Physics 4C. Chapter 35: Conceptual Questions: 2, 8, 12 Problems: 9, 21, 25, 26, 39, 40, 55, 72, 82, 83, 93

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1 Physics 4C Solutions to Chapter 35 HW Chapter 35: Conceptual Questions:, 8, 1 Problems: 9, 1, 5, 6, 39, 40, 55, 7, 8, 83, 93 Question 35- (a) increase (b) 1λ Question 35-8 (a) 300 nm (b) exactly out of phase Question 35-1 (a) no (b) (0) = 0 (c) L Problem 35-9 (a) We wish to set Eq equal to 1/, since a half-wavelength phase difference is equivalent to a π radians difference. Thus, L min l 60 nm = = n n b g b g = 1550nm = 155. μ m. (b) Since a phase difference of 3 (wavelengths) is effectively the same as what we required in part (a), then L = 3l = 3Lmin = 3155 = 465 bn n1g b. μmg. μm. Problem 35-1 The maxima of a two-slit interference pattern are at angles θ given by d sin θ = mλ, where d is the slit separation, λ is the wavelength, and m is an integer. If θ is small, sin θ may be replaced by θ in radians. Then, dθ = mλ. The angular separation of two maxima associated with different wavelengths but the same value of m is Δθ = (m/d)(λ λ 1 ), and their separation on a screen a distance D away is

2 md Δy Dtan Δθ DΔθ l l1 d L 310 = M b. mg O 9 9 P =. c m mh m = = L NM N The small angle approximation tan Δθ Δθ (in radians) is made. Q O QP b g 5 m. Problem 35-5 Let the distance in question be x. The path difference (between rays originating from S 1 and S and arriving at points on the x > 0 axis) is F I HG K J l, d + x x = m+ 1 where we are requiring destructive interference (half-integer wavelength phase differences) and m = 0,1,,. After some algebraic steps, we solve for the distance in terms of m: d bm + 1gl x =. m + 1 l 4 To obtain the largest value of x, we set m = 0: ( 3.00λ) b g d λ λ 3 x0 = = = 8.75λ = 8.75(900 nm) = nm = 7.88μm. λ 4 λ 4 Problem 35-6 (a) We use Eq to find d: d sinθ = mλ d = (4)(450 nm)/sin(90 ) = 1800 nm. For the third order spectrum, the wavelength that corresponds to θ = 90 is λ = d sin(90 )/3 = 600 nm. Any wavelength greater than this will not be seen. Thus, 600 nm < θ 700 nm are absent. (b) The slit separation d needs to be decreased. (c) In this case, the 400 nm wavelength in the m = 4 diffraction is to occur at 90. Thus

3 This represents a change of d new sinθ = mλ d new = (4)(400 nm)/sin(90 ) = 1600 nm. Δd = d d new = 00 nm = 0.0 µm. Problem For constructive interference, we use Eq : For the smallest value of L, let m = 0: λ 64nm L0 = = = 117nm = μ m. n ( ) b nl= m+ 1 gλ. (b) For the second smallest value, we set m = 1 and obtain ( ) 1+ 1 λ 3λ L = = = 3L = m = 0.35μm. 1 0 n n ( μ ) Problem The incident light is in a low index medium, the thin film of acetone has somewhat higher n = n, and the last layer (the glass plate) has the highest refractive index. To see very little or no reflection, the condition 1 L= ( m+ ) l where m= n 0, 1,, must hold. This is the same as Eq which was developed for the opposite situation (constructive interference) regarding a thin film surrounded on both sides by air (a very different context than the one in this problem). By analogy, we expect Eq to apply in this problem to reflection maxima. A more careful analysis such as that given in Section 35-7 bears this out. Thus, using Eq with n = 1.5 and λ = 700 nm yields L = 0, 80nm, 560nm, 840nm,110nm, for the first several m values. And the equation shown above (equivalent to Eq ) gives, with λ = 600 nm, L = 10nm,360nm,600nm,840nm,1080nm, for the first several m values. The lowest number these lists have in common is L = 840 nm.

4 Problem The index of refraction of oil is greater than that of the air, but smaller than that of the water. Let the indices of refraction of the air, oil and water be n 1, n, and n 3, respectively. Since n1 < n and n < n 3, there is a phase change of π rad from both surfaces. Since the second wave travels an additional distance of L, the phase difference is π φ = ( L) λ where λ = λ / n is the wavelength in the oil. The condition for constructive interference is π ( L) = m π, λ or λ L= m, m= 0,1,,... n (a) For m =1,,..., maximum reflection occurs for wavelengths ( )( ) nl nm l = 1104nm, 55nm, 368nm... m = m = We note that only the 55 nm wavelength falls within the visible light range. (b) Maximum transmission into the water occurs for wavelengths for which reflection is a minimum. The condition for such destructive interference is given by F 1I l L= m+ HG K J l = n 4nL m + 1 which yields λ = 08 nm, 736 nm, 44 nm for the different values of m. We note that only the 44-nm wavelength (blue) is in the visible range, though we might expect some red contribution since the 736 nm is very close to the visible range. Note: A light ray reflected by a material changes phase by π rad (or 180 ) if the refractive index of the material is greater than that of the medium in which the light is traveling. Otherwise, there is no phase change. Note that refraction at an interface does not cause a phase shift. Problem 35-7 We apply Eq to both scenarios: m = 4001 and n = n air, and m = 4000 and n = n vacuum = :

5 b g b g l l L = 4001 and L = n air Since the L factor is the same in both cases, we set the right hand sides of these expressions equal to each other and cancel the wavelength. Finally, we obtain b g. n air = = We remark that this same result can be obtained starting with Eq (which is developed in the textbook for a somewhat different situation) and using Eq to eliminate the L/λ term. Problem 35-8 We apply Eq to both wavelengths and take the difference: L L 1 1 N1 N = = L. λ1 λ λ1 λ We now require N 1 N = 1 and solve for L: L = = = = λ1 λ nm nm μ nm 91 m. Problem (a) The path length difference between Rays 1 and is 7d d = 5d. For this to correspond to a half-wavelength requires 5d = λ/, so that d = 50.0 nm. (b) The above requirement becomes 5d = λ/n in the presence of the solution, with n = Therefore, d = 36. nm. Problem The condition for a minimum in the two-slit interference pattern is d sin θ = (m + ½)λ, where d is the slit separation, λ is the wavelength, m is an integer, and θ is the angle made by the interfering rays with the forward direction. If θ is small, sin θ may be approximated by θ in radians. Then, θ = (m + ½)λ/d, and the distance from the minimum to the central fringe is 1 Dλ y = Dtanθ Dsinθ Dθ = m+, d where D is the distance from the slits to the screen. For the first minimum m = 0 and for the tenth one, m = 9. The separation is

6 1 Dλ 1 Dλ 9Dλ Δ y = 9 + =. d d d We solve for the wavelength: 3 3 dδy c mhc18 10 mh 7 l = = = m = 600 nm. 9D m c h Note: The distance between two adjacent dark fringes, one associated with the integer m and the other associated with the integer m + 1, is Δy = Dθ = Dλ/d.