtained for the amplitude grating, but the diffraction factor is calculated with the optical path difference yi sin(θ + ε); see Figure 3.20. The intensity is then
I (θ) {[sin(πd sin{θ + ε}/λ)]/[πd sin{θ + ε}/λ]}2 |
(3.62) |
· {[sin(πNa sin θ/λ)]/N sin(πa sin θ)/λ]}2. |
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The three graphs in FileFig 3.12 demonstrate the dependence on the step angle ε. When ε −.25 we see that the maximum of the diffraction factor is lined up with the first-order maximum of the interference factor. The zeroth order of the interference factor is at the first minimum of the diffraction factor and is suppressed. For the particular wavelength the grating is said to be blazed. For slightly different wavelengths, the first order would be displaced within the envelope of the diffraction factor. Considering a source emitting many wavelengths, the first order of a range of wavelengths is distributed over the range determined by the diffraction factor.
When changing the step angle ε, one has for ε .25 that the zeroth order of the diffraction factor is shifted to the −1 order of the interference factor (see the second graph of FileFig 3.12). For ε −.52, the zeroth order of the diffraction factor is at the +2 order of the interference factor (third graph of FileFig 3.12). For other values of (between these values, one finds that the maxima of the interference factor are not lined up with the maxima of the diffraction factor.
FileFig 3.12 (D12FAELGRS)
Graphs of the intensity of diffraction on an echelette grating. Three graphs are shown for three values of ε. There is only one interference factor and three diffraction factors. We see that the diffraction factor is lined up with the first maximum of the interference factor for ε −0.25. The zero order of the interference factor is at the first minimum of the diffraction factor and is suppressed. If one applies a source emitting many wavelengths, the first-order of a narrow band of different wavelengths is “filling” the area of the diffraction factor. For the wavelength for which both maxima coincide, the grating is said to be “blazed.” When choosing ε 0.25 the zeroth order of the diffraction factor is shifted to the −1 order of the interference factor. For ε −0.52 we have the zeroth order
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of the diffraction factor at the +2 order of the interference factor. We may choose other values of ε and find a misalignment.
D12FAELGRS
Diffraction on an Echelette Grating
The graphs for the three different values of . D(θ) is the diffraction factor, I (θ) the interference factor, and P (θ) the product. The angle in radians of the echelette is . Diffraction angle θ in radians, wavelength λ, width of openings d, and separation of openings a in mm. N is the number of lines. All parameters are defined globally above the graph.
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θ : −.301, −1.299 . . . 1.3 |
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D1(θ) : |
sin π · dλ · (sin(θ + 1)) |
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sin π · a · (sin(θ)) · Nλ |
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π · λ · sin(θ + 1) |
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sin π · dλ · sin(θ + 2) |
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sin π · dλ · (sin(θ + 3)) |
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π · dλ · sin(θ + 2)) |
π · dλ · sin(θ + 3) |
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P 1(θ) : D1(θ) · I (θ) |
P 2(θ) : D2(θ) · I (θ) |
P 3(θ) : D3(θ) · I (θ) |
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1 ≡ −0.25 |
d ≡ 37 |
λ ≡ 10 |
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N ≡ 20. |
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2 ≡ 0.25. |
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3 ≡ 0.52. |
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Application 3.12. Use FileFig 3.12 with d |
0.04, a 0.04, λ 0.01, and |
N 20. |
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1.Start from ε 0 and use for ε positive numbers until the negative first-order of I (θ) is at the peak of D(θ). Make another graph using a negative value for ε until the positive first-order of I (θ) is at the peak of D(θ).
2.Make three graphs with nine plots, starting with P 1(θ), D1(θ), and I 1(θ) for
λ1 0.015, then P 2(θ), D2(θ), and I 2(θ) for λ2 0.01, and P 3(θ), D3(θ), and λ3(θ) for λ3 0.007. Find a value of ε such that the interference peak of the positive first order of λ2 is at the maximum of the diffraction factor. The grating is now “blazed” for the wavelength λ2 0.01. Make an extra plot only of P 1(θ) to P 3(θ) and see that for the wavelengths λ1 and λ3 more than one peak appears. These wavelengths do not have maximum intensity and two orders would be observed. The echelette grating would be less efficient.
3.4.6 Resolution
Gratings are used in spectrometers to obtain the spectrum of the incident light. Let us consider incident light containing a spectrum of just two different wavelengths λ1 and λ2. As an example let us consider the sodium D lines at 5890 and 5896 A◦ in the yellow part of the visible spectrum, emitted when the Sodium atom is at a high temperature. One may observe these lines when salt falls into a gas flame and one sees yellow light. The wavelength difference of the two lines is just 6◦ and we now discuss the question of what has to be required of a grating to observe the two separated lines. We consider two lines of wavelength differenceλ λ2 − λ1. The diffraction angles in the first order will have maxima at sin θ1 λ1/a and sin θ2 λ2/a (Figure 3.21a). Assuming λ1 < λ2 we have to determine how large λ λ2 − λ1 must be in order to recognize the two spectral lines separately. If λ is large enough, we can recognize these two lines (Figure 3.21b on the left) and if λ is too small, we would have just one peak
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FIGURE 3.21 (a) Closely positioned main maxima of the diffraction patterns produced by the wavelengths λ1 and λ2 with λ λ2 − λ1; (b) (left) images of two point sources sufficiently separated; (right) image of two point sources at the limit of resolution. (From M. Cagnet, M. Francon, and J.C. Thrieer, Atlas of Optical Phenomena, Springer-Verlag, Heidelberg, 1962.)
