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Principles of Instrumental Analysis, 6th ed. Chapter 7
 
 3
7-10. From Equation 7-5, d = λn/2n. If first-order interference is used, one end of the wedge 
would have a thinckness d of 
 d = 700 nm × 1/(2 × 1.32) = 265 nm or 0.265 μm. 
 This thickness would also transmit second-order radiation of 700/2 = 350 nm, which 
would be absorbed by the glass plates supporting the wedge. 
 The other end of the wedge should have a thickness of 
 d = 400 × 1/(2 × 1.32) = 1.52 nm or 0.152 μm 
 Thus, a layer should be deposited with is 0.265 μm on one end and which tapers linearly 
over 10.0 cm to 0.152 μm at the other end. 
7-11. The dispersion of glass for visible radiation is considerably greater than that for fused 
silica or quartz (see Figure 6-9). 
7-12. nλ = d (sin i + sin r) (Equation 7-6) 
 d = nλ/(sin i + sin r) = 1 × 400 nm/(sin 45 + sin 5) = 400 nm/(0.707 + 0.087) = 503.7 nm 
 lines/mm = 61 line nm 10 = 1985
503.7 nm mm
× 
7-13. For first-order diffraction, Equation 7-13 takes the form 
 λ/Δλ = nN = 1 × 15.0 mm × 84.0 lines/mm = 1260 
 In order to obtain the resolution in units of cm–1, we differentiate the equation λ = 1/ν 
 2 2
1 1 = or = d
d
λ λ
ν ν ν
Δ
Δ ν
 
 Thus, 
 Δλ = 2/ν νΔ 
 Substituting for λ and Δλ in the first equation gives