M36 YOUN7066 13 ISM c36 tủ tài liệu training

36 DIFFRACTION 36.1 IDENTIFY: Use y = x tan θ to calculate the angular position θ of the first minimum The minima are mλ , m = ±1, ± 2,… First minimum means m = and sin θ1 = λ /a and a λ = a sin θ1 Use this equation to calculate λ located by Eq (36.2): sin θ = SET UP: The central maximum is sketched in Figure 36.1 EXECUTE: tan θ1 = y1 = x tan θ1 y1 = x 1.35 × 10−3 m = 0.675 × 10−3 2.00 m θ1 = 0.675 × 10−3 rad Figure 36.1 λ = a sin θ1 = (0.750 × 10−3 m)sin(0.675 × 10−3 rad) = 506 nm EVALUATE: θ1 is small so the approximation used to obtain Eq (36.3) is valid and this equation could have been used 36.2 IDENTIFY: The angle is small, so ym = x SET UP: y1 = 10.2 mm xλ xλ (0.600 m)(5.46 × 10−7 m) ⇒a= = = 3.21 × 10−5 m a y1 10.2 × 10−3 m EVALUATE: The diffraction pattern is observed at a distance of 60.0 cm from the slit mλ IDENTIFY: The dark fringes are located at angles θ that satisfy sin θ = , m = ±1, ± 2, … a SET UP: The largest value of sin θ is 1.00 EXECUTE: 36.3 mλ a y1 = EXECUTE: (a) Solve for m that corresponds to sin θ = 1: m = a = 0.0666 × 10−3 m 585 × 10−9 m value m can have is 113 m = ±1, ± 2, …, ±113 gives 226 dark fringes λ = 113.8 The largest ⎛ 585 × 10−9 m ⎞ = ±0.9926 and θ = ±83.0° (b) For m = ±113, sin θ = ±113 ⎜ ⎜ 0.0666 × 10−3 m ⎟⎟ ⎝ ⎠ EVALUATE: When the slit width a is decreased, there are fewer dark fringes When a < λ there are no dark fringes and the central maximum completely fills the screen © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 36-1 36-2 36.4 Chapter 36 mλ is accurate The a distance between the two dark fringes on either side of the central maximum is y1 IDENTIFY and SET UP: λ /a is very small, so the approximate expression ym = R λR (633 × 10−9 m)(3.50 m) = 2.95 × 10−3 m = 2.95 mm y1 = 5.90 mm a 0.750 × 10−3 m EVALUATE: When a is decreased, the width y1 of the central maximum increases EXECUTE: 36.5 y1 = = IDENTIFY: The minima are located by sin θ = SET UP: a = 12.0 cm x = 8.00 m 36.6 mλ a ⎛ 9.00 cm ⎞ ⎛λ⎞ EXECUTE: The angle to the first minimum is θ = arcsin ⎜ ⎟ = arcsin ⎜ ⎟ = 48.6° a ⎝ ⎠ ⎝ 12.00 cm ⎠ So the distance from the central maximum to the first minimum is just y1 = x tan θ = (8.00 m) tan(48.6°) = ± (9.07 m) EVALUATE: 2λ /a is greater than 1, so only the m = minimum is seen IDENTIFY: The angle that locates the first diffraction minimum on one side of the central maximum is v λ given by sin θ = The time between crests is the period T f = and λ = a T f SET UP: The time between crests is the period, so T = 1.0 h v 800 km/h 1 = = 1.0 h −1 λ = = = 800 km T 1.0 h f 1.0 h −1 800 km (b) Africa-Antarctica: sin θ = and θ = 10.2° 4500 km 800 km and θ = 12.5° Australia-Antarctica: sin θ = 3700 km EVALUATE: Diffraction effects are observed when the wavelength is about the same order of magnitude as the dimensions of the opening through which the wave passes IDENTIFY: We can model the hole in the concrete barrier as a single slit that will produce a single-slit diffraction pattern of the water waves on the shore SET UP: For single-slit diffraction, the angles at which destructive interference occurs are given by sin θ m = mλ /a, where m = 1, 2, 3, … EXECUTE: (a) f = 36.7 EXECUTE: (a) The frequency of the water waves is f = 75.0 −1 = 1.25 s −1 = 1.25 Hz, so their wavelength is λ = v/f = (15.0 cm/s)/(1.25 Hz) = 12.0 cm At the first point for which destructive interference occurs, we have tan θ = (0.613 m)/(3.20 m) ⇒ θ = 10.84° a sin θ = λ and a = λ / sin θ = (12.0 cm)/(sin 10.84°) = 63.8 cm (b) First find the angles at which destructive interference occurs sin θ = 2λ /a = 2(12.0 cm)/(63.8 cm) → θ = ±22.1° sin θ3 = 3λ /a = 3(12.0 cm)/(63.8 cm) → θ3 = ±34.3° sin θ = 4λ /a = 4(12.0 cm)/(63.8 cm) → θ = ±48.8° sin θ5 = 5λ /a = 5(12.0 cm)/(63.8 cm) → θ5 = ±70.1° EVALUATE: These are large angles, so we cannot use the approximation that θ m ≈ mλ /a 36.8 mλ applies a SET UP: The width of the central maximum is y1, so y1 = 3.00 mm IDENTIFY: The angle is small, so ym = x © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Diffraction EXECUTE: (a) y1 = (b) a = 36-3 xλ xλ (2.50 m)(5.00 × 10−7 m) ⇒a= = = 4.17 × 10−4 m a y1 3.00 × 10−3 m xλ (2.50 m)(5.00 × 10−5 m) = = 4.17 × 10−2 m = 4.2 cm y1 3.00 × 10−3 m xλ (2.50 m)(5.00 × 10−10 m) = = 4.17 × 10−7 m y1 3.00 × 10−3 m EVALUATE: The ratio a/λ stays constant, so a is smaller when λ is smaller IDENTIFY and SET UP: v = f λ gives λ The person hears no sound at angles corresponding to diffraction minima The diffraction minima are located by sin θ = mλ /a, m = ±1, ± 2,… Solve for θ EXECUTE: λ = v/f = (344 m/s)/(1250 Hz) = 0.2752 m; a = 1.00 m m = ±1, θ = ±16.0°; m = ±2, θ = ±33.4°; m = ±3, θ = ±55.6°; no solution for larger m EVALUATE: λ /a = 0.28 so for the large wavelength sound waves diffraction by the doorway is a large effect Diffraction would not be observable for visible light because its wavelength is much smaller and λ /a IDENTIFY: Compare E y to the expression E y = Emax sin(kx − ωt ) and determine k, and from that (c) a = 36.9 36.10 calculate λ f = c/λ The dark bands are located by sin θ = mλ a SET UP: c = 3.00 × 108 m/s The first dark band corresponds to m = 2π 2π 2π EXECUTE: (a) E = Emax sin(kx − ωt ) k = ⇒λ = = = 5.24 × 10−7 m λ k 1.20 × 107 m −1 fλ =c⇒ f = c λ = (b) a sin θ = λ a = 3.0 × 108 m/s 5.24 × 10−7 m λ sin θ = = 5.73 × 1014 Hz 5.24 × 10−7 m = 1.09 × 10−6 m sin 28.6° (c) a sin θ = mλ (m = 1, 2, 3, …) sin θ = ±2 λ a = ±2 5.24 × 10−7 m 1.09 × 10−6 m and θ = ±74° mλ is greater than so only the first and second dark bands appear a IDENTIFY and SET UP: sinθ = λ /a locates the first minimum y = x tan θ EXECUTE: tan θ = y/x = (36.5 cm)/(40.0 cm) and θ = 42.38° EVALUATE: For m = 3, 36.11 36.12 a = λ / sin θ = (620 × 10−9 m)/(sin 42.38°) = 0.920 μ m EVALUATE: θ = 0.74 rad and sin θ = 0.67, so the approximation sin θ ≈ θ would not be accurate IDENTIFY: Calculate the angular positions of the minima and use y = x tan θ to calculate the distance on the screen between them (a) SET UP: The central bright fringe is shown in Figure 36.12a EXECUTE: The first minimum is located by λ 633 × 10−9 m sin θ1 = = = 1.809 × 10−3 a 0.350 × 10−3 m θ1 = 1.809 × 10−3 rad Figure 36.12a y1 = x tan θ1 = (3.00 m) tan(1.809 × 10−3 rad) = 5.427 × 10−3 m w = y1 = 2(5.427 × 10−3 m) = 1.09 × 10−2 m = 10.9 mm © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 36-4 Chapter 36 (b) SET UP: The first bright fringe on one side of the central maximum is shown in Figure 36.12b EXECUTE: w = y2 − y1 y1 = 5.427 × 10−3 m (part (a)) sin θ = 2λ = 3.618 × 10−3 a θ = 3.618 × 10−3 rad y2 = x tan θ = 1.085 × 10−2 m Figure 36.12b w = y2 − y1 = 1.085 × 10−2 m − 5.427 × 10−3 m = 5.4 mm 36.13 EVALUATE: The central bright fringe is twice as wide as the other bright fringes mλ IDENTIFY: The minima are located by