FIGURE 3.22 Main maximum and side maxima of the diffraction pattern obtained with a grating having N lines and using the wavelength λ1.
(Figure 3.21b on the right). To get a quantitative value for the case where the two lines are just separated, we use the Rayleigh criterion. Which states that the two maxima of a diffraction pattern are considered separated when the maximum of one is at the position of the minimum of the other.
From Figure 3.22 we have for the distance from maximum to minimum of the diffraction pattern of a grating for any order
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N N a a
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and obtain |
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λ λ1/N. |
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Calling λ2 just λ and considering the mth diffraction order, we have |
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λ/ λ mN. |
(3.65) |
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Since λν velocity of the light in the medium, where ν is the frequency, one |
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also has |
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ν/ ν mN. |
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The ratio λ/ λ or ν/ ν is called the resolving powers of the grating.
The graph in FileFig 3.13 shows two diffracted waves of wavelength λ1 and λ2 on a grating. The separation of their maxima depends on N and one may change N to find the separation in agreement with the Rayleigh criterion. Also seen is the dependence of the resolution on the diffraction order. The first graph in FileFig 3.14 shows the diffraction pattern of two round openings of radius a. One is at zero and the other at distance b. One can determine b using the Rayleigh criterion. The second graph shows the two spectral lines in three dimensions.
For our example of the sodium D lines we have from Eq. (3.66) about 1000 for mN. In other words, in first order we need 1000 lines of the grating. If the grating has a periodicity constant of about 1 micron, the width of the grating would be one millimeter. If one could use a larger grating of 10 cm then the number of lines would be 100,000 and in first order, for a wavelength of 5000A,˚λ would be 0.02A˚ .
FileFig 3.13 (D13FAGRRES)
Graph of the diffraction on an amplitude grating for two wavelengths λ1 0.0005, λ2 0.0006, openings d 0.0001, periodicity constant a 0.002, and number of lines N 6.
D13FAGRRES is only on the CD.
Application 3.13.
1.Make three choices of λ2 and determine the N value so that the maximum of λ2 is on the minimum of λ1.
2.Choose λ1 and λ2 and determine the N value necessary to resolve the two maxima.
3.Consider higher orders m. Choose a pair of lines so close together that they are not resolved in first-order. Determine the order for resolution.
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FileFig 3.14 (D14FARES3DS)
Determination of Rayleigh distance and graph of the 3-D-diffraction pattern of two round apertures of radius a and distance b.
D14FARES3DS
Determination of the Wavelength difference for Two Peaks, Resolved According to the Rayleigh Criterion
We call the distance between the maxima b, radius of apertures a, distance between the apertures d, coordinate on the observation screen R, wavelength λ, and distance from aperture to screen X.
1. Determination of Rayleigh distance
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X ≡ 4000 |
R : 0, .1 . . . 50 |
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g1(R) : |
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Distance b is assumed to be b ≡ 24.5.
2. 3-D Graph of pattern of two round apertures at distance b
i : 0 . . . N |
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xi : (−40) + 2.0001 · i |
yj : (−40) + 2.0001 · j λ ≡ .0005 |
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N ≡ 60 X : 4000 |
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RR(x, y) : (x)2 + (y)2 |
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gg2(x, y) : |
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Mi,j : g2(xi , yj ) + gg2(xi , yj ).
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3.Calculation of wavelength difference corresponding to b
The diffraction angle is calculated from b/X θ. The grating is made of round apertures of diameter a and spaced at distance d. From the grating formula we have for the wavelength difference λ d θ or λ (d/X)b.
For d : .1, λ : d · Xb , λ 6.125 × 10−4.
Application 3.14.
1.Convert to a surface plot and make b such that the maximum of one of the peaks is on the minimum of the other. Then go back to the contour plot.
2.We have used the same wavelength for both peaks. After determining the wavelength resolution, introduce the corresponding wavelengths into the two expressions of the Bessel function and estimate the change.
3.Assume that the angle difference for resolution is given for order m and find the wavelength difference.
3.5BABINET’S THEOREM
Babinet’s Theorem tells us that two complementary screens will give us the same diffraction pattern at the observation screen. Complementary screens are two screens, one of which is transparent at areas where the other is opaque (Figure 3.23). If the two complementary screens are placed together in the same plane, no aperture will exist.