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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81 Problem 82 Problem 83 Problem 84 Problem 85 Problem 86 Problem 87 Problem 88 Problem 89 Problem 90 Problem 91 Problem 92 Problem 93 Problem 94 Problem 95 Problem 96 Problem 97 Problem 98 Problem 99 Problem 100

Problem 37 Hard Difficulty

(a) Show that a 30,000-line-per-centimeter grating will not produce a maximum for visible light. (b) What is the longest wavelength for which it does produce a first-order maximum?
(c) What is the greatest number of lines per centimeter a diffraction grating can have and produce a complete second- order spectrum for visible light?

Answer

A. $=333.3 \mathrm{nm}$
B. 333 $\mathrm{nm}$
C. $N=6.58 \times 10^{3}$

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Physics 103

College Physics for AP® Courses

Chapter 27

Wave Optics

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Wave Optics

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Problem 6
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Problem 10
Problem 11
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Problem 13
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Problem 16
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Problem 18
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Problem 24
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Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
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Problem 33
Problem 34
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Problem 36
Problem 37
Problem 38
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Problem 40
Problem 41
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Problem 49
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Problem 100

Video Transcript

in about a first. We need to clear the slit separation. D uh, in once in the future we have, Ah, 30,000 lines. So 0.1 millimeters. You I 30,000 off. This gives us a D to be 3.33 times 10 to the power mine. A seven meter, which is 333.3 in enemy too. We knows the sine theta is equal to em, Linda, for the longest prevalent. Ah, Sign that. Ah, physical one and the m E report to one. Therefore d will be equal to London, which is 333.3 in enemy what would be the longest we have? A lentil MBA is equal to the slits Aggression Smith separation Mission RC to get the largest number off lines person to be true and still would produce a complete spectrum. We want the smallest slit separation that allows the longest prevalent off visible light to produce the second northern next month. So second order maximum well, the next moon So the Lambda Max lend a maximum will be equal to 7 69 meter, uh, or second order in musical too. And signing that icicle to one, so d So we do tear See part here. So D is it cool to to lambda maximum Ah, which is equal to two times off. 7 60 Nanami, Jerk. Excuse us. Um, 1.52 times 10 to the power minus seven meter. Then we can find the number off lines and bye. One line lying birdie that is equal to one divided by 1.52 guns tempted power minus seven tons, one meter divided by 1000 100 Son to meet you. So a disc use us the number of lines to be just write the result here, that is and to be 6.58 times 10 to the power, uh, three lines. So lines three lines person to meet you person do each year. So this this number of lines we have in a person to be true

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Irina Lyublinskaya, Gregg Wolfe, Douglas Ingram , Liza Pujji

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