sin θ = For part (b) apply Eq (36.7) a SET UP: For the first minimum, m = The intensity at θ = is I mλ mλ λ = sin 90.0° = = = Thus a = λ = 580 nm = 5.80 × 10−4 mm a a a (b) According to Eq (36.7), EXECUTE: (a) sinθ = 2 I ⎧ sin[π a(sin θ )/λ ] ⎫ ⎧ sin[π (sin π /4)] ⎫ =⎨ ⎬ =⎨ ⎬ = 0.128 I ⎩ π a (sin θ )/λ ⎭ ⎩ π (sin π /4) ⎭ I ⎧ sin[(π /2)(sin π /4)] ⎫ =⎨ ⎬ = 0.65 As a/λ I ⎩ (π /2)(sin π /4) ⎭ decreases, the screen becomes more uniformly illuminated EVALUATE: If a = λ /2, for example, then at θ = 45°, 36.14 ⎛ sin( β /2) ⎞ 2π IDENTIFY: I = I ⎜ a sin θ ⎟ β= λ ⎝ β /2 ⎠ SET UP: The angle θ is small, so sin θ ≈ tan θ ≈ y/x EXECUTE: β = 2π a λ (a) y = 1.00 × 10−3 m: sin θ ≈ β = 2π a y 2π (4.50 × 10−4 m) = y = (1520 m −1 ) y λ x (6.20 × 10−7 m)(3.00 m) (1520 m −1 )(1.00 × 10−3 m) = 0.760 2 ⎛ sin( β /2) ⎞ ⎛ sin (0.760) ⎞ ⇒ I = I0 ⎜ ⎟ = I0 ⎜ ⎟ = 0.822 I β /2 ⎝ 0.760 ⎠ ⎝ ⎠ (b) y = 3.00 × 10−3 m: β = (1520 m −1)(3.00 × 10−3 m) = 2.28 2 ⎛ sin( β /2) ⎞ ⎛ sin (2.28) ⎞ ⇒ I = I0 ⎜ ⎟ = I0 ⎜ ⎟ = 0.111I ⎝ 2.28 ⎠ ⎝ β /2 ⎠ (c) y = 5.00 × 10−3 m: β = (1520 m −1)(5.00 × 10−3 m) = 3.80 2 ⎛ sin( β /2) ⎞ ⎛ sin (3.80) ⎞ ⇒ I = I0 ⎜ ⎟ = I0 ⎜ ⎟ = 0.0259 I ⎝ 3.80 ⎠ ⎝ β /2 ⎠ EVALUATE: The first minimum occurs at y1 = λx (6.20 × 10−7 m)(3.00 m) = 4.1 mm The distances in 4.50 × 10−4 m parts (a) and (b) are within the central maximum y = 5.00 mm is within the first secondary maximum a = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Diffraction 36.15 36-5 (a) IDENTIFY: Use Eq (36.2) with m = to locate the angular position of the first minimum and then use y = x tan θ to find its distance from the center of the screen SET UP: The diffraction pattern is sketched in Figure 36.15 sin θ1 = λ a = 540 × 10−9 m 0.240 × 10−3 m = 2.25 × 10−3 θ1 = 2.25 × 10−3 rad Figure 36.15 y1 = x tan θ1 = (3.00 m) tan(2.25 × 10−3 rad) = 6.75 × 10−3 m = 6.75 mm (b) IDENTIFY and SET UP: Use Eqs (36.5) and (36.6) to calculate the intensity at this point EXECUTE: Midway between the center of the central maximum and the first minimum implies y = (6.75 mm) = 3.375 × 10−3 m y 3.375 × 10−3 m = = 1.125 × 10−3 ; θ = 1.125 × 10−3 rad x 3.00 m The phase angle β at this point on the screen is tan θ = 2π ⎛ 2π ⎞ a sin θ = (0.240 × 10−3 m)sin(1.125 × 10−3 rad) = π ⎟ ⎝ λ ⎠ 540 × 10−9 m β =⎜ 2 ⎛ sin β /2 ⎞ −6 ⎛ sin π /2 ⎞ Then I = I ⎜ ⎟ = (6.00 × 10 W/m ) ⎜ ⎟ ⎝ π /2 ⎠ ⎝ β /2 ⎠ ⎛ 4⎞ I = ⎜ ⎟ (6.00 × 10−6 W/m ) = 2.43 × 10−6 W/m ⎝π ⎠ 36.16 EVALUATE: The intensity at this point midway between the center of the central maximum and the first minimum is less than half the maximum intensity Compare this result to the corresponding one for the two-slit pattern, Exercise 35.22 IDENTIFY: The intensity on the screen varies as the light spreads out (diffracts) after passing through the single slit ⎡ sin( β /2) ⎤ 2π where β = SET UP: I = I ⎢ a sin θ ⎥ λ ⎣ β /2 ⎦ EXECUTE: β = 2π λ 2π ⎛ ⎞ a sin θ = ⎜ (0.0290 × 10−3 m)sin1.20o = 7.852 rad ⎝ 486 × 10−9 m ⎟⎠ 36.17 ⎡ sin( β /2) ⎤ −5 −6 ⎡ sin(3.926 rad) ⎤ I = I0 ⎢ ⎥ = (4.00 × 10 W/m ) ⎢ 3.926 rad ⎥ = 1.29 × 10 W/m β /2 ⎣ ⎦ ⎣ ⎦ EVALUATE: The intensity is less than 1/30 of the intensity of the light at the slit IDENTIFY and SET UP: Use Eq (36.6) to calculate λ and use Eq (36.5) to calculate I θ = 3.25°, β = 56.0 rad, a = 0.105 × 10−3 m ⎛ 2π ⎞ (a) EXECUTE: β = ⎜ ⎟ a sin θ so ⎝ λ⎠ λ= 2π a sin θ β = 2π (0.105 × 10−3 m)sin 3.25° = 668 nm 56.0 rad © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 36-6 Chapter 36 ⎛ ⎞ ⎛ sin β /2 ⎞ [sin(28.0 rad)]2 = 9.36 × 10−5 I (b) I = I ⎜ ⎟ = I ⎜⎜ ⎟⎟ (sin( β /2)) = I β (56 rad) ⎝ β /2 ⎠ ⎝ ⎠ EVALUATE: At the first minimum β = 2π rad and at the point considered in the problem β = 17.8π rad, so the point is well outside the central maximum Since β is close to mπ with m = 18, this point is near one of the minima The intensity here is much less than I 36.18 IDENTIFY: Use β = 2π a λ sin θ to calculate β SET UP: The total intensity is given by drawing an arc of a circle that has length E0 and finding the length of the chord which connects the starting and ending points of the curve Ep 2π a 2π a λ EXECUTE: (a) β = = π From Figure 36.18a, π = E0 ⇒ E p = E0 sin θ = λ λ 2a π 4I ⎛2⎞ The intensity is I = ⎜ ⎟ I = 20 = 0.405I This agrees with Eq (36.5) π π ⎝ ⎠ 2π a 2π a λ (b) β = = 2π From Figure 36.18b, it is clear that the total amplitude is zero, as is the sin θ = λ λ a intensity This also agrees with Eq (36.5) Ep 2π a 2π a 3λ sin θ = (c) β = = 3π From Figure 36.18c, 3π = E0 ⇒ E p = E0 The intensity is 3π λ λ 2a ⎛ 2⎞ I = ⎜ ⎟ I = I This agrees with Eq (36.5) ⎝ 3π ⎠ 9π EVALUATE: In part (a) the point is midway between the center of the central maximum and the first minimum In part (b) the point is at the first maximum and in (c) the point is approximately at the location of the first secondary maximum The phasor diagrams help illustrate the rapid decrease in intensity at successive maxima Figure 36.18 36.19 IDENTIFY: The space between the skyscrapers behaves like a single slit and diffracts the radio waves SET UP: Cancellation of the waves occurs when a sin θ = mλ , m = 1, 2, 3, …, and the intensity of the ⎛ sin β /2 ⎞ π a sin θ waves is given by I ⎜ ⎟ , where β /2 = λ ⎝ β /2 ⎠ EXECUTE: (a) First find the wavelength of the waves: λ = c/f = (3.00 × 108 m/s)/(88.9 MHz) = 3.375 m For no signal, a sin θ = mλ m = 1: sinθ1 = (1)(3.375 m)/(15.0 m) ⇒ θ l = ±13.0° m = 2: sinθ = (2)(3.375 m)/(15.0 m) ⇒ θ = ±26.7° m = 3: sinθ3 = (3)(3.375 m)/(15.0 m) ⇒ θ3 = ±42.4° m = 4: sinθ = (4)(3.375 m)/(15.0 m) ⇒ θ = ±64.1° © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Diffraction 36-7 ⎛ sin β /2 ⎞ π a sin θ π (15.0 m)sin(5.00°) , where β /2 = (b) I ⎜ = = 1.217 rad ⎝ β /2 ⎟⎠ λ 3.375 m 36.20 ⎡ sin(1.217 rad) ⎤ I = (3.50 W/m ) ⎢ ⎥ = 2.08 W/m ⎣ 1.217 rad ⎦ EVALUATE: The wavelength of the radio waves is very long compared to that of visible light, but it is still considerably shorter than the distance between the buildings IDENTIFY: The net intensity is the product of the factor due to single-slit diffraction and the factor due to double slit interference φ⎞ ⎛ sin β /2 ⎞ ⎛ SET UP: The double-slit factor is I DS = I ⎜ cos ⎟ and the single-slit factor is ISS = ⎜ ⎟ ⎝ 2⎠ ⎝ β /2 ⎠ EXECUTE: (a) d sin θ = mλ ⇒ sin θ = mλ /d sin θ1 = λ /d , sin θ = 2λ /d , sinθ3 = 3λ /d , sin θ = 4λ /d (b) At the interference bright fringes, cos φ /2 = and β /2 = At θ1, sin θ1 = λ /d , so β /2 = π a sin θ π (d/3)sin θ = λ λ π (d/3)(λ /d ) = π /3 The intensity is therefore λ 2 φ ⎞ ⎛ sin β /2 ⎞ ⎛ ⎛ sin π /3 ⎞ I1 = I ⎜ cos ⎟ ⎜ ⎟ = I (1) ⎜ ⎟ = 0.684 I ⎠ ⎝ β /2 ⎠ ⎝ ⎝ π /3 ⎠ At θ , sin θ = 2λ /d , so β /2 = π ( d/3)(2λ /d ) = 2π /3 Using the same procedure as for θ1, we have λ ⎛ sin 2π /3 ⎞ I = I (1) ⎜ ⎟ = 0.171 I ⎝ 2π /3 ⎠ At θ3 , we get β /2 = π , which gives I = since sin π = ⎛ sin 4π /3 ⎞ At θ , sin θ = 4λ /d , so β /2 = 4π /3, which gives I = I ⎜ = 0.0427 