ch11

# ch11 - CHAPTER 11 PROBLEM Solutions 11.1 Consider a...

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CHAPTER 11 PROBLEM Solutions 11.1. Consider a photodetector operating in the neighborhood of 60 GHz. How many cycles of green light are there in a single cycle of 60 GHz? Can a photodetector be used to follow the oscillations of the electromagnetic field associated with this light? For the RF signal, 9 1 1 17 60 10 RF RF T ps v = = = × For the optical signal, the wavelength is about 500 nm, from Figure 11.1. The period is 9 8 1 500 10 0.0017 3 10 op op T ps v c λ - × = = = = × The RF period is 17 10000 0.0017 = times longer, so 10,000 cycles of optical light fit into one cycle of the RF. Therefore even the fastest detectors can only measure a time average intensity compared to the oscillation of light. 11.2. Light with wavelength λ = 700 nm is incident on a sample of GaAs. a) Where in the spectrum does this radiation lie? From Figure 11.1, this is at the red edge of visible light. b) At what depth is the incident flux (neglecting Fresnel loss) reduced to 10% of its value at the surface? 1%? To find the absorption coefficient from Figure 11.4, we must convert this wavelength into energy. using the Golden Rule: E ( e V ) = 1 . 2 4 λ ( μ m ) = 1 . 2 4 0 . 7 0 0 = 1 . 7 7 e V From Figure 11.4, α =2.1 × 10 4 cm -1 . The decay in intensity is given by Anderson & Anderson Solutions Chapter 11 1 11/19/2007

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F L ( x ) = F L ( 0 ) e - α x or ( 29 4 1 2.1 10 ( ) 0.1 (0) cm x L L F x e F - - × = = x 1 0 % = - l n 0 . 1 ( 29 2 . 1 × 1 0 4 c m - 1 = 1 . 1 × 1 0 - 4 c m = 1 . 1 μ m For 1% remaining flux, ( 29 4 1% 4 1 ln 0.01 2.2 10 2.2 2.1 10 x cm m cm μ - - = - = × = × c) The color is changed to orange. Now how deep does the light penetrate (to the 10% level)? From Figure 11.1, orange light has a wavelength of about 600 nm, or a photon energy of 1.24/0.6=2.07eV. From Figure 11.4, at this energy the absorption coefficient of GaAs increases to 5.3 × 10 4 cm -1 , and x 1 0 % = - l n 0 . 1 ( 29 5 . 3 × 1 0 4 c m - 1 = 0 . 4 3 μ m 11.3. a) Calculate the Fresnel reflection at normal incidence for light going from air to glass ( n =1.5). R = 1 - 1 . 5 ( 29 2 ( 1 + 1 . 5 ) 2 = 0 . 5 2 2 . 5 2 = 0 . 0 4 or 4% is reflected. b) Explain why you can see into a store window and see your reflection at the same time, but at night looking out your window from a lighted room you can only see your reflection?” When you look into a store window, you see 4% of the sunlight reflected from you and off the surface of the glass, but you also see the light from the merchandise being transmitted through the window. To your eye both images are superimposed. Anderson & Anderson Solutions Chapter 11 2 11/19/2007
At night, there is very little light from the far side of the window, so your 4% reflection of the room light is the only thing you can see. 11.4. Show that Equation (11.8) follows from Equation (11.6). Equation (11.6) is: 1 q d J n ( x ) d n + G L ( x ) - n p τ n = 0 But we know ( 29 ( 29 ( ) 1 ( ) 1 x L L Li G x R F x R F e α α α - = - = - and we also know J n = q D n d n d x Using n=n p , 2 2 p n n d n dJ qD dx dx = Therefore ( 29 2 2 1 1 0 p p x n Li n d n n qD R F e q dx α α τ - + - - = ( 29 2 2 0 2 2 1 1 0 p p p x n Li n d n d n n qD R F e q dx dx α α τ - + + - - = But since n p0 is a constant, this reduces to ( 29 2 2 1 1 0 p p x n Li n d n n qD R F e q dx α α τ - + - - =

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