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Chapter 20: Q4P (page 525)

Which variables increase the resolution of a grating? Which variables increase the dispersion of the grating? How is the blaze angle chosen to optimize a grating for a particular wavelength?

Short Answer

Expert verified

The number of grooves that are illuminated and the diffraction order increases the resolution of a grating.

Step by step solution

01

Step: 1 Define the grating:

Diffraction grating is a series of clusters used to separate an incident wave into its wavelength by directly separating the magnitude of the diffraction.

02

Step: 2 The variables which increase the resolution of a grating:

The number of grooves that are illuminated and the diffraction order increases the resolution of a grating.

03

Step: 3 The variables which increase the dispersion of the grating:

The diffraction order increase the dispersion of the grating, while the spacing between lines in the grating is inversely proportional to dispersion of the grating.

04

Step: 4 The blaze angle chosen to optimize a grating:

The blaze angle chosen to optimize a grating for a particular wavelength is selected to give specular reflection at desired diffraction angle.

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Most popular questions from this chapter

A measurement with a signal-to-noise ratio of 100 / 1 can be thought of as a signal, S, with 1 % uncertainty, e. That is, the measurement is $S \pm e = 100 \pm 1$ .

(a) Use the rules for propagation of uncertainty to show that, if you add two such signals, the result is total signal $= 200 \pm \sqrt 2$ , giving a signal-to-noise ratio of $200 / \sqrt 2 = 141 / 1$ .

(b) Show that if you add four such measurements, the signal-tonoise ratio increases to 200 / 1.

(c) Show that averaging n measurements increases the signal-to-noise ratio by a factor of $\sqrt{n}$ compared with the value for one measurement.

The exitance (power per unit area per unit wavelength) from a blackbody (Box 20-1) is given by the Planck distribution: $M_{λ} = \frac{2 {πhc}^{2}}{λ^{5}} (\frac{1}{e^{hc / Akt} - 1})$ where $λ$ is wavelength, Tis temperature K), his Planck’s constant, Cis the speed of light, and kis Boltzmann’s constant. The area under each curve between two wavelengths in the blackbody graph in Box 20-1is equal to the power per unit area $(W / m^{2})$ emitted between those two wavelengths. We find the area by integrating the Planck function between wavelengths $λ_{1}$ and $λ_{2}$

role="math" localid="1664862982839" $Powe remitted = \int_{λ_{1}}^{λ_{2}} M_{λ} dλ$

For a narrow wavelength range, $∆ λ_{1}$ the value of $M_{λ}$ is nearly constant and the power emitted is simply the product $M_{λ} ∆ λ_{2}$ .

(a) Evaluate $M_{h} at λ = 2.00 μm$ and at $λ = 10.00 μm at T = 1000 K$

(b) Calculate the power emitted per square meter atin the intervalby evaluating the product

(c) Repeat part (b) for the interval $9.99 to 10.01 μm$

(d) The quantity $M_{2} μ_{m} / M_{1} 0 μm$ is the relative exitance at the two wavelengths. Compare the relative exitance at these two wave-lengths atwith the relative exitance at 100K. What does your answer mean?

Visible light strikes the surface of benzene at a $45 °$ angle. At what angle does the light ray pass through the benzene?

(a) In the cavity ring-down measurement at the opening of this chapter, absorbance is given by $A = \frac{L}{cln 10} (\frac{1}{τ} - \frac{1}{τ_{0}})$ whereis the length of the triangular path in the cavity, Cis the speed of light, Tis the ring-down lifetime with sample in the cavity, and $T_{0}$ is the ring-down lifetime with no sample in the cavity. Ring-down lifetime is obtained by fitting the observed ring-down signal intensityto an exponential decay of the form $l = l_{0} e^{- yτ}$ , whereis the initial intensity and t is time. A measurement ofis made at a wavelength absorbed by the molecule. The ring-down lifetime for 21.0-cm-1 along empty cavity is $18.52 μs$ and $18.52 μs$ for a cavity containing.role="math" localid="1664865078479" ${CO}_{2}$ Find the absorbance of ${CO}_{2}$ at this wavelength.

(b) The ring-down spectrum below arises from ${}^{13}{CH}_{4}$ and ${}^{12}{CH}_{4}$ from $1.9 ppm (vol / yol)$ of methane in outdoor air at 0.13. The spectrum arises from individual rotational transitions of the ground vibrational state to a second excited C-H)vibrational state of the molecule. (i) Explain what quantity is plotted on the ordinate ( Y-axis). (ii) The peak for ${}^{12}{CH}_{4}$ is at $6046.9546 {cm}^{- 1}$ . What is the wavelength of this peak in? What is the name of the spectral region where this peak is found?

The interferometer mirror of a Fourier transform infrared spectrophotometer travels pm1 cm .

(a) How many centimeters is the maximum retardation, $∆$ ?

(b) State what is meant by resolution.

(c) What is the approximate resolution $({cm}^{‐ 1})$ of the instrument?

(d) At what retardation interval, $∆$ , must the interferogram be sampled (converted into digital form) to cover a spectral range of 0 to 2000 cm^-1 ?

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