1074_Physics ProblemsTechnical Physics

1074_Physics ProblemsTechnical Physics - Chapter 38 P38.17...

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Unformatted text preview: Chapter 38 P38.17 By Rayleigh’s criterion, two dots separated center-to-center by 2.00 mm would overlap when *P38.19 θ min = Thus, P38.18 415 L= λ D = 1.22 = θ min λ d = 1.22 . L D e je j 2.00 × 10 −3 m 4.00 × 10 −3 m dD = = 13.1 m . 1.22λ 1.22 500 × 10 −9 m e 1.22 5.00 × 10 −7 1.00 × 10 −5 e j m= j 6.10 cm The concave mirror of the spy satellite is probably about 2 m in diameter, and is surely not more than 5 m in diameter. That is the size of the largest piece of glass successfully cast to a precise shape, for the mirror of the Hale telescope on Mount Palomar. If the spy satellite had a larger mirror, its manufacture could not be kept secret, and it would be visible from the ground. Outer space is probably closer than your state capitol, but the satellite is surely above 200-km altitude, for reasonably low air friction. We find the distance between barely resolvable objects at a distance of 200 km, seen in yellow light through a 5-m aperture: y λ = 1.22 L D b ga fFGH 6 × 10m m IJK = 3 cm 5 −7 y = 200 000 m 1.22 (Considering atmospheric seeing caused by variations in air density and temperature, the distance b ga between barely resolvable objects is more like 200 000 m 1 s 1° rad fFGH 3 600 s IJK FGH π180° IJK = 97 cm .) Thus the snooping spy satellite cannot see the difference between III and II or IV on a license plate. It cannot count coins spilled on a sidewalk, much less read the date on them. P38.20 1.22 λ D = d L λ= D = 2.10 m d = 1.22 c = 0.020 0 m f L = 9 000 m b0.020 0 mgb9 000 mg = 2.10 m λ a2.00 mf = a10.0 mf 105 m P38.21 θ min = 1.22 P38.22 L = 88.6 × 10 9 m , D = 0.300 m , λ = 590 × 10 −9 m D = 1.22 λ = θ min = 2.40 × 10 −6 rad (a) 1.22 (b) d = θ min L = 213 km D 0.244 rad = 14.0° ...
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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