I ⎝ 4π /3 ⎟⎠ 36.21 (c) Since d = 3a, every third interference maximum is missing (d) In Figure 36.12c in the textbook, every fourth interference maximum at the sides is missing because d = 4a EVALUATE: The result in this problem is different from that in Figure 36.12c in the textbook because in this case d = 3a, so every third interference maximum at the sides is missing Also the “envelope” of the intensity function decreases more rapidly here than in Figure 36.12c in the text because the first diffraction minimum is reached sooner, and the decrease in intensity from one interference maximum to the next is faster for a = d/3 than for a = d/4 (a) IDENTIFY and SET UP: The interference fringes (maxima) are located by d sin θ = mλ , with ⎛ sin β /2 ⎞ m = 0, ± 1, ± 2, … The intensity I in the diffraction pattern is given by I = I ⎜ ⎟ , with ⎝ β /2 ⎠ ⎛ 2π ⎞ a sin θ We want m = ±3 in the first equation to give θ that makes I = in the second equation ⎝ λ ⎟⎠ β =⎜ ⎛ 2π ⎞ ⎛ 3λ ⎞ EXECUTE: d sin θ = mλ gives β = ⎜ ⎟ a ⎜ ⎟ = 2π (3a/d ) ⎝ λ ⎠ ⎝ d ⎠ sin β /2 = so β = 2π and then 2π = 2π (3a/d ) and (d/a ) = β /2 (b) IDENTIFY and SET UP: Fringes m = 0, ± 1, ± are within the central diffraction maximum and the m = ±3 fringes coincide with the first diffraction minimum Find the value of m for the fringes that coincide with the second diffraction minimum I = says © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 36-8 Chapter 36 EXECUTE: Second minimum implies β = 4π ⎛ 2π ⎞ ⎛ 2π ⎞ ⎛ mλ ⎞ = 2π m(a/d ) = 2π (m/3) a sin θ = ⎜ ⎟ a ⎜ ⎝ λ ⎟⎠ ⎝ λ ⎠ ⎝ d ⎟⎠ β =⎜ Then β = 4π says 4π = 2π (m/3) and m = Therefore the m = ±4 and m = +5 fringes are contained 36.22 36.23 within the first diffraction maximum on one side of the central maximum; two fringes EVALUATE: The central maximum is twice as wide as the other maxima so it contains more fringes IDENTIFY and SET UP: Use Figure 36.14b in the textbook There is totally destructive interference between slits whose phasors are in opposite directions EXECUTE: By examining the diagram, we see that every fourth slit cancels each other EVALUATE: The total electric field is zero so the phasor diagram corresponds to a point of zero intensity The first two maxima are at φ = and φ = π , so this point is not midway between two maxima (a) IDENTIFY and SET UP: If the slits are very narrow then the central maximum of the diffraction pattern for each slit completely fills the screen and the intensity distribution is given solely by the two-slit interference The maxima are given by d sin θ = mλ so sin θ = mλ /d Solve for θ EXECUTE: 1st order maximum: m = 1, so sin θ = 2nd order maximum: m = 2, so sin θ = λ d = 580 × 10−9 m 0.530 × 10−3 m = 1.094 × 10−3 ; θ = 0.0627° 2λ = 2.188 × 10−3 ; θ = 0.125° d ⎛ sinβ /2 ⎞ (b) IDENTIFY and SET UP: The intensity is given by Eq (36.12): I = I cos (φ /2) ⎜ ⎟ Calculate ⎝ β /2 ⎠ φ and β at each θ from part (a) ⎛ 2π d ⎞ ⎛ 2π d ⎞ ⎛ mλ ⎞ EXECUTE: φ = ⎜ = 2π m, so cos (φ /2) = cos (mπ ) = sin θ = ⎜ ⎝ λ ⎟⎠ ⎝ λ ⎟⎠ ⎜⎝ d ⎟⎠ (Since the angular positions in part (a) correspond to interference maxima.) ⎛ 2π a ⎞ ⎛ 2π a ⎞⎛ mλ ⎞ ⎛ 0.320 mm ⎞ β =⎜ ⎟ sin θ = ⎜ ⎟⎜ ⎟ = 2π m( a/d ) = m 2π ⎜ ⎟ = m(3.794 rad) ⎝ λ ⎠ ⎝ λ ⎠⎝ d ⎠ ⎝ 0.530 mm ⎠ ⎛ sin(3.794/2)rad ⎞ 1st order maximum: m = 1, so I = I (1) ⎜ = 0.249 I ⎝ (3.794/2)rad ⎟⎠ ⎛ sin 3.794 rad ⎞ 2nd order maximum: m = 2, so I = I (1) ⎜ = 0.0256 I ⎝ 3.794 rad ⎟⎠ EVALUATE: The first diffraction minimum is at an angle θ given by sin θ = λ /a so θ = 0.104° The first order fringe is within the central maximum and the second order fringe is inside the first diffraction maximum on one side of the central maximum The intensity here at this second fringe is much less than I 36.24 IDENTIFY: The intensity at the screen is due to a combination of single-slit diffraction and double-slit interference φ ⎞ ⎡ sin(β /2) ⎤ 2π d 2π ⎛ , where φ = SET UP: I = I ⎜ cos ⎟ ⎢ sin θ and β = a sin θ ⎝ ⎠ ⎣ β /2 ⎥⎦ λ λ EXECUTE: tan θ = φ= β= 2π d λ 2π a λ sin θ = sin θ = 9.00 × 10−4 m = 1.200 × 10−3 θ is small, so sin θ ≈ tan θ 0.750 m 2π (0.640 × 10−3 m) 568 × 10−9 m 2π (0.434 × 10−3 m) 568 × 10 −9 m (1.200 × 10−3 ) = 8.4956 rad (1.200 × 10−3 ) = 5.7611 rad ⎡ sin 2.8805 rad ⎤ −7 I = (5.00 × 10−4 W/m )(cos 4.2478 rad)2 ⎢ ⎥ = 8.06 × 10 W/m 2.8805 ⎣ ⎦ EVALUATE: The intensity as decreased by a factor of almost a thousand, so it would be difficult to see the light at the screen © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Diffraction 36.25 36-9 IDENTIFY and SET UP: The phasor diagrams are similar to those in Figure 36.14 in the textbook An interference minimum occurs when the phasors add to zero EXECUTE: (a) The phasor diagram is given in Figure 36.25a Figure 36.25a There is destructive interference between the light through slits and and between and (b) The phasor diagram is given in Figure 36.25b Figure 36.25b There is destructive interference between the light through slits and and between and (c) The phasor diagram is given in Figure 36.25c Figure 36.25c 36.26 There is destructive interference between light through slits and and between and EVALUATE: Maxima occur when φ = 0, 2π , 4π , etc Our diagrams show that there are three minima between the maxima at φ = and φ = 2π This agrees with the general result that for N slits there are N − minima between each pair of principal maxima IDENTIFY: A double-slit bright fringe is missing when it occurs at the same angle as a double-slit dark fringe SET UP: Single-slit diffraction dark fringes occur when a sin θ = mλ , and double-slit interference bright fringes occur when d sin θ = m′λ EXECUTE: (a) The angle at which the first bright fringe occurs is given by tan θ1 = (1.53 mm)/(2500 mm) ⇒ θ1 = 0.03507° d sin θ1 = λ and d = λ /(sin θ1) = (632.8 nm)/ sin(0.03507°) = 0.00103 m = 1.03 mm (b) The double-slit interference bright fringe is just cancelled by the 1st diffraction dark fringe, so sin θdiff = λ /a and sin θinterf = 7λ /d The angles are equal, so λ /a = 7λ /d → a = d/7 = (1.03 mm)/7 = 0.148 mm th 36.27 EVALUATE: We can generalize that if d = na, where n is a positive integer, then every n th double-slit bright fringe will be missing in the pattern mλ IDENTIFY: The diffraction minima are located by sin θ = d and the two-slit interference maxima are a miλ The third bright band is missing because the first order single-slit minimum occurs located by sin θ = d at the same angle as the third order double-slit maximum cm , so θ = 1.91° SET UP: The pattern is sketched in Figure 36.27 tan θ = 90 cm © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 36-10 Chapter 36 λ 500 nm = = 1.50 × 104 nm = 15.0 μ m (width) sin θ sin1.91° 3λ 3(500 nm) = = 4.50 × 104 nm = 45.0 μ m (separation) Double-slit bright fringe: d sin θ = 3λ and d = sin θ sin1.91° EVALUATE: Note that d/a = 3.0 EXECUTE: Single-slit dark spot: