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

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Unformatted text preview: Ch. 36 p. 1 CHAPTER 36 – Diffraction and Polarization 1. We find the angle to the first minimum from sin θ 1min = m λ / a = (1)(680 × 10 –9 m)/(0.0345 × 10 –3 m) = 0.0197, so θ 1min = 1.13°. Thus the angular width of the central diffraction peak is ? θ 1 = 2 θ 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 θ 1min = m λ ; (3.00 × 10 –6 m) sin (18.5°) = (1) λ , which gives λ = 9.52 × 10 –7 m = 952 nm . 3. For constructive interference from the single slit, the path difference is a sin θ = ( m + ! ) λ , m = 1, 2, 3, … . For the first fringe away from the central maximum, we have (3.50 × 10 –6 m) sin θ 1 = ( * )(550 × 10 –9 m), which gives θ 1 = 13.7°. We find the distance on the screen from y 1 = L tan θ 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 θ = ( m + ! ) λ , 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 tan θ 1min = ! (? y 1 )/ L = sin θ 1min . The condition for the first minimum is a sin θ 1min = ! a ? y 1 / L = λ . If we form the ratio of the expressions for the two wavelengths, we get ? y 1 b /? y 1 a = λ b / λ 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 θ 1min = m λ ; a max sin 90° = (1) λ , or a max = λ . ( 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 sin θ 1min = m λ / a = (1)(400 × 10 –9 m)/(0.0655 × 10 –3 m) = 6.11 × 10 –3 , so θ 1min = 0.350°. We find the distance on the screen from y 1 = L tan θ 1 = (3.50 m) tan 0.350° = 2.14 × 10 –2 m = 2.14 cm....
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## This homework help was uploaded on 03/26/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

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

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