Solution_Set_03

# Solution_Set_03 - Î = 600 nm = 600 x 10-9 m then Î± =(1.25...

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SOLUTION SET 3 Physics 2022 – Spring 2008 1. F = 2 f M = F / f = 2 f / f = 2 2. Fraction = Area (cage) / Area (mirror) = π R 2 (cage) / π R 2 (mirror) = ( 1 / 5 ) 2 = 1 / 25 = 0.04 3. LGA Area LGA(1) (300 m) 2 LGA(2) (50 m) 2 LGA(1) / LGA(2) = (300 / 50) 2 = (6) 2 = 36 times 4. LGA(1) / LGA(2) = (Diameter / 1) 2 = Diameter 2 = 1000 m 2 Diameter = 31.6 m 5a. Light Gathering Power = Area (Subaru) / Area (Hubble) = π R 2 (Subaru) / π R 2 (Hubble) = (8.3 m / 2.4 m) 2 = 12.0 X 5b. Subaru has a larger aperture, so it can detect fainter objects. Hubble is outside the Earth’s atmosphere, so it achieves a higher resolution. 6a. Magnification = F / f = 2 m / 9mm = 2000 mm / 9mm = 222 X 6b. Magnification = F / f = 2000 mm / 20 mm = 100 X 6c. Magnification = F / f = 2000 mm / 55 mm = 36.4 X 6d. α = 2.5 x 10 5 λ / D = 2.5 x 10 5 / 0.20 λ = 1.25 x 10 6 λ where λ and D are in meters

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Unformatted text preview: Î» = 600 nm = 600 x 10-9 m then Î± = (1.25 x 10 6 )(600 x 10-9 ) = 0.75 arcsec 7a. D = Î± d = (0.1 arcsec) [ Ï€ / (180 x 60 x 60)] (6.28 x 10 8 km) = 304.5 km Hubbleâ€™s resolution on Io. 7b. D = Î± d = (1.0 arcmin) [ Ï€ / (180 x 60)] (3.844 x 10 5 km) = 111.8 km Human eyeâ€™s resolution on the Moon. 8a. sin Î± = D / d D = d sin Î± = (7 x 10 7 ly) sin ( 0.1 / 3600 ) = 34 ly 8b. d = D / sin Î± = (1.8 x 10-5 km) sin ( 0.1 / 3600 ) = 37 km 9. Î¸ = 250,000 Î» / D Î» 1 = 5 x 10-7 m D 1 = 4 m Î» 2 = 10-4 m Î» 1 / D 1 = Î» 2 / D 2 D 2 = D 1 Î» 2 / Î» 1 = (4 m) (10-4 m) / (5 x 10-7 m) = 800 m 10. Î¸ = 250,000 Î» / D 10-3 arcsec = 250,000 (550 x 10-9 m) / D D = 250,000 (550 x 10-9 m) / (10-3 arcsec) D = 137.5 m Problems 2, 3, 6a, and 7a are each worth 1 point. Problem 9 is worth 2 points....
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Solution_Set_03 - Î = 600 nm = 600 x 10-9 m then Î± =(1.25...

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