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Consider a planar waveguide used for attenuated total reflection measurement of a film coated on one surface of the waveguide. For a given angle of incidence, the sensitivity of attenuated total reflectance increases as the thickness of the waveguide decreases. Explain why. (The waveguide can be less than1ฮผmthick.)

Short Answer

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The rise in sensitivity of attenuated total reflectance as the waveguide thickness decreases.

Step by step solution

01

Definition of Reflectance and waveguide

  • The proportion of perpendicularly incident light reflected from a component relative to that reflected from a standard of established reflectance is known as reflectance.
  • A device (such as a duct, coaxial cable, or glass fibre) for confining and directing the propagation of electromagnetic waves (such as light), particularly a metal tube for channelling ultrahigh-frequency waves.
02

Determine the sensitivity of attenuated total reflectance increase as the thickness of the waveguide decreases

  • It is necessary to explain why the sensitivity of attenuated total reflectance increases as the waveguide thickness decreases.
  • The number of reflections inside increases as sensitivity increases, and the waveguide expands as well. Because each reflection has a certain amount of attenuation.
  • The number of reflections is inversely proportional to the waveguide at a constant angle of incidence.
  • As the waveguide shrinks, the number of reflections grows.

  • The rise in sensitivity of attenuated total reflectance as the waveguide thickness decreases was explained.

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Most popular questions from this chapter

Consider a reflection grating operating with an incident angle of 408 in Figure 20-7. (a) How many lines per centimeter should be etched in the grating if the first-order diffraction angle for 600 nm (visible) light is to be 2308? (b) Answer the same question for 1 000 cm21 (infrared) light.

Show that a grating with 103 grooves/cm provides a dispersion of 5.88 per mm of wavelength if n 5 1 and f 5 108 in Equation 20-4.

The exitance (power per unit area per unit wavelength) from a blackbody (Box 20-1) is given by the Planck distribution: Mฮป=2ฯ€hc2ฮป5(1ehc/Akt-1)where ฮปis wavelength, Tis temperature K), his Planckโ€™s constant, Cis the speed of light, and kis Boltzmannโ€™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/m2)emitted between those two wavelengths. We find the area by integrating the Planck function between wavelengthsฮป1andฮป2

role="math" localid="1664862982839" Poweremitted=โˆซฮป1ฮป2Mฮปdฮป

For a narrow wavelength range, โˆ†ฮป1the value of Mฮปis nearly constant and the power emitted is simply the product Mฮปโˆ†ฮป2.

(a) Evaluate Mhatฮป=2.00ฮผmand atฮป=10.00ฮผmatT=1000K

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

(c) Repeat part (b) for the interval9.99to10.01ฮผm

(d) The quantityM2ฮผm/M10ฮผmis 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?

Refer to the Fourier transform infrared spectrum in Figure 20-33.

(a) The interferogram was sampled at retardation intervals of1.2260ร—10-4cm. What is the theoretical wavenumber range (0 to ?) of the

spectrum?

(b) A total of 4 096 data points were collected from ฮด=-โˆ†toฮด=+โˆ†. Compute the value ofโˆ†, the maximum retardation.

(c) Calculate the approximate resolution of the spectrum.

(d) The interferometer mirror velocity is given in the figure caption. How many microseconds elapse between each datum?

(e) How many seconds were required to record each interfero gram once?

(f) What kind of beam splitter is typically used for the region 400 to 4 000cm-1? Why is the region below 400cm-1not observed?

(a) What resolution is required for a diffraction grating to resolve wavelengths of 512.23 and 512.26 nm? (b) With a resolution of 104, how close in nm is the closest line to 512.23 nm that can barely be resolved? (c) Calculate the fourth-order resolution of a grating that is 8.00 cm long and is ruled at 185 lines/mm. (d) Find the angular dispersion (ฯ•โˆ†) between light rays with wavelengths of 512.23 and 512.26 nm for first-order diffraction (n 5 1) and thirtieth-order diffraction from a grating with 250 lines/mm and f 5 3.08.

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