a sin θ = λ and a = Figure 36.27 36.28 36.29 IDENTIFY: The maxima are located by d sin θ = mλ SET UP: The order corresponds to the values of m EXECUTE: First-order: d sin θ1 = λ Fourth-order: d sin θ = 4λ d sin θ 4λ = , sin θ = 4sin θ1 = 4sin 8.94° and θ = 38.4° d sin θ1 λ EVALUATE: We did not have to solve for d IDENTIFY and SET UP: The bright bands are at angles θ given by d sin θ = mλ Solve for d and then solve for θ for the specified order EXECUTE: (a) θ = 78.4° for m = and λ = 681 nm, so d = mλ / sin θ = 2.086 × 10−4 cm The number of slits per cm is 1/d = 4790 slits/cm (b) 1st order: m = 1, so sin θ = λ /d = (681 × 10−9 m)/(2.086 × 10−6 m) and θ = 19.1° 2nd order: m = 2, so sin θ = 2λ /d and θ = 40.8° (c) For m = 4, sin θ = 4λ /d is greater than 1.00, so there is no 4th-order bright band 36.30 36.31 EVALUATE: The angular position of the bright bands for a particular wavelength increases as the order increases IDENTIFY: The bright spots are located by d sin θ = mλ SET UP: Third-order means m = and second-order means m = mλ mλ mλ EXECUTE: = d = constant, so r r = v v sin θ r sin θ v sin θ ⎛ m ⎞⎛ λ ⎞ ⎛ ⎞⎛ 400 nm ⎞ sin θ v = sin θ r ⎜ v ⎟⎜ v ⎟ = (sin 65.0°) ⎜ ⎟⎜ ⎟ = 0.345 and θ v = 20.2° ⎝ ⎠⎝ 700 nm ⎠ ⎝ mr ⎠⎝ λr ⎠ EVALUATE: The third-order line for a particular λ occurs at a larger angle than the second-order line In a given order, the line for violet light (400 nm) occurs at a smaller angle than the line for red light (700 nm) IDENTIFY and SET UP: Calculate d for the grating Use Eq (36.13) to calculate θ for the longest wavelength in the visible spectrum and verify that θ is small Then use Eq (36.3) to relate the linear separation of lines on the screen to the difference in wavelength ⎛ ⎞ cm = 1.111 × 10−5 m EXECUTE: (a) d = ⎜ ⎝ 900 ⎟⎠ For λ = 700 nm, λ /d = 6.3 × 10−2 The first-order lines are located at sin θ = λ /d ; sin θ is small enough for sin θ ≈ θ to be an excellent approximation (b) y = xλ /d , where x = 2.50 m The distance on the screen between first-order bright bands for two different wavelengths is Δy = x (Δλ)/d , so Δλ = d ( Δy )/x = (1.111 × 10−5 m)(3.00 × 10−3 m)/(2.50 m) = 13.3 nm EVALUATE: The smaller d is (greater number of lines per cm) the smaller the Δλ that can be measured © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Diffraction 36.32 36.33 IDENTIFY: The maxima are located by d sin θ = mλ SET UP: 350 slits/mm ⇒ d = = 2.86 × 10−6 m 3.50 × 105 m −1 ⎛ 4.00 × 10−7 m ⎞ ⎛λ ⎞ EXECUTE: (a) m = 1: θ 400 = arcsin ⎜ ⎟ = arcsin ⎜ ⎟⎟ = 8.05° −6 ⎜ ⎝d ⎠ ⎝ 2.86 × 10 m ⎠ ⎛ 7.00 × 10−7 m ⎞ ⎛λ⎞ θ 700 = arcsin ⎜ ⎟ = arcsin ⎜ ⎟⎟ = 14.18° Δθ1 = 14.18° − 8.05° = 6.13° −6 ⎜ ⎝d ⎠ ⎝ 2.86 × 10 m ⎠ ⎛ 3(4.00 × 10−7 m) ⎞ ⎛ 3λ ⎞ (b) m = 3: θ 400 = arcsin ⎜ ⎟ = arcsin ⎜ = 24.8° ⎜ 2.86 × 10−6 m ⎟⎟ ⎝ d ⎠ ⎝ ⎠ ⎛ 3(7.00 × 10−7 m) ⎞ ⎛ 3λ ⎞ θ 700 = arcsin ⎜ ⎟ = arcsin ⎜ = 47.3° Δθ1 = 47.3° − 24.8° = 22.5° ⎜ 2.86 × 10−6 m ⎟⎟ ⎝ d ⎠ ⎝ ⎠ EVALUATE: Δθ is larger in third order IDENTIFY: Knowing the wavelength of the light and the location of the first interference maxima, we can calculate the line density of the grating SET UP: The line density in lines/cm is 1/d , with d in cm The bright spots are located by d sin θ = mλ , m = 0, ± 1, ± 2, … EXECUTE: (a) d = (b) sin θ = 36.34 36-11 mλ (1)(632.8 × 10−9 m) = = 2.07 × 10−6 m = 2.07 × 10−4 cm = 4830 lines/cm d sin θ sin17.8° ⎛ 632.8 × 10−9 m ⎞ mλ = m⎜ = m(0.3057) For m = ±2, θ = ±37.7° For m = ±3, θ = ±66.5° ⎜ 2.07 × 10−6 m ⎟⎟ d ⎝ ⎠ EVALUATE: The angles are large, so they are not equally spaced; 37.7° ≠ 2(17.8°) and 66.5° ≠ 3(17.8°) IDENTIFY: The maxima are located by d sin θ = mλ SET UP: 5000 slits/cm ⇒ d = = 2.00 × 10−6 m 5.00 × 105 m −1 d sin θ (2.00 × 10−6 m)sin13.5° = = 4.67 × 10−7 m m ⎛ 2(4.67 × 10−7 m) ⎞ ⎛ mλ ⎞ (b) m = 2: θ = arcsin ⎜ ⎟⎟ = 27.8° ⎟ = arcsin ⎜⎜ −6 ⎝ d ⎠ ⎝ 2.00 × 10 m ⎠ EVALUATE: Since the angles are fairly small, the second-order deviation is approximately twice the firstorder deviation IDENTIFY: The maxima are located by d sin θ = mλ SET UP: 350 slits/mm ⇒ d = = 2.86 × 10−6 m 3.50 × 105 m −1 EXECUTE: (a) λ = 36.35 ⎛ m(5.20 × 10−7 m) ⎞ ⎛ mλ ⎞ EXECUTE: θ = arcsin ⎜ ⎟⎟ = arcsin((0.182) m) ⎟ = arcsin ⎜⎜ −6 ⎝ d ⎠ ⎝ 2.86 × 10 m ⎠ m = 1: θ = 10.5°; m = 2: θ = 21.3°; m = 3: θ = 33.1° EVALUATE: The angles are not precisely proportional to m, and deviate more from being proportional as the angles increase 36.36 IDENTIFY: The resolution is described by R = λ Δλ = Nm Maxima are located by d sin θ = mλ SET UP: For 500 slits/mm, d = (500 slits/mm) −1 = (500,000 slits/m) −1 EXECUTE: (a) N = λ mΔλ = 6.5645 × 10−7 m 2(6.5645 × 10−7 m − 6.5627 × 10−7 m) = 1820 slits © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 36-12 Chapter 36 ⎛ mλ ⎞ (b) θ = sin −1 ⎜ ⇒ θ1 = sin −1 ((2)(6.5645 × 10−7 m)(500,000 m −1 )) = 41.0297° and ⎝ d ⎟⎠ θ = sin −1((2)(6.5627 × 10−7 m)(500,000 m −1 )) = 41.0160° Δθ = 0.0137° EVALUATE: d cosθ dθ = λ /N , so for 1820 slits the angular interval Δθ between each of these maxima and the first adjacent minimum is Δθ = 36.37 λ 6.56 × 10−7 m = 0.0137° This is the (1820)(2.0 × 10−6 m)cos 41° same as the angular separation of the maxima for the two wavelengths and 1820 slits is just sufficient to resolve these two wavelengths in second order IDENTIFY: The resolving power depends on the line density and the width of the grating SET UP: The resolving power is given by R = Nm = = λ /Δλ EXECUTE: (a) R = Nm = (5000 lines/cm)(3.50 cm)(1) = 17,500 Nd cos θ = (b) The resolving power needed to resolve the sodium doublet is R = λ /Δλ = (589 nm)/(589.59 nm – 589.00 nm) = 998 so this grating can easily resolve the doublet (c) (i) R = λ /Δλ Since R = 17,500 when m = 1, R = × 17,500 = 35,000 for m = Therefore Δλ = λ /R = (587.8 nm)/35,000 = 0.0168 nm λmin = λ + Δλ = 587.8002 nm + 0.0168 nm = 587.8170 nm (ii) λmax = λ − Δλ = 587.8002 nm − 0.0168 nm = 587.7834 nm EVALUATE: (iii) Therefore the range of resolvable wavelengths is 587.7834 nm < λ < 587.8170 nm 36.38 EXECUTE: 36.39 λ = Nm Δλ IDENTIFY and SET UP: N= λ mΔλ = 587.8002 nm 587.8002 = = 3302 slits (587.9782 nm − 587.8002 nm) 0.178 3302 slits N = = 2752 1.20 cm 1.20 cm cm EVALUATE: A smaller number of slits would be needed to resolve these two lines in higher order IDENTIFY and SET UP: The maxima occur at angles θ given by Eq (36.16), 2d sin θ = mλ , where d is the spacing between adjacent atomic planes Solve for d EXECUTE: Second order says m = 2(0.0850 × 10−9 m) mλ = = 2.32 × 10−10 m = 0.232 nm 2sin θ 2sin 21.5° EVALUATE: Our result is similar to d calculated in Example 36.5 IDENTIFY: The maxima are given by 2d sin θ = mλ , m = 1, 2, … d= 36.40 SET UP: d = 3.50 × 10−10 m EXECUTE: (a) m = and λ = 2d sin θ = 2(3.50 × 10−10 m)sin15.0° = 1.81 × 10−10 m = 0.181 nm This is an m x ray ⎛ 1.81 × 10−10 m ⎞ ⎛ λ ⎞ (b) sin θ = m ⎜ ⎟ = m(0.2586) m = 2: θ = 31.1° m = 3: θ = 50.9° The equation ⎟ = m ⎜⎜ −10 m] ⎟⎠ ⎝ 2d ⎠ ⎝ 2[3.50 × 10 36.41 doesn’t have any solutions for m > EVALUATE: In this problem λ /d = 0.52 IDENTIFY: The crystal behaves like a diffraction grating SET UP: The maxima are at angles θ given by 2d sin θ = mλ , where d = 0.440 nm 2d sinθ = 2(0.440 nm)sin 39.4° = 0.559 nm EVALUATE: The result is a reasonable x ray wavelength EXECUTE: m = λ = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Diffraction 36.42 IDENTIFY: Apply sin θ = 1.22 SET UP: θ = (1/60)° λ D 36-13 1.22λ 1.22(5.5 × 10−7 m) = = 2.31 × 10−3 m = 2.3 mm sin θ sin(1/60)° EVALUATE: The larger the diameter the smaller the angle that can be resolved EXECUTE: D = 36.43 IDENTIFY: Apply sin θ = 1.22 SET UP: θ = W , where W = 28 km and h = 1200 km θ is small, so sin θ ≈ θ h EXECUTE: D = 36.44 λ D 1.22λ 1.2 × 106 m h = 1.22λ = 1.22(0.036 m) = 1.88 m sin θ W 2.8 × 104 m EVALUATE: D must be significantly larger than the wavelength, so a much larger diameter is needed for microwaves than for visible wavelengths IDENTIFY: The diameter D of the mirror determines the resolution SET UP: The resolving power is θ res = 1.22 EXECUTE: The same θ res means that 36.45 λ1 D1 = λ D λ2 D2 D2 = D1 ⎛ 550 × 10−9 m ⎞ λ2 = (8000 × 103 m) ⎜ = 220 m ⎜ 2.0 × 10−2 m ⎟⎟ λ1 ⎝ ⎠ EVALUATE: The Hubble telescope has an aperture of 2.4 m, so this would have to be an enormous optical telescope! IDENTIFY and SET UP: The angular size of the first dark ring is given by sin θ1 = 1.22λ /D (Eq 36.17) Calculate θ1, and then the diameter of the ring on the screen is 2(4.5 m) tan θ1 36.46 ⎛ 620 × 10−9 m ⎞ EXECUTE: sin θ1 = 1.22 ⎜ = 0.1022; θ1 = 0.1024 rad ⎜ 7.4 × 10−6 m ⎟⎟ ⎝ ⎠ The radius of the Airy disk (central bright spot) is r = (4.5 m) tan θ1 = 0.462 m The diameter is 2r = 0.92 m = 92 cm EVALUATE: λ /D = 0.084 For this small D the central diffraction maximum is broad IDENTIFY: Rayleigh’s criterion limits the angular resolution SET UP: Rayleigh’s criterion is sin θ ≈ θ = 1.22λ /D EXECUTE: (a) Using Rayleigh’s criterion sin θ ≈ θ = 1.22λ /D = (1.22)(550 nm)/(135/4 mm) = 1.99 × 10 –5 rad On the bear this angle subtends a distance x θ = x/R and x = Rθ = (11.5 m)(1.99 × 10 –5 rad) = 2.29 × 10 –4 m = 0.23 mm (b) At f/22, D is 4/22 times as large as at f/4 Since θ is proportional to 1/D, and x is proportional to θ , x is 1/(4/22) = 22/4 times as large as it was at f/4 x = (0.229 mm)(22/4) = 1.3 mm 36.47 EVALUATE: A wide-angle lens, such as one having a focal length of 28 mm, would have a much smaller opening at f/2 and hence would have an even less resolving ability IDENTIFY and SET UP: Resolved by Rayleigh’s criterion means angular separation θ of the objects equals 1.22λ /D The angular separation θ of the objects is their linear separation divided by their distance from the telescope 250 × 103 m EXECUTE: θ = , where 5.93 × 1011 m is the distance from earth to Jupiter Thus 5.93 × 1011 m θ = 4.216 × 10−7 1.22λ 1.22(500 × 10−9 m) λ and D = = = 1.45 m Then = 1.22 D 4.216 ì 107 â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 36-14 Chapter 36 EVALUATE: This is a very large telescope mirror The greater the angular resolution the greater the diameter the lens or mirror must be 36.48 IDENTIFY: Rayleigh’s criterion says θ res = 1.22 SET UP: D = 7.20 cm θ res = EXECUTE: 36.49 λ D y , where s is the distance of the object from the lens and y = 4.00 mm s λ (4.00 × 10−3 m)(7.20 × 10−2 m) y yD = 1.22 s = = = 429 m s D 1.22λ 1.22(550 × 10−9 m) EVALUATE: The focal length of the lens doesn’t enter into the calculation In practice, it is difficult to achieve resolution that is at the diffraction limit IDENTIFY and SET UP: Let y be the separation between the two points being resolved and let s be their λ y distance from the telescope Then the limit of resolution corresponds to 1.22 = D s EXECUTE: (a) Let the two points being resolved be the opposite edges of the crater, so y is the diameter of the crater For the moon, s = 3.8 × 108 m y = 1.22λ s/D Hubble: D = 2.4 m and λ = 400 nm gives the maximum resolution, so y = 77 m Arecibo: D = 305 m and λ = 0.75 m; y = 1.1 × 106 m (b) s = yD Let y ≈ 0.30m (the size of a license plate) 1.22λ s = (0.30 m)(2.4 m)/[(1.22)(400 × 10−9 m)] = 1500 km EVALUATE: D/λ is much larger for the optical telescope and it has a much larger resolution even though the diameter of the radio telescope is much larger 36.50 IDENTIFY: Apply sin θ = 1.22 λ D SET UP: θ is small, so sin θ ≈ θ Smallest resolving angle is for short-wavelength light (400 nm) EXECUTE: θ ≈ 1.22 D 10,000 mi = (1.22) 400 × 10−9 m 10,000 mi = 9.61 × 10−8 rad θ = , where R is the distance to R 5.08 m 16,000 km = 1.7 × 1011 km 9.6 × 10−8 rad EVALUATE: This is less than a light year, so there are no stars this close IDENTIFY: We can apply the equation for single-slit diffraction to the hair, with the thickness of the hair replacing the thickness of the slit the star R = 36.51 λ θ = λ 36.52 SET UP: The dark fringes are located by sin θ = m The first dark fringes are for m = ±1 y = R tan θ is a the distance from the center of the screen From the center to one minimum is 2.61 cm y 2.61 cm λ 632.8 × 10−9 m EXECUTE: tan θ = = = = 30.2 μ m = 0.02088 so θ = 1.20° a = R 125 cm sin θ sin1.20° EVALUATE: Although the thickness of human hairs can vary considerably, 30 μ m is a reasonable thickness IDENTIFY: If the apparatus of Exercise 36.4 is placed in water, then all that changes is the wavelength λ λ → λ′ = n SET UP: For y x, the distance between the two dark fringes on either side of the central maximum is D′ = y′ Let D = y be the separation of 5.91 × 10−3 m found in Exercise 36.4 xλ ′ xλ D 5.91 × 10−3 m = = = = 4.44 × 10−3 m = 4.44 mm 1.33 a an n EVALUATE: The water shortens the wavelength and this decreases the width of the central maximum EXECUTE: y′1 = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Diffraction 36.53 36-15 IDENTIFY: In the single-slit diffraction pattern, the intensity is a maximum at the center and zero at the dark spots At other points, it depends on the angle at which one is observing the light SET UP: Dark fringes occur when sin θ m = mλ /a, where m = 1, 2, 3, …, and the intensity is given by ⎛ sin β /2 ⎞ π a sin θ I0 ⎜ ⎟ , where β /2 = β λ /2 ⎝ ⎠ EXECUTE: (a) At the maximum possible angle, θ = 90°, so mmax = ( a sin 90°)/λ = (0.0250 mm)/(632.8 nm) = 39.5 Since m must be an integer and sin θ must be ≤ 1, mmax = 39 The total number of dark fringes is 39 on each side of the central maximum for a total of 78 (b) The farthest dark fringe is for m = 39, giving sin θ39 = (39)(632.8 nm)/(0.0250 mm) ⇒ θ39 = ±80.8° (c) The next closer dark fringe occurs at sinθ38 = (38)(632.8 nm)/(0.0250 mm) ⇒ θ38 = 74.1° The angle midway these two extreme fringes is (80.8° + 74.1°)/2 = 77.45°, and the intensity at this angle is ⎛ sin β /2 ⎞ π a sin θ π (0.0250 mm)sin(77.45°) I = I0 ⎜ = = 121.15 rad, which gives ⎟ , where β /2 = λ 632.8 nm ⎝ β /2 ⎠ 36.54 36.55 ⎡ sin(121.15 rad) ⎤ −4 I = (8.50W/m ) ⎢ ⎥ = 5.55 × 10 W/m ⎣ 121.15 rad ⎦ EVALUATE: At the angle in part (c), the intensity is so low that the light would be barely perceptible IDENTIFY: The two holes behave like double slits and cause the sound waves to interfere after they pass through the holes The motion of the speakers causes a Doppler shift in the wavelength of the sound SET UP: The wavelength of the sound that strikes the wall is λ = λ0 − vsTs , and destructive interference first occurs where sin θ = λ /2 EXECUTE: (a) First find the wavelength of the sound that strikes the openings in the wall λ = λ0 − vsTs = v/f s − vs /f s = (v − vs )/f s = (344 m/s − 80.0 m/s)/(1250 Hz) = 0.211 m Destructive interference first occurs where d sin θ = λ /2, which gives d = λ /(2 sin θ ) = (0.211 m)/(2 sin 11.4°) = 0.534 m (b) λ = v/f = (344 m/s)/(1250 Hz) = 0.275 m sinθ = λ /2d = (0.275 m) /[2(0.480 m)] → θ = ±16.7° EVALUATE: The moving source produces sound of shorter wavelength than the stationary source, so the angles at which destructive interference occurs are smaller for the moving source than for the stationary source IDENTIFY and SET UP: sin θ = λ /a locates the first dark band In the liquid the wavelength changes and this changes the angular position of the first diffraction minimum λliquid ⎛ sin θliquid ⎞ λ sin 21.6° EXECUTE: sin θair = air ; sin θliquid = λliquid = λair ⎜ = 0.5953λair ⎟ = λair a a sin θ sin 38.2° air ⎠ ⎝ λliquid = λair /n (Eq 33.5), so n = λair /λliquid = 36.56 λair 0.5953λair = 1.68 EVALUATE: Light travels faster in air and n must be > 1.00 The smaller λ in the liquid reduces θ that located the first dark band 1 IDENTIFY: d = , so the bright fringes are located by sin θ = λ N N 1 SET UP: Red: sin λ R = 700 nm Violet: sin λ V = 400 nm N N sin θ R sin(θ V + 21.0°) EXECUTE: (a) = θ R − θ V = 21.0° → θ R = θ V + 21.0° = Using a trig identity sin θ V sin θ V from Appendix B gives sin θ V cos 21.0° + cosθ V sin 21.0° = 7/4 cos 21.0° + cot θ V sin 21.0° = 7/4 sin θ V tan θ V = 0.4390 ⇒ θ V = 23.7° and θ R = θ V + 21.0° = 23.7° + 21.0° = 44.7° Then sin θ R = 700 nm N © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 36-16 Chapter 36 sin θ R sin 44.7° = = 1.00 × 106 lines/m = 1.00 × 104 lines/cm 700 nm 700 × 10−9 m (b) The spectrum begins at 23.7° and ends at 44.7° EVALUATE: As N is increased, the angular range of the visible spectrum increases (a) IDENTIFY and SET UP: The angular position of the first minimum is given by a sin θ = mλ (Eq 36.2), with m = The distance of the minimum from the center of the pattern is given by y = x tan θ gives N = 36.57 sin θ = λ a = 540 × 10−9 m 0.360 × 10−3 m = 1.50 × 10−3 ; θ = 1.50 × 10−3 rad y1 = x tan θ = (1.20 m) tan(1.50 × 10−3 rad) = 1.80 × 10−3 m = 1.80 mm (Note that θ is small enough for θ ≈ sin θ ≈ tan θ , and Eq (36.3) applies.) (b) IDENTIFY and SET UP: Find the phase angle β where I = I /2 Then use Eq (36.6) to solve for θ and y = x tan θ to find the distance EXECUTE: Eq (36.5) gives that I = I when β = 2.78 rad βλ ⎛ 2π ⎞ a sin θ (Eq (36.6)), so sin θ = ⎝ λ ⎟⎠ 2π a β =⎜ y = x tan θ ≈ x sin θ ≈ βλ x (2.78 rad)(540 × 10−9 m)(1.20 m) = = 7.96 × 10−4 m = 0.796 mm 2π a 2π (0.360 × 10−3 m) EVALUATE: The point where I = I /2 is not midway between the center of the central maximum and the first minimum; see Exercise 36.15 36.58 ⎛ sin γ ⎞ IDENTIFY: I = I ⎜ ⎟ The maximum intensity occurs when the derivative of the intensity function ⎝ γ ⎠ with respect to γ is zero SET UP: d sin γ d ⎛1⎞ = cos γ ⎜ ⎟=− dγ ⎝ γ ⎠ dγ γ ⎛ sin γ ⎞ ⎛ cos γ sin γ ⎞ dI d ⎛ sin γ ⎞ cos γ sin γ = I0 − ⎟⎟ = − ⇒ γ cos γ = sin γ ⇒ γ = tan γ ⎜ ⎟ = 2⎜ ⎟ ⎜⎜ dγ dγ ⎝ γ ⎠ γ γ γ γ ⎠ γ ⎝ ⎠⎝ (b) The graph in Figure 36.58 is a plot of f (γ ) = γ − tan γ When f (γ ) equals zero, there is an intensity EXECUTE: maximum Getting estimates from the graph, and then using trial and error to narrow in on the value, we find that the three smallest γ -values are γ = 4.49 rad 7.73 rad, and 10.9 rad EVALUATE: γ = is the central maximum The three values of γ we found are the locations of the first three secondary maxima The first four minima are at γ = 3.14 rad, 6.28 rad, 9.42 rad, and 12.6 rad The maxima are between adjacent minima, but not precisely midway between them Figure 36.58 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Diffraction 36.59 36-17 IDENTIFY and SET UP: Relate the phase difference between adjacent slits to the sum of the phasors for all 2π d 2π dθ when θ is small and sin θ ≈ θ slits The phase difference between adjacent slits is φ = sin θ ≈ λ λφ Thus θ = 2π d λ EXECUTE: A principal maximum occurs when φ = φ max = m2π , where m is an integer, since then all the phasors add The first minima on either side of the m th principal maximum occur when φ =φ ± = m 2π ± (2π /N ) and the phasor diagram for N slits forms a closed loop and the resultant phasor ⎛ λ ⎞ is zero The angular position of a principal maximum is θ = ⎜ φ The angular position of the ⎝ 2π d ⎟⎠ max ⎛ λ ⎞ ± φ adjacent minimum is θ ±min = ⎜ ⎝ 2π d ⎟⎠ 2π ⎞ λ ⎛ λ ⎞⎛ ⎛ ⎞ ⎛ 2π ⎞ + =θ +⎜ =θ + φ ⎝ 2π d ⎟⎠ ⎜⎝ max N ⎟⎠ ⎝ 2π d ⎟⎠ ⎜⎝ N ⎟⎠ Nd θ +min = ⎜ 2π ⎞ λ ⎛ λ ⎞⎛ φ − =θ − ⎝ 2π d ⎟⎠ ⎜⎝ max N ⎟⎠ Nd θ −min = ⎜ 2λ , as was to be shown Nd EVALUATE: The angular width of the principal maximum decreases like 1/N as N increases IDENTIFY: Heating the plate causes it to expand, which widens the slit The increased slit width changes the angles at which destructive interference occurs (2.75 × 10−3 /2) SET UP: First minimum is at angle θ given by tan θ = Therefore, θ is small and the 0.620 mλ xλ equation ym = x is accurate The width of the central maximum is w = The change in slit width a a is Δa = aαΔT xλ w w ⎛ da ⎞ EXECUTE: dw = xλ ⎜ − ⎟ = − da = − da Therefore, Δw = − Δa The equation for thermal ⎝ a ⎠ a a a The angular width of the principal maximum is θ 36.60 + − − θ = expansion says Δa = aαΔT , so Δw = − wαΔT = −(2.75 mm)(2.4 × 10−5 K −1)(500 K) = −0.033 mm When the temperature of the plate increases, the width of the slit increases and the width of the central maximum decreases EVALUATE: The fractional change in the width of the slit is (0.033 mm)/(2.75 mm) = 1.2% This is small, 36.61 but observable IDENTIFY and SET UP: Draw the specified phasor diagrams There is totally destructive interference between two slits when their phasors are in opposite directions EXECUTE: (a) For eight slits, the phasor diagrams must have eight vectors The diagrams for each specified value of φ are sketched in Figure 36.61a In each case the phasors all sum to zero (b) The additional phasor diagrams for φ = 3π /2 and 3π /4 are sketched in Figure 36.61b 3π 5π 7π For φ = ,φ = , and φ = , totally destructive interference occurs between slits four apart For 4 3π φ = , totally destructive interference occurs with every second slit EVALUATE: At a minimum the phasors for all slits sum to zero © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 36-18 Chapter 36 Figure 36.61 36.62 IDENTIFY: The wavelength of the helium spectral line from the receding galaxy will be different from the spectral line on earth due to the Doppler shift in the light from the galaxy ⎛ λ ⎞ 2λgalaxy 2λ sin θgalaxy = sin θ lab ⎜ lab ⎟ The Doppler SET UP: d sin θ = mλ sin θlab = lab sin θgalaxy = ⎜ ⎟ d d ⎝ λgalaxy ⎠ formula says f R = c−v 1 c−v c = Since the lab is the receiver R and fS Using f = , we have λ λR λS c + v c+v the galaxy is the source S, this becomes λlab = λgalaxy EXECUTE: sin θgalaxy = sin θ lab c+v c−v c+v 2.998 × 108 m/s + 2.65 × 107 m/s which gives = sin(18.9o ) c−v 2.998 × 108 m/s − 2.65 × 107 m/s θ galaxy = 20.7o 36.63 EVALUATE: The galaxy is moving away, so the wavelength of its light will be lengthened, which means that the angle should be increased compared to the angle from light on earth, as we have found IDENTIFY and SET UP: The condition for an intensity maximum is d sin θ = mλ , m = 0, ± 1, ± 2,… Third order means m = The longest observable wavelength is the one that gives θ = 90° and hence θ = 1 m = 1.087 × 10−6 m EXECUTE: 9200 lines/cm so 9.2 × 105 lines/m and d = 9.2 × 10 d sin θ (1.087 × 10−6 m)(1) = = 3.6 × 10−7 m = 360 nm m EVALUATE: The longest wavelength that can be obtained decreases as the order increases IDENTIFY and SET UP: As the rays first reach the slits there is already a phase difference between adjacent 2π d sin θ ′ This, added to the usual phase difference introduced after passing through the slits, yields slits of λ= 36.64 λ the condition for an intensity maximum For a maximum the total phase difference must equal 2π m 2π d sin θ 2π d sin θ ′ EXECUTE: + = 2π m ⇒ d (sin θ + sin θ ′ ) = mλ λ (b) 600 slits/mm ⇒ d = λ 6.00 × 105 m −1 = 1.67 × 10−6 m For θ ′ = θ °, m = 0: θ = arcsin(0) = ⎛ 6.50 × 10−7 m ⎞ ⎛λ⎞ m = 1: θ = arcsin ⎜ ⎟ = arcsin ⎜ = 22.9° ⎜ 1.67 × 10−6 m ⎟⎟ ⎝d⎠ ⎝ ⎠ ⎛ 6.50 × 10−7 m ⎞ ⎛ λ⎞ m = −1: θ = arcsin ⎜ − ⎟ = arcsin ⎜ − = −22.9° ⎜ 1.67 × 10−6 m ⎟⎟ ⎝ d⎠ ⎝ ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Diffraction 36-19 For θ ′ = 20.0°, m = 0: θ = arcsin(− sin 20.0°) = −20.0° 36.65 ⎛ 6.50 × 10−7 m ⎞ sin 20.0° ⎟ = 2.71° m = 1: θ = arcsin ⎜ ⎜ 1.67 × 10−6 m ⎟ ⎝ ⎠ − ⎛ 6.50 × 10 m ⎞ sin 20.0° ⎟ = −47.0° m = −1: θ = arcsin ⎜ − ⎜ 1.67 × 10−6 m ⎟ ⎝ ⎠ EVALUATE: When θ ′ > 0, the maxima are shifted downward on the screen, toward more negative angles mλ IDENTIFY: The maxima are given by d sin θ = mλ We need sin θ = ≤ in order for all the visible d wavelengths to be seen SET UP: For 650 slits/mm ⇒ d = = 1.53 × 10−6 m 6.50 × 105 m −1 2λ1 3λ = 0.52; m = 3: = 0.78 d d λ λ λ λ2 = 7.00 × 10−7 m: m = 1: = 0.46; m = 2: = 0.92; m = 3: = 1.37 So, the third order does not d d d contain the violet end of the spectrum, and therefore only the first- and second-order diffraction patterns contain all colors of the spectrum EVALUATE: θ for each maximum is larger for longer wavelengths EXECUTE: λ1 = 4.00 × 10−7 m: m = 1: 36.66 IDENTIFY: Apply sin θ = 1.22 λ1 d = 0.26; m = 2: λ D Δx SET UP: θ is small, so sin θ ≈ , where Δx is the size of the detail and R = 7.2 × 108 ly R ly = 9.41 × 1012 km λ = c/f EXECUTE: sinθ = 1.22 λ D ≈ Δx 1.22λ R (1.22)cR (1.22)(3.00 × 105 km/s)(7.2 × 108 ly) ⇒ Δx = = = = 2.06 ly R D Df (77.000 × 103 km)(1.665 × 109 Hz) 12 (9.41 × 10 km/ly)(2.06 ly) = 1.94 × 1013 km Δx is very small Still, R is very large and Δx is many R orders of magnitude larger than the diameter of the sun IDENTIFY and SET UP: Add the phases between adjacent sources EXECUTE: (a) d sin θ = mλ Place 1st maximum at ∞ or θ = 90° d = λ If d < λ , this puts the first maximum “beyond ∞ ” Thus, if d < λ there is only a single principal maximum (b) At a principal maximum when δ = 0, the phase difference due to the path difference between adjacent EVALUATE: λ = 18 cm λ /D is very small, so 36.67 ⎛ d sin θ ⎞ This just scales 2π radians by the fraction the wavelength is of the path slits is Φpath = 2π ⎜ ⎝ λ ⎟⎠ difference between adjacent sources If we add a relative phase δ between sources, we still must maintain a total phase difference of zero to keep our principal maximum 2π d sin θ ⎛ δλ ⎞ Φpath ± δ = ⇒ = ±δ or θ = sin −1 ⎜ ⎟ λ ⎝ 2π d ⎠ 0.280 m = 0.0200 m (count the number of spaces between 15 points) Let θ = 45° Also recall (c) d = 14 f λ = c, so δ max = ± 2π (0.0200 m)(8.800 × 109 Hz)sin 45° = ±2.61 radians (3.00 × 108 m/s) EVALUATE: δ must vary over a wider range in order to sweep the beam through a greater angle © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 36-20 36.68 Chapter 36 IDENTIFY: The wavelength of the light is smaller under water than it is in air, which will affect the resolving power of the lens, by Rayleigh’s criterion SET UP: The wavelength under water is λ = λ0 /n, and for small angles Rayleigh’s criterion is θ = 1.22λ /D EXECUTE: (a) In air the wavelength is λ0 = c/f = (3.00 × 108 m/s)/(6.00 × 1014 Hz) = 5.00 × 10 –7 m In water the wavelength is λ = λ0 /n = (5.00 × 10−7 m) /1.33 = 3.76 × 10−7 m With the lens open all the way, we have D = f/2.8 = (35.0 mm)/2.80 = (0.0350 m)/2.80 In the water, we have sin θ ≈ θ = 1.22λ /D = (1.22)(3.76 × 10 −7 m)[(0.0350 m)/2.80] = 3.67 × 10−5 rad Calling w the width of the resolvable detail, we have θ = w/x → w = xθ = (2750 mm)(3.67 × 10−5 rad) = 0.101 mm (b) θ = 1.22λ /D = (1.22)(5.00 × 10−7 m) /[(0.0350 m)/2.80] = 4.88 × 10−5 rad w = xθ = (2750 mm)(4.88 × 10−5 rad) = 0.134 mm 36.69 EVALUATE: Due to the reduced wavelength underwater, the resolution of the lens is better under water than in air IDENTIFY: The diameter D of the aperture limits the resolution due to diffraction, by Rayleigh’s criterion y λ SET UP: Rayleigh’s criterion says that θ res = 1.22 D = 4.00 mm θ res = , where s is the altitude and D s y = 65.0 m EXECUTE: Combining two equations above gives s= 36.70 y λ = 1.22 s D yD (65.0 m)(4.00 × 10−3 m) = = 3.87 ×105 m = 387 km 1.22λ 1.22(550×10−9 m) EVALUATE: This is comparable to the altitude of the Hubble telescope IDENTIFY: The resolution of the eye is limited because light diffracts as it passes through the pupil The size of the pupil determines the resolution SET UP: The smallest angular separation that can be resolved is θ res = 1.22 λ D The angular size of the object is its height divided by its distance from the eye 50 × 10−6 m = 2.0 × 10−4 rad EXECUTE: (a) The angular size of the object is θ = 25 × 10−2 m ⎛ 550 × 10−9 m ⎞ λ θ res = 1.22 = 1.22 ⎜ = 3.4 × 10−4 rad θ < θ res so the object cannot be resolved ⎜ 2.0 × 10−3 m ⎟⎟ D ⎝ ⎠ y (b) θ res = and y = sθ res = (25 cm)(3.4 × 10−4 rad) = 8.5 × 10−3 cm = 85 μm s (c) θ = θ res = 3.4 × 10−4 rad = 0.019° = 1.1 This is very close to the