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solutions_chapter26

# Solve(a first maximum so(b for for the equation has

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Unformatted text preview: 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 . d 5 m l 2 sin u 5 1 1 21 0.173 3 10 2 9 m 2 2 1 sin 22.4° 2 5 0.227 nm. m 5 1. m 5 1, 2, 3, c u 2 d sin u 5 m l , 26-10 Chapter 26 26.50. Set Up: Solve: Same means 26.51. Set Up: The angular size of the object is its height divided by its distance from the eye. Solve: and Diffraction doesn’t play a significant role. 26.52. Set Up: The angular size of the object is its height divided by its distance from the eye. Solve: (a) The angular size of the object is and the object cannot be resolved. (b) and (c) This is very close to the experimental value of 1 min. (d) Diffraction is more important. 26.53. Set Up: Take the width of a typical license plate to be about 30 cm. Solve: The angular size of the object is Setting gives Reflect: A mirror of this diameter (less than one meter) is feasible....
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Solve(a First maximum so(b For For the equation has no...

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