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Example: Diamond in air

Solution: At the critical angle,. and. Rearranging the equation. Substitution gives. Example: Diamond in air. What is the critical angle  c for light passing from diamond ( n 1 = 2.41) into air ( n 2 = 1)?. Light Interactions with Solids. Absorbed: I A. Reflected : I R.

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Example: Diamond in air

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  1. Solution: At the critical angle, and Rearranging the equation Substitution gives Example: Diamond in air • What is the critical angle c for light passing from diamond (n1 = 2.41) into air (n2 = 1)?

  2. Light Interactions with Solids Absorbed: IA Reflected: IR Transmitted: IT Incident: I0 Scattered: IS • Optical classification of materials: Transparent Adapted from Fig. 21.10, Callister 6e. (Fig. 21.10 is by J. Telford, with specimen preparation by P.A. Lessing.) Translucent Opaque polycrystalline porous polycrystalline dense single crystal • Incident light is reflected, absorbed, scattered, and/or transmitted:

  3. Optical Properties of Metals: Absorption • Absorption of photons by electron transitions: Energy of electron unfilled states DE= h required! Incident photon of energy h filled states freq. Planck’s constant of (6.63 x 10-34J/s) Adapted from Fig. 21.4(a), Callister & Rethwisch 8e. incident light • Unfilled electron states are adjacent to filled states • Near-surface electrons absorb visible light.

  4.  = absorption coefficient, cm-1 = sample thickness, cm = incident light intensity = transmitted light intensity Light Absorption The amount of light absorbed by a material is calculated using Beer’s Law Rearranging and taking the natural log of both sides of the equation leads to

  5. Reflection of Light for Metals Energy of electron IR unfilled states “conducting” electron photon emitted from metal surface Electron transition filled states • Electron transition from an excited state produces a photon. Adapted from Fig. 21.4(b), Callister & Rethwisch 8e.

  6. Reflection of Light for Metals (cont.) Reflectivity = IR/I0 is between 0.90 and 0.95. Metal surfaces appear shiny Most of absorbed light is reflected at the same wavelength Small fraction of light may be absorbed Color of reflected light depends on wavelength distribution Example: The metals copper and gold absorb light in blue and green => reflected light has gold color 7

  7. Example: For Diamond n = 2.41 Reflectivity of Nonmetals • For normal incidence and light passing into a solid having an index of refraction n:

  8. Scattering of Light in Polymers • For highly amorphous and pore-free polymers • Little or no scattering • These materials are transparent • Semicrystalline polymers • Different indices of refraction for amorphous and crystalline regions • Scattering of light at boundaries • Highly crystalline polymers may be opaque • Examples: • Polystyrene (amorphous) – clear and transparent • Low-density polyethylene milk cartons – opaque

  9. Selected Light Absorption in Semiconductors Energy of electron Examples of photon energies: unfilled states blue light: h = 3.1 eV red light: h = 1.8 eV Egap incident photon energy hn filled states Adapted from Fig. 21.5(a), Callister & Rethwisch 8e. Absorption of light of frequency  by by electron transition occurs if hn > Egap • If Egap < 1.8 eV, all light absorbed; material is opaque (e.g., Si, GaAs) • If Egap > 3.1 eV, no light absorption; material is transparent and colorless (e.g., diamond) • If 1.8 eV < Egap < 3.1 eV, partial light absorption; material is colored

  10. (b) Redoing this computation for Si which has a band gap of 1.1 eV Computations of Minimum Wavelength Absorbed (a) What is the minimum wavelength absorbed by Ge, for which Eg= 0.67 eV? Solution: Note: the presence of donor and/or acceptor states allows for light absorption at other wavelengths.

  11. Color of Nonmetals • Example 2:Ruby = Sapphire (Al2O3) + (0.5 to 2) at% Cr2O3 -- Sapphire is transparent and colorless (Eg > 3.1 eV) -- adding Cr2O3 : • alters the band gap • blue and orange/yellow/green light is absorbed • red light is transmitted • Result: Ruby is deep red in color 80 sapphire 70 Transmittance (%) ruby 60 50 l (= c/)(m) wavelength, 40 0.3 0.5 0.7 0.9 Adapted from Fig. 21.9, Callister & Rethwisch 8e. (Fig. 21.9 adapted from "The Optical Properties of Materials" by A. Javan, Scientific American, 1967.) • Color determined by the distribution of wavelengths: -- transmitted light -- re-emitted light from electron transitions • Example 1: Cadmium Sulfide (CdS), Eg = 2.4 eV -- absorbs higher energy visible light (blue, violet) -- color results from red/orange/yellow light that is transmitted

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