solutions_chapter26

# M 5 2 l 5 700 nm u 5 90 l u d sin u 5 m l l u v 5 202

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m 5 2. l 5 700 nm u 5 90° l . u d sin u 5 m l , l u v 5 20.2°. sin u v 5 sin u r 1 m v m r 21 l v l r 2 5 1 sin 65.0° 2 1 2 3 21 400 nm 700 nm 2 5 0.345 m r l r sin u r 5 m v l v sin u v . m l sin u 5 d 5 constant, m 5 2. m 5 3 d sin u 5 m l . u 5 66.8°. sin u 5 m l d 5 3 1 631 3 10 2 9 m 2 2.06 3 10 2 6 m m 5 3. 6 3 2 6 2, 6 1, 1 m 5 0, 0 m 0 m 5 d sin u l 5 1 2.06 3 10 2 6 m 21 1.00 2 631 3 10 2 9 m 5 3.3. sin u 5 1.00 d 5 1 485 lines / mm 5 2.06 3 10 2 3 mm 5 2.06 3 10 2 6 m. 0 sin u 0 6 2, c . 6 1, m 5 0, d sin u 5 m l , u a 2 u b 5 8.8°. u b 5 22.88°. sin u b 5 1 2 21 486 3 10 2 9 m 2 2.50 3 10 2 6 m b u a 5 31.65°. sin u a 5 1 2 21 656 3 10 2 9 m 2 2.50 3 10 2 6 m a m 5 2. u a 2 u b 5 4.0°. u b 5 11.21°. sin u b 5 1 1 21 486 3 10 2 9 m 2 2.50 3 10 2 6 m 5 0.1944 b u a 5 15.21°. sin u a 5 1 1 21 656 3 10 2 9 m 2 2.50 3 10 2 6 m 5 0.2624 a m 5 1. d 5 1.00 3 10 2 2 m 4.00 3 10 3 5 2.50 3 10 2 6 m. sin u 5 m l d . u u 5 38.2° sin u 5 1 2 21 516 3 10 2 9 m 2 1.67 3 10 2 6 m 5 0.618 l 5 d sin u m 5 1 1.67 3 10 2 6 m 21 sin 18.0° 2 1 5 516 nm Interference and Diffraction 26-9

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26.45. Set Up: Maxima are located by where is the angle of incidence and d is the separation of crystal planes. Solve: (a) First maximum, so (b) For For the equation has no solution. Reflect: Since is not small, we find that for and there are only two angles for which constructive interference occurs. 26.46. Set Up: The maxima are given by Solve: (a) and This is an x ray. (b) The equation doesn’t have any solutions for 26.47. Set Up: Rayleigh’s criterion says where s is the distance of the object from the lens and Solve: Reflect: The focal length of the lens doesn’t enter into the calculation. In practice, it is difficult to achieve resolution that is at the diffraction limit. 26.48. Set Up: The diffraction is by a single slit of width a , not by a circular aperture. The diffraction minima are located by and Rayleigh’s criterion is where s is the distance of the object from the lens and Solve: 26.49. Set Up: Rayleigh’s criterion for circular apertures says Solve: (a) (b) (c) (d) (e) u res 5 1.22 1 2.0 3 10 2 2 m 35 3 10 3 m 2 5 7.0 3 10 2 7 rad 5 1 4.0 3 10 2 5 2 ° u res 5 1.22 1 6.0 3 10 2 2 m 300 m 2 5 2.4 3 10 2 4 rad 5 0.014° u res 5 1.22 1 550 3 10 2 9 m 10.0 m 2 5 6.7 3 10 2 8 rad 5 1 3.8 3 10 2 6 2 ° u res 5 1.22 1 550 3 10 2 9 m 2.4 m 2 5 2.8 3 10 2 7 rad 5 1 1.6 3 10 2 5 2 ° u res 5 1.22 1 550 3 10 2 9 m 50 3 10 2 3 m 2 5 1.3 3 10 2 5 rad 5 1 7.4 3 10 2 4 2 ° u res 5 1.22 l D . s 5 ya l 5 1 2.50 m 21 0.350 3 10 2 3 m 2 600 3 10 2 9 m 5 1.46 km y s 5 l a . y 5 2.50 m. u res 5 y s , u res 5 l a . sin u 5 m l a s 5 yD 1.22 l 5 1 4.00 3 10 2 3 m 21 7.20 3 10 2 2 m 2 1.22 1 550 3 10 2 9 m 2 5 429 m. y s 5 1.22 l D . y 5 4.00 mm. u res 5 y s , D 5 7.20 cm. u res 5 1.22 l D . m . 3. u 5 50.9°. m 5 3: u 5 31.1°. m 5 2: sin u 5 m 1 l 2 d 2 5 m 1 1.81 3 10 2 10 m 2 3 3.50 3 10 2 10 m 4 2 5 m 1 0.2586 2 . l 5 2 d sin u m 5 2 1 3.50 3 10 2 10 m 2 sin 15.0° 5 1.81 3 10 2 10 m 5 0.181 nm. m 5 1 d 5 3.50 3 10 2 10 m. m 5 1, 2, c 2 d sin u 5 m l , m 5 2.6 m l 2 d 5 1 l d l d 5 0.713 nm 0.227 nm 5 0.762. m . 2 u 5 49.6°. m 5 2, sin u 5 m l 2 d 5 m 1 0.173 3 10 2 9 m 2 3 0.227 3 10 2 9 m 4 2 5 1 0.381 2 m .

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