1021_PartUniversity Physics Solution

1021_PartUniversity Physics Solution - 36-4 Chapter 36...

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36-4 Chapter 36 36.14. IDENTIFY: 2 0 sin( /2) /2 II β ⎛⎞ = ⎜⎟ ⎝⎠ . 2 sin a π θ λ = . SET UP: The angle is small, so sin tan / yx θθ ≈≈ . EXECUTE: 4 1 7 2 2 2 (4.50 10 m) sin (1520 m ) . (6.20 10 m)(3.00 m) aa y yy x πππ βθ λλ × =≈ = = × (a) 13 3 (1520 m )(1.00 10 m) 1.00 10 m : 0.760. 22 y −− × = = 2 2 00 0 sin( 2) sin(0.760) 0.822 20 . 7 6 0 I I == = (b) 3 (1520 m )(3.00 10 m) 3.00 10 m : 2.28. y × = = 2 2 0 sin( 2) sin(2.28) 0.111 . . 2 8 I I == = (c) 3 (1520 m )(5.00 10 m) 5.00 10 m : 3.80. y × = = 2 2 0 sin( 2) sin(3.80) 0.0259 . 23 . 8 0 I I EVALUATE: The first minimum occurs at 7 1 4 (6.20 10 m)(3.00 m) 4.1 mm 4.50 10 m x y a × = × . The distances in parts (a) and (b) are within the central maximum. 5.00 mm y = is within the first secondary maximum. 36.15. (a) IDENTIFY: Use Eq.(36.2) with 1 m = to locate the angular position of the first minimum and then use tan = to find its distance from the center of the screen. SET UP: The diffraction pattern is sketched in Figure 36.15. 1 sin a 9 3 3 540 10 m 2.25 10 0.240 10 m × × 3 1 2.25 10 rad Figure 36.15 33 11 tan (3.00 m)tan(2.25 10 rad) 6.75 10 m 6.75 mm × = ×= (b) IDENTIFY and SET UP: Use Eqs.(36.5) and (36.6) to calculate the intensity at this point. EXECUTE: Midway between the center of the central maximum and the first minimum implies 3 1 (6.75 mm) 3.375 10 m. 2 y × 3 3.375 10 m tan 1.125 10 ; 1.125 10 rad 3.00 m y x × = = The phase angle at this point on the screen is 9 sin (0.240 10 m)sin(1.125 10 rad) . a ππ θπ × × = × Then 2 2 62 0 sin / 2 sin (6.00 10 W/m ) βπ × 2 4 2.43 10 W/m . I = × EVALUATE: The intensity at this point midway between the center of the central maximum and the first minimum is less than half the maximum intensity. Compare this result to the corresponding one for the two-slit pattern, Exercise 35.23.
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Diffraction 36-5 36.16. IDENTIFY: In the single-slit diffraction pattern, the intensity is a maximum at the center and zero at the dark spots. At other points, it depends on the angle at which one is observing the light. SET UP: Dark fringes occur when sin θ m = m λ / a , where m = 1, 2, 3, …, and the intensity is given by 2 0 sin / 2 /2 I β ⎛⎞ ⎜⎟ ⎝⎠ , where sin a πθ = . EXECUTE: (a) At the maximum possible angle, = 90°, so m max = ( a sin90°)/ = (0.0250 mm)/(632.8 nm) = 39.5 Since m must be an integer and sin must be 1, m max = 39. The total number of dark fringes is 39 on each side of the central maximum for a total of 78. (b) The farthest dark fringe is for m = 39, giving sin 39 = (39)(632.8 nm)/(0.0250 mm) 39 = ±80.8° (c) The next closer dark fringe occurs at sin 38 = (38)(632.8 nm)/(0.0250 mm) 38 = 74.1°. The angle midway these two extreme fringes is (80.8° + 74.1°)/2 = 77.45°, and the intensity at this angle is I = 2 0 sin / 2 I , where sin (0.0250 mm)sin(77.45 ) 632.8 nm a π ° == = 121.15 rad, which gives () 2 2 sin(121.15 rad) 8.50 W/m 121.15 rad I = = 5.55 × 10 -4 W/m 2 EVALUATE: At the angle in part (c), the intensity is so low that the light would be barely perceptible. 36.17. IDENTIFY and SET UP: Use Eq.(36.6) to calculate and use Eq.(36.5) to calculate I . 3.25 , 3 56.0 rad, 0.105 10 m. a × (a) EXECUTE: 2 sin a = so 3 2s i n 2 ( 0 . 1 0 5 1 0 m ) s i n 3 . 2 5 668 nm 56.0 rad a ×° = (b) 2 22 5 00 0 0 sin / 2 4 4 (sin( /2)) [sin(28.0 rad)] 9.36 10 / 2 (56.0 rad) II I I I ββ = = × EVALUATE: At the first minimum 2 = rad and at the point considered in the problem 17.8 = rad, so the point is well outside the central maximum. Since
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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1021_PartUniversity Physics Solution - 36-4 Chapter 36...

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