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# 38 - Kapoor(mk9499 homework38 Turner(60230 This print-out...

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Kapoor (mk9499) – homework38 – Turner – (60230) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Light of wavelength 580 nm illuminates two slits of width 0 . 03 mm and separation 0 . 18 mm. How many interference maxima fall within the full width of the central diffraction maxi- mum? Correct answer: 11. Explanation: Let : λ = 580 nm , d = 0 . 18 mm , and w = 0 . 03 mm . The angle of the first diffraction minimum to the width w of the slits of the diffraction grating is sin θ 1 = λ w , and the angle corresponding to the m th inter- ference maxima is sin θ m = m λ d . We require θ 1 = θ m , so λ w = m λ d m = d w . Therefore the number of fringes in the central maximum is N = 2 m - 1 = 2 parenleftbigg d w parenrightbigg - 1 = 2 parenleftbigg 0 . 18 mm 0 . 03 mm parenrightbigg - 1 = 11 . 002 (part 2 of 2) 10.0 points What is the ratio of the intensity of the third interference maximum to the side of the cen- terline (not counting the center interference maximum) to the intensity of the center in- terference maximum? Correct answer: 0 . 405285. Explanation: The phase difference between the lights at centerline and at third interference maximum is φ = 2 π λ w sin θ 3 = 2 π λ w 3 λ d = 6 π w d , so the ratio of the intensities is I 3 I 0 = sin φ 2 φ 2 2 = sin 6 π w 2 d 6 π w 2 d 2 = sin 3 π w d 3 π w d 2 = sin 3 π (0 . 03 mm) 0 . 18 mm 3 π (0 . 03 mm) 0 . 18 mm 2 = 0 . 405285 . 003 10.0 points Using a conventional two-slit apparatus with light of wavelength 593 nm, 29 bright fringes per centimeter are observed on a screen 4 m away. What is the slit separation? Correct answer: 6 . 8788 mm. Explanation:

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Kapoor (mk9499) – homework38 – Turner – (60230) 2 Let : N = 29 cm 1 = 2900 m 1 , λ = 593 nm = 5 . 93 × 10 7 m , and L = 4 m . The distance on the screen to the m th and ( m + 1) th bright fringe is Δ y = ( m + 1) λ L d - m λ L d = λ L d . Then the number of fringes per unit length is N = 1 Δ y = d λ L and the slit separation is d = λ L N = (5 . 93 × 10 7 m) (4 m) (2900 m 1 ) × 1000 mm m = 6 . 8788 mm . 004 10.0 points Two coherent microwave sources that produce waves of wavelength 1 cm are in the xy plane, one on the y axis at y = 9 cm and the other at x = 3 cm, y = 8 cm. If the sources are in phase, find the differ- ence in phase between the two waves from these sources at the origin. Correct answer: 164 . 159 . Explanation: Let : vectorx 1 = ( x 1 , y 1 ) = (0 , 9 cm) , vectorx 2 = ( x 2 , y 2 ) = (3 cm , 8 cm) , and λ = 1 cm . The path difference between the distances from these two sources to the origin is Δ r = r 1 - r 2 = radicalBig x 2 1 + y 2 1 - radicalBig x 2 2 + y 2 2 = 9 cm - radicalBig (3 cm) 2 + (8 cm) 2 = 0 . 455996 cm , so the difference in phase is δ = Δ r λ 360 = (0 . 455996 cm) (360 ) 1 cm = 164 . 159 . 005 10.0 points Hint: Use a small angle approximation; e.g. , sin θ = tan θ .
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38 - Kapoor(mk9499 homework38 Turner(60230 This print-out...

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