Ch36Word - Ch. 36 p. 1 CHAPTER 36 Diffraction and...

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Ch. 36 p. 1 CHAPTER 36 – Diffraction and Polarization 1. We find the angle to the first minimum from s i n θ 1min = m λ / a = (1)(680 × 10 –9 m)/(0.0345 × 10 –3 m) = 0.0197, so 1min = 1.13°. Thus the angular width of the central diffraction peak is ? 1 = 2 1min = 2(1.13) = 2.26°. 2. The angle from the central maximum to the first minimum is 18.5°. We find the wavelength from a sin 1min = m ; ( 3 . 0 0 × 10 –6 m) sin (18.5°) = (1) , which gives = 9.52 × 10 –7 m = 952 nm . 3. For constructive interference from the single slit, the path difference is a sin = ( m + ! ) , m = 1, 2, 3, … . For the first fringe away from the central maximum, we have ( 3 . 5 0 × 10 –6 m) sin 1 = ( * )(550 × 10 –9 m), which gives 1 = 13.7°. We find the distance on the screen from y 1 = L tan 1 = (10.0 m) tan 13.7° = 2.4 m . 4. The angle from the central maximum to the first bright fringe is 19°. For constructive interference from the single slit, the path difference is a sin = ( m + ! ) , m = 1, 2, 3, … . For the first fringe away from the central maximum, we have a sin (19°) = ( * )(689 × 10 –9 m), which gives a = 3.17 × 10 –6 m = 3.2 µ m . 5. Because the angles are small, we have t a n 1min = ! (? y 1 )/ L = sin 1min . The condition for the first minimum is a sin 1min = ! a ? y 1 / L = . If we form the ratio of the expressions for the two wavelengths, we get ? y 1 b /? y 1 a = b / a ; ? y 1 b /(8.0 cm) = (400 nm)/(550 nm), which gives ? y 1 b = 5.8 cm . 6. ( a ) There will be no diffraction minima if the angle for the first minimum is greater than 90°. Thus the limiting condition is a sin 1min = m ; a max sin 90° = (1) , or a max = . ( b ) Visible light has wavelengths from 400 nm to 750 nm, so the maximum slit width for no diffraction minimum for all of these wavelengths is the one for the smallest wavelength: 400 nm . 7. We find the angle to the first minimum from s i n 1min = m / a = (1)(400 × 10 –9 m)/(0.0655 × 10 –3 m) = 6.11 × 10 –3 , so 1min = 0.350°. We find the distance on the screen from y 1 = L tan 1 = (3.50 m) tan 0.350° = 2.14 × 10 –2 m = 2.14 cm. Thus the width of the peak is ? y 1 = 2 y 1 = 2(2.14 cm) = 4.28 cm .
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Ch. 36 p. 2 8. The path-length difference between the top and bottom of the slit for the incident wave is a sin θ i . The path-length difference between the top and bottom of the slit for the diffracted wave is a sin . When the net path-length difference is a multiple of a wavelength, there will be an even number of segments of the wave which will have a path-length difference of λ /2; there will be minima given by ( a sin i ) – ( a sin ) = m , m = ± 1, ± 2, … , or s i n = sin 30° – ( m / a ), where m = ± 1, ± 2, … . When = 30°, the net path-length difference is zero, and there will be constructive interference. There is a “central maximum” at 30° to the normal . 9. ( a ) If we consider the slit made up of N wavelets of amplitude ? E 0 , the total amplitude at the central maximum, where they are all in phase, is N ? E 0 . Doubling the size of the slit doubles the number of wavelets and thus the total amplitude. Because I ~ E 0 2 , the intensity at the central maximum is increased by a factor of 4.
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch36Word - Ch. 36 p. 1 CHAPTER 36 Diffraction and...

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