Optical Properties of Solids Second Edition
Mark Fox Oxford University Press, 2010 SOLUTIONS TO EXERCISES These notes contain detailed solutions to the Exercises at the end of each chapter of the book, for the benefit of class instructors. Please note that figures within the solutions are numbered consecutively from the start of the document (e.g. Fig. 1) in order to distinguish them from the figures in the book, which have an additional chapter label (e.g. Fig. 1.1). A similar convention applies to the labels of tables. The author would be very grateful if mistakes that are discovered in the solutions would be communicated to him. He is also very appreciative of comments about the text and/or the Exercises. He may be contacted at the following address: Department of Physics and Astronomy University of Sheffield Hicks Building Sheffield, S3 7RH United Kingdom. email:
[email protected] c Mark Fox 2010 °
Chapter 1 Introduction (1.1) Glass is transparent in the visible spectral region and hence we can assume α = κ = 0. The reflectivity is calculated by inserting n = 1.51 and κ = 0 into eqn 1.29 to obtain R = 0.041. The transmission is calculated from eqn 1.9 with R = 0.041 to obtain T = 92%. (1.2) From Table 1.4 we read that the refractive indices of fused silica and dense flint glass are 1.46 and 1.746 respectively. The reflectivities are then calculated from eqn 1.29 to be 0.035 and 0.074 respectively, with κ = 0 in both cases because the glass is transparent. We thus find that the reflectivity of dense flint glass is larger than that of fused silica by a factor of 2.1. This is why cut–glass products made from dense flint glass have a sparkling appearance. (1.3) We first use eqns 1.25 and 1.26 to convert ²˜r to n ˜ , giving n = 3.01 and κ = 0.38. We then proceed as in Example 1.2. This gives: v = c/n = 9.97 × 107 m s−1 , α = 4πκ/λ = 9.6 × 106 m−1 , R = [(n − 1)2 + κ2 ]/[(n + 1)2 + κ2 ] = 25.6%. (1.4) The anti–reflection coating prevents losses at the air–semiconductor interface, and 90% of the light is absorbed when exp(−αl) = 0.1 at the operating wavelength. With α = 1.3 × 105 m−1 at 850 nm, we then find l = 1.8 × 10−5 m = 18 µm. (1.5) We are given n = 3.68 and we can use eqn 1.19 to work out κ = αλ/4π = 0.083. We then use eqn 1.29 to find R = 0.328. Since αl = 2.6, we do not need to consider multiple reflections and we can just use eqn 1.8 to find the transmission. This gives: T = (1 − 0.328)2 exp(−1.3 × 2) = 0.034. The optical density is calculated from eqn 1.11 as 0.434 × 1.3 × 2 = 1.1. (1.6) 99.8% absorption in 10 m means exp(−αl) = 0.002, and hence α = 0.62 m−1 . We use eqn 1.19 to find κ = αλ/4π = 3.5 × 10−8 . We thus have n ˜ = 1.33 + i 3.5 × 10−8 . The real and imaginary parts of ²˜r are found from eqns 1.23 and 1.24 respectively, and we thus obtain ²˜r = 1.77+i 9.2×10−8 . (1.7) The filter appears yellow and so it must transmit red and green light, but not blue. The filter must therefore have absorption at blue wavelengths. (1.8) (a) In the incoherent limit, we just add the intensities of the beams. The intensities of the beams transmitted after multiple reflections are shown
1
incident light R1 I0
R2 e-al
I0(1-R1)
I0(1-R1)e-a l
I0R1
I0(1-R1)R2e
I0(1-R1)R2e-a l I0(1-R1) R2R1e-3a l
I0(1-R1)2R2e-2a l
I0(1-R1)2 R22R1 e-4a l
transmitted light
-2a l
I0(1-R1) R2R1e-2a l I0(1-R1) R22R1e-4a l
I0(1-R1) (1-R2) e-a l
I0(1-R1) R22R1e-3a l I0(1-R1) R22R12e-5a l
I0(1-R1) R22R12e-4a l
I0(1-R1) (1-R2) R2R1 e-3a l
I0(1-R1) R23R12e-5a l
reflected light
I0(1-R1) (1-R2) R22R12 e-5a l Figure 1: Multiple reflections in the incoherent limit, as considered in Exercise 1.8. in Fig. 1. The transmitted intensity is given by: It
= I0 (1 − R1 )(1 − R2 )e−αl + I0 (1 − R1 )(1 − R2 )R1 R2 e−3αl + I0 (1 − R1 )(1 − R2 )R12 R22 e−5αl + · · · ¡ ¢ = I0 (1 − R1 )(1 − R2 )e−αl 1 + R1 R2 e−2αl + (R1 R2 )2 e−4αl + · · · , ∞ X = I0 (1 − R1 )(1 − R2 )e−αl (R1 R2 e−2αl )k , k=0
= I0 (1 − R1 )(1 − R2 )e where we used the identity transmissivity is thus: T =
−αl
1 , 1 − R1 R2 e−2αl
P∞ k=0
xk = 1/(1 − x) in the last line. The
It (1 − R1 )(1 − R2 )e−αl = . I0 1 − R1 R2 e−2αl
(b) We have an air-medium-air situation, and so it will be the case that R1 = R2 ≡ R. We need to compare the exact formulae given in eqns 1.6 and 1.9 with the approximate one that neglects multiple reflections given in eqn 1.8. (i) With α = 0 the extinction coefficient κ will also be zero. We then calculate the reflectivity from eqn 1.29 to be: R=
2.42 (3.4 − 1)2 = = 0.30 . (3.4 + 1)2 4.42 2
