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hw9 - 24.4 Use the geometry in Figure 24.8 and the notation...

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Unformatted text preview: 24.4 Use the geometry in Figure 24.8 and the notation of Example 24.1 — both on page 1107 of the text From the geometry, I ' 7.7x10‘3 tend: a = W = 1.2x 10"3 :> 9: 1.2 x 10—3 rad. E 6.50 In Now use the equation for locating maxima for double slit interference —— Equation 24.2 on page 1107 of the text. In this case m = 1, d = 0.50 nun, and 6 r: 1.2 ><10‘3 rad, so dame _ (0.50 x 10-3 rnrn)sin{1.2 x 10“3 red) m 1 = 6.0 X 10—7111. dsindxmA :u A: 24.7 , a) _ The wavelength is c 3.00x108m/s n, ' 2 A_;H 5.0x1014Hz —6.[}><10 meetiflxli} nm. The range or" Visible light wavelengths is from about 4.00 nm to 700 nm ,-so this iight is within the Visibie hand of wavelengths. ‘It would appear to our eyes as a shade of red. b) VVe’d iike a singie cosine term expression for the superposition of two waves of the same amplitude. Apply the trigonometric identity , ‘ + a — cosa+cosfi=2cos(a2fi)cos< 95) .4 from the inside of the front cover of the text, with or : wt «in 6 and 6 = wt. Then Acos(wt + 6) +Acos(wt) = A (2cm (wt+:+wt> cos (wt +:— 0%)) 6 3 A (Ecos (wt + E) cos (g)) m 2Acos (-3) cos (wt + g) The amplitude of the superposition is the coefficient in front of the oscillating cosine term, so 5 {1) amplitude of the Superposition : 2A cos (5) . . - 1T _ , ' ' , Notice that if 6 m '7T rad, then the cosine becomes cos (-5 rad) = 0. This is the case of completely destructive interference. At the other extreme, if 6 2: 0 rad , then the amplitude is 2A, and the “interference” is COmpletely constructive. In this probiem we are given that A z 10 V/m and 6 = 0.60 rad, so ' 0.60 d amplitude of the superposition 2 2A cos (g) = 2(10 V/rn) cos (__Traw) : 19 V/m. ._: 924 __ ,. 7 CHAPTER 24. PHYSICAL OPTICS :c Accordino' to E ' _ . . - . l c: gtKLtTlon 24 1 on page 1105, the phase difference 6, path difference Act, and wave length A are reiated by 6 : 727;“ fad, so H A6 (5.0 X 10—? m)(0.60 rad) =.-_ M— 27rrad z “—51% =5” 10-8111 zarnm. d) - In‘this case, the difference 5 in angle is the angular frequency w multiplied bv the time differepCe At 5 2 wAt = (277 rad)yAt. ' So, :5 2 0.60 rad (27f radhx (2n rad )5.0 x 1014112 *Af: :1.9x10—155. 24.11- a) From Equation 24.3 on page 1110 of the text, the first diffraction minimum from a single slit aperture occurs where usin 191 : A, 50 . A 632.8 x 10“9 m 811191 : — m _—=. —3 =_ 1—3 -:- 0‘ a 0.10x10“3m 63X“) =>91 63x0 rad 035 13) From Equation 24.5 on page 1111 of the text, the first diffraction minimum from a Circular aperture occurs where osin 61 = 1.220/\, SO 1'22“ a w m 7.7 x 10'3 22> a = 7.7 x 10—3 rad : 044°. a 0.10 x 10—3 m $111 61 = So, the first minimum from a circular slit aperture occurs at an angle that is about 20% greater than the angle at which the first minimum occurs with a single slit aperture. 932 - , ' CHAPTER 24. PHYSICAL OPTICS 24.25 Use the grating equation, Equation 24.8 on page 1116 of the text: dsinfim : m/\ =2: sint’m = mg. For the first order, m m 1, so _ 488 x 104’ m ' 511161 2 1(m) I 0.24 => 61 = 0.24 rad =140. For the second order, m = 2, so 488 X 10”9 m '9 :2 ~— 51” 2 (2.0x10‘6m ) = 0.49 2:; 191 = 0.51 rad _—‘_ 29°, 24.26 Use the grating equation, Equation 24.8 on page 1116 of the text, dsin 6 2 mA. The fourth order peak for 488 nm light has the same at and 6’ as the third order peak for the Aunkmwn light. Hence claim 6 : 4(488 nm) and d sin 6 :: 3Aunkno,,n 4 488 X 10‘9 :::> Aunkncmm : ——“—"”‘( 3 In) 3 653. nm . 24.37 11 24.20'on page 1121 of the text, which gives the wavelength A; of light in a medium other a) Use Equatio . _ ength A 111 a vacuum: than a vacuum as'a function of the index of refraction m of that medium and the wavel ,\ a A _W:492m 1—711“ 1.50 d - in} Use Equation 24.17 on page 1120 of the text, which relates the speed c of light in a vacuum to its speed 1) and the index of refraction n of -a medium other than a vacuum: 8 n22 :> v22: 3.00:1; m/s =2.00x1{}sm/s. v n Sflgjjenih: $31: refllected £20m thie first interface and the ray reflected from the second interface (will _ se (3 ange o 71‘ re. on reflection because the ' b ' . . . . . . 4 y 0th reflect from a medium with a la - index of refraction. Hence, this factor 18 1rrelevant. For minimum thickness, the difference in pcth 255%? a r rays reflected from the first ' t f- the coating. 50} in er ace and from the second should be half the wavelength of the light within V 24.51 The Brewster angle 63 is found from Equation 24.31 on page 1131 of the text, tanafizflxlézlfi 22> 93:580. n; 1.0 Referring to Figure 24.48 on page 1131 of the text, we see that the Brewster angle and angle of refraction 6'2 satisfy 93 + 90° + 62 2 180°, so 62 : 90° e 93 2 90° h 58° 2 32°. ...
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