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Answers & Additional Problems

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    75543
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    Optical Design

    Exercise Answers

    For 1 nm resolution, use the 100 mm focal length, 1st order diffraction, which gives 8.3 nm/mm and a slit width of 125 microns.

    For 0.01 nm resolution, use the 2 m focal length, 2nd order diffraction, which gives 0.415 nm/mm and a slit width of 25 microns.

    Problems

    1. An emission instrument has a resolution of 0.1 nm. Would emission from calcium ion at 393.366 nm interfere with emission from neutral ruthenium at 393.176 nm? What assumptions have you made?
    2. For the Cu 324.7 nm emission line, would an instrument with 0.1 nm resolution reveal evidence of self-reversal and, if so, how?
    3. For the Cu 324.7 nm emission line, would an instrument with 0.001 nm resolution reveal evidence of self-reversal and, if so, how?
    4. Complete the grating selection for the echelle system discussed in "Plane grating, Crossed with prism or grating, Multiple detector" by going to a grating catalog, finding a combination of groove density and blaze angle that gives the desired dispersion and free spectral range, and then choosing a first-order cross-dispersion grating to provide the desired order separation.

    Answers

    1. Ca emission is centered at 393.366 nm, while Ru emission is centered at 393.176 nm. The line separation is 393.366 – 393.176 = 0.190 nm. Since spectral resolution is 0.1 nm, one should be able to look at the center of one line while not overlapping the central wavelength of the other line. However, emission lines are not infinitely narrow. How wide might these two lines be? First consider Doppler broadening only. T in an ICP is ~ 5000 K. M for Ca is 40, while for Ru it is 101. ∆λD = 7.16×10-7 λ (T/M)1/2 = 7.16×10-7 * 393.2 (5000/M)1/2 nm = 0.02 / M1/2 nm, with M either 40 Dalton or 101 Dalton. Either way, the Doppler width is much less than 0.1 nm (M>1 ⇒ M1/2>1 ⇒ 0.02/M1/2<0.02 nm). What about lifetime broadening? There's no need to use "rule of thumb" information here. Go to the NIST atomic structure website, physics.nist.gov/cgi-bin/ASD/lines1.pl, and learn that the transition probability for the Ca line is 1.4×108 s-1. The transition occurs at c/λ = 2.997×108 m s-1/3.9366×10-7 m = 7.61×1014 Hz. ∆λ/λ = ∆υ/υ = 1.4×108 /7.61×1014 = 1.8×10-7, or ∆λL = 393.366*1.8×10-7 = 7.2×10-5 nm. No transition probability data is available for the Ru line, but one would not expect a significantly wider line for an allowed transition. Clearly, unless there's a collisional component to the line width, the spectrometer could resolve the two lines. But what about collisional broadening? Ca+ has a single outer shell electron and so its transitions are subject to Stark broadening. A search at NIST (http://physics.nist.gov/cgi-bin/ASBi...earch_form.cgi) for the Ca II spectrum, search term "broadening" gets 15 citations. There are several possible sources for information. Choosing M. S. Dimitrijević and S. Sahal-Bréchot , "Stark Broadening of Ca II Spectral Lines," J. Quant. Spectrosc. Radiat. Trans. 49, 157-164 (1993) and selecting an electron density of 1016 cm-3, Table 3 indicates that a line width ~ 0.004 to 0.01 nm might be observed. Figure 1 of G. F. Larson and V. A. Fassel, "Line Broadening and Radiative Recombination Background Interferences in Inductively Coupled Plasma-Atomic Emission Spectroscopy," Appl. Spectrosc. 33, 592-599 (1979) shows that, for a Lorentzian line, the half-width is only part of the story. There is significant emission over a range of about 3 nm! Thus, ignoring any specifics about the Ru line, the wings of the Ca line will emit far beyond one full-width at half-maximum, and the Ca WILL influence the signal at the Ru wavelength no matter what the resolution of the spectrometer!
    2. With resolution of 0.1 nm and typical atomic emission lines having widths ~ 0.01 nm or less, the details of line shape will be obscured. What will be clear is that the working curve will become nonlinear at high concentrations, since some of the emitted light will be reabsorbed, then lost. Eventually the slope of the working curve increases only as (concentration)1/2.
    3. With resolution of 0.001 nm and typical atomic emission lines having widths ~ 0.01 nm or less, the details of line shape will be apparent. The line's self-reversal will be evident, and the line will show a dip at its center. The intensity in the wings of the line will continue to increase in proportion to concentration, while the intensity at line center will fall according to Beer's Law superimposed on intensity that grows linearly with concentration, Iobs = constant*concentration*(e-εb concentration). For small concentration, that's a linear increase. For high concentration, intensity decreases. For k = 0.01, C e-kC looks like this:

      ans3.PNG

    4. From the discussion, we know we want roughly 100 grooves mm-1 near a 70° blaze. Let's try 3 gratings:

      79 gpmm, 63°

      98.7 gpmm, 63°

      110 gpmm, 64°

      As in the examples in the text, we first assume Littrow mode, running at the blaze angle, and find the order in which 325 nm appears, the order in which 200 nm appears, the number of intervening orders, and the focal length to give the desired 0.003 nm/25 µm dispersion. Since \(\dfrac{dλ}{dx} = \dfrac{d\cos β}{nf} = \dfrac{dλ\cos β}{f2d\sin β_{blaze}} = \dfrac{λ}{2f\tan θ_{blaze}}\), \(f = \dfrac{325nm}{2 \tan θ_{blaze}×0.12 nm/mm} = \dfrac{1354mm}{\tan θ_{blaze}}\)

      gpmm d (nm) blaze = 2dsinθB (nm) n(325 nm) n(200 nm) ∆orders (200 to 325 nm) f (mm)
      79 12658 22557 69.4 → 69 112.78→113 44 690
      98.7 10132 18055 55.55→ 56 90.28→ 90 34 690
      110 9091 16342 50.28→ 50 81.7 → 82 32 660

      We wanted 50 orders in this range; 44 comes closest. 690 mm is a non-standard focal length. At a slight sacrifice in resolution, we can get better spectral coverage, so choose f = 5/8 m = 625 mm. For cross-dispersion, we want 200 nm to 325 nm spread out over ½ of the CCD = 12.5 mm, so 125 nm/12.5 mm = 10 nm/mm. In turn, the blaze angle for such dispersion is 10 nm/mm = 325 nm/(2*625 mm*tan θblaze) or θblaze = 1.49°. That's a very low blaze – almost no blaze at all. In Littrow mode, assuming first order, 325 nm = 2 d sin(1.49°) giving d = 6249 nm or 160 gpmm.

      That's as far as the problem requires answers. But let's proceed to see where various wavelengths appear and what coverage the system will actually have. Plausibly, the CCD should observe 200 nm to 450 nm (since 325 nm is half-way through the wavelength range). Let's compute the (x,y) coordinates in pixels for each nm wavelength. Decide that 325 nm shows up at (0,0). This means that the range of pixels is (-500,-500) to (500,500) for a megapixel CCD. Ignoring changes in dispersion for the cross-disperser, we know we have 10 nm/mm or 10 nm/40 pixels or 0.25 nm/order, with 69th order close to (0,0) pixels. Actually, from the aqua highlighted row in the table, 69th order is centered at 326.9 nm (22557 nm/69). Round all positions to the nearest pixel. Example calculation:

      200 nm. n = 112.78, so use 113. Center of 113th order = 22557 nm/113 = 199.62 nm. 200 nm is thus 0.38 nm to the blue of the center of the order. Dispersion needs to take into account that the observed wavelength is not exactly at the blaze angle. = d(sin α + sin β) = d(sinθB + sin β). Recognizing that we're near the center of an order, even if not quite on top of it, n(λ − λorder center) = d(sin θB + sin β) − 2d sin θB = d(sin β − sin θB). Rearranging, n(λλorder center) + d sinθB = d sin β. Further substituting gives d sin θB = d sin β. Solve for β. \(β = \sin^{-1}\left(\dfrac{nλ}{d} - \sin θ_B \right )\). That's β at the observed wavelength, so the mean value for β during dispersion is (β + θB)/2. Sticklers will say we should integrate the dispersion relationship, but for current purposes, the average β is good enough. \(\dfrac{dλ}{dx} \sim \dfrac{∆λ}{∆x} = \dfrac{d \cos β}{nf}\) so \(∆x = (λ − λ_{order\: center})\dfrac{nf}{d \cos β}\). Then divide ∆x by 25 µm to get the number of pixels offset from order center. For 200 nm, ∆λ = -0.38 nm, n = 113, β = 63.435° so the average β for dispersion is (63.435+63)/2 = 63.217°. ∆x = (-0.38 nm)*113*625 mm/(12658 nm cos(63.217°) = - 4.705 mm. That's -188.2 pixels. Rounded, -188 pixels. In the cross-dispersion direction, 200 nm is -125 nm / 10 nm/mm = -12.5 mm = -500 pixels. Put this all into Excel, and here's what comes out:

      ans4.png

      Each dot represents an integer wavelength 200 – 450 nm. At a dispersion of 0.003 nm/pixel * 690 mm/625 mm = 0.0033 nm/pixel, about ¼ of all wavelengths fall outside the 1000 by 1000 pixel size of a square CCD. Many CCD makers have 1000 by 500 pixel chips or other, similarly oblate geometries. Having plotted the above, I find it quite unsatisfactory. Where do the orders fall, and what range of wavelengths fall in the 1000 by 1000 range of the chip? First, let's recompute, running the computation backwards. Start at ±500 pixels for each order 50 to 113. 500 pixels = 12.5 mm. \(∆x = (λ − λ_{order\: center})\dfrac{nf}{d \cos β}\) so \(λ=\dfrac{d {∆x} \cos β}{nf}+λ_{order\: center}\). But β is linked to λ: \(nλ = d(\sin α + \sin β) = d(\sin θ_B + \sin β)\). \(λ = \dfrac{d{∆x}\cos \left(\sin^{−1}\left(\dfrac{nλ}{d} − \sin θ_B \right ) \right )}{nf} + λ_{order\: center}\). This is a successive approximations calculation. Approximate β = 63°, knowing d and n find λ, plug λ into the last equation to find the next approximation, and repeat to consistency. At high orders, the minimum for order n is less than the maximum for order n+1, so the same wavelength is visible at the left of one order and the right of another. The full spreadsheet is attached. Below 100th order, there are gaps in the spectrum per the above plot.

      Order λmin (nm) λmax (nm) Order λmin (nm) λmax (nm)
      113 198.563 200.598 76 295.232 298.258
      112 200.336 (overlap 0.27 nm) 202.389 75 299.168 (gap 0.91 nm) 302.235
      100 224.376 226.676 63 356.153 359.803
      99 226.643 (overlap 0.033 nm) 228.966 62 361.897 (gap 2.094 nm) 365.607
      98 228.955 (overlap 0.011 nm) 231.302 50 448.752 453.352

    This page titled Answers & Additional Problems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Alexander Scheeline & Thomas M. Spudich via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.