experimental value of 36.71 (d) Diffraction is more important EVALUATE: We could not see any clearer if our retinal cells were much smaller than they are now because diffraction is more important in limiting the resolution of our vision IDENTIFY: The liquid reduces the wavlength of the light (compared to its value in air), and the scratch causes light passing through it to undergo single-slit diffraction SET UP: sin θ = λ a EXECUTE: tan θ = , where λ is the wavelength in the liquid n = λair λ (22.4/2) cm and θ = 20.47o 30.0 cm λ = a sin θ = (1.25 × 10−6 m)sin 20.47o = 4.372 × 10−7 m = 437.2 nm n = EVALUATE: λair 612 nm = = 1.40 λ 437.2 nm n > 1, as it must be, and n = 1.40 is reasonable for many transparent films © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Diffraction 36.72 IDENTIFY: Apply sin θ = 1.22 36-21 λ D Δx SET UP: θ is small, so sin θ ≈ , where Δx is the size of the details and R is the distance to the earth R ly = 9.41 × 1015 m EXECUTE: (a) R = DΔx (6.00 × 106 m)(2.50 × 105 m) = = 1.23 × 1017 m = 13.1 ly 1.22λ (1.22)(1.0 × 10−5 m) 1.22λ R (1.22)(1.0 × 10−5 m)(4.22 ly)(9.41 × 1015 m/ly) = = 4.84 × 108 km This is about 10,000 D 1.0 m times the diameter of the earth! Not enough resolution to see an earth-like planet! Δx is about times the distance from the earth to the sun (1.22)(1.0 × 10−5 m)(59 ly)(9.41 × 1015 m/ly) = 1.13 × 106 m = 1130 km (c) Δx = 6.00 × 106 m (b) Δx = Δx 1130 km = = 8.19 × 10−3 ; Δx is small compared to the size of the planet Dplanet 1.38 × 105 km 36.73 EVALUATE: The very large diameter of Planet Imager allows it to resolve planet-sized detail at great distances IDENTIFY and SET UP: Follow the steps specified in the problem EXECUTE: (a) From the segment dy′, the fraction of the amplitude of E0 that gets through is ⎛ dy ′ ⎞ ⎛ dy ′ ⎞ E0 ⎜ sin(kx − ωt ) ⇒ dE = E0 ⎜ ⎝ a ⎟⎠ ⎝ a ⎟⎠ (b) The path difference between each little piece is E dy′ y′ sin θ ⇒ kx = k ( D − y′ sin θ ) ⇒ dE = sin(k ( D − y′ sin θ ) − ωt ) This can be rewritten as a E0 dy′ dE = (sin(kD − ωt )cos(ky′ sin θ ) + sin( ky′ sin θ )cos( kD − ωt )) a (c) So the total amplitude is given by the integral over the slit of the above a/2 E a/2 ⇒E=∫ dE = ∫ dy′ (sin( kD − ωt ) cos( ky′ sin θ ) + sin(ky′ sin θ )cos(kD − ωt )) − a/2 a − a/ But the second term integrates to zero, so we have: a/2 E= a/ E0 ⎡⎛ sin(ky ′ sin θ ) ⎞ ⎤ sin(kD − ωt )∫ dy ′ (cos( ky ′ sin θ )) = E0 sin(kD − ωt ) ⎢⎜ ⎟⎥ − a/2 a ⎣⎝ ka sin θ /2 ⎠ ⎦ − a/2 ⎛ sin( ka(sin θ )/2) ⎞ ⎛ sin(π a(sin θ )/λ ) ⎞ ⇒ E = E0 sin(kD − ω x ) ⎜ ⎟ = E0 sin(kD − ω x) ⎜ ⎟ ka (sin θ )/2 ⎝ ⎠ ⎝ π a (sin θ )/λ ) ⎠ sin[…] At θ = 0, = ⇒ E = E0 sin( kD − ω x) […] 2 ⎛ sin(ka (sin θ )/2) ⎞ ⎛ sin( β /2) ⎞ 2 (d) Since I ∝ E ⇒ I = I ⎜ ⎟ = I0 ⎜ ⎟ , where we have used I = E0 sin (kx − ωt ) ka (sin θ )/2 β /2 ⎝ ⎠ ⎝ ⎠ EVALUATE: The same result for I (θ ) is obtained as was obtained using phasors 36.74 IDENTIFY and SET UP: Follow the steps specified in the problem EXECUTE: (a) Each source can be thought of as a traveling wave evaluated at x = R with a maximum amplitude of E0 However, each successive source will pick up an extra phase from its respective ⎛ d sin θ ⎞ pathlength to point P φ = 2π ⎜ which is just 2π , the maximum phase, scaled by whatever fraction ⎝ λ ⎟⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 36-22 Chapter 36 the path difference, d sin θ , is of the wavelength, λ By adding up the contributions from each source (including the accumulating phase difference) this gives the expression provided (b) ei ( kR −ωt + nφ ) = cos( kR − ωt + nφ ) + i sin(kR − ωt + nφ ) The real part is just cos(kR − ωt + nφ ) So, ⎡ N −1 ⎤ N −1 Re ⎢ ∑ E0ei ( kR −ω x + nφ ) ⎥ = ∑ E0 cos(kR − ω x + nφ ) (Note: Re means “the real part of….”) But this is just ⎣ n =0 ⎦ n =0 E0 cos(kR − ωt ) + E0 cos( kR − ωt + φ ) + E0 cos( kR − ωt + 2φ ) + + E0 cos( kR − ωt + ( N − 1)φ ) N −1 N −1 N −1 ∞ N −1 n =0 n =0 n =0 n =0 n =0 (c) ∑ E0 ei ( kR −ωt + nφ ) = E0 ∑ e−iωt e+ikR einφ = E0ei ( kR −ωt ) ∑ einφ ∑ einφ = ∑ (eiφ )n But recall xN −1 ∑ xn = Putting everything together: x −1 n =0 N −1 N −1 ∑ E0ei ( kR −ωt + nφ ) = E0ei ( kR −ωt + ( N −1)φ /2) (eiNφ /2 − e−iNφ /2 ) n =0 (eiφ /2 − e−iφ /2 ) ⎡ cos Nφ /2 + sin Nφ /2 − cos Nφ /2 + i sin Nφ /2 ⎤ = E0 [cos(kR − ωt + ( N − 1)φ /2) + i sin(kR − ωt + ( N − 1)φ /2)] ⎢ ⎥ cos φ /2 + i sin φ /2 − cos φ /2 + i sin φ /2 ⎣ ⎦ Taking only the real part gives ⇒ E0 cos(kR − ωt + ( N − 1)φ /2) (d) I = E av = I sin ( N φ /2) sin (φ /2) definition of I ) I ∝ (The cos term goes to in the time average and is included in the E02 EVALUATE: (e) N = I = I I ′0 ∝ E02 but for us I ∝ 36.75 sin( Nφ /2) = E sin φ /2 sin (2φ /2) sin φ /2 = I (2sin φ / 2cos φ /2) sin φ /2 φ = I cos Looking at Eq (35.9), E02 I ′0 = IDENTIFY and SET UP: From Problem 36.74, I = I sin ( Nφ /2) sin φ /2 Use this result to obtain each result specified in the problem ⎛ N /2 ⎞ cos( Nφ /2) sin ( Nφ /2) EXECUTE: (a) lim I → Use 1’Hôpital’s rule: lim = lim ⎜ = N So ⎟ φ →0 φ →0 sin φ /2 φ →0 ⎝ 1/2 ⎠ cos(φ /2) lim I = N I φ →0 (b) The location of the first minimum is when the numerator first goes to zero at The width of the central maximum goes like 2φmin , so it is proportional to N 2π φmin = π or φmin = N N Nφ = nπ where n is an integer, the numerator goes to zero, giving a minimum in intensity 2nπ This is true assuming that the denominator doesn’t go to zero That is, I is a minimum wherever φ = N (c) Whenever as well, which occurs when φ = mπ , where m is an integer When both go to zero, using the result from part(a), there is a maximum That is, if n is an integer, there will be a maximum N © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Diffraction (d) From part (c), if 36-23 n is an integer we get a maximum Thus, there will be N − minima (Places where N n is not an integer for fixed N and integer n.) For example, n = will be a maximum, but N n = 1, 2…, N − will be minima with another maximum at n = N (e) Between maxima sin ( Nφ /2) sin φ /2 π 3π ⎛ ⎞ is a half-integer multiple of π ⎜ i.e., , etc ⎟ and if N is odd then 2 ⎝ ⎠ φ → 1, so I → I EVALUATE: These results show that the principal maxima become sharper as the number of slits is increased © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... earth to the sun (1.22)(1.0 × 10−5 m)(59 ly)(9.41 × 1015 m/ly) = 1 .13 × 106 m = 1130 km (c) Δx = 6.00 × 106 m (b) Δx = Δx 1130 km = = 8.19 × 10−3 ; Δx is small compared to the size of the planet... 41.0160° Δθ = 0. 0137 ° EVALUATE: d cosθ dθ = λ /N , so for 1820 slits the angular interval Δθ between each of these maxima and the first adjacent minimum is Δθ = 36.37 λ 6.56 × 10−7 m = 0. 0137 ° This... m/s)/(88.9 MHz) = 3.375 m For no signal, a sin θ = mλ m = 1: sinθ1 = (1)(3.375 m)/(15.0 m) ⇒ θ l = 13. 0° m = 2: sinθ = (2)(3.375 m)/(15.0 m) ⇒ θ = ±26.7° m = 3: sinθ3 = (3)(3.375 m)/(15.0 m) ⇒ θ3