p37_019 - 4 km . This distance is greater than the diameter...

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19. (a) We use the Rayleigh criteria. If L is the distance from the observer to the objects, then the smallest separation D they can have and still be resolvable is D = R ,where θ R is measured in radians. The small angle approximation is made. Thus, D = 1 . 22 d = 1 . 22(8 . 0 × 10 10 m)(550 × 10 9 m) 5 . 0 × 10 3 m =1 . 1 × 10 7 m=1 . 1 × 10
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Unformatted text preview: 4 km . This distance is greater than the diameter of Mars; therefore, one part of the planets surface cannot be resolved from another part. (b) Now d = 5 . 1 m and D = 1 . 22(8 . 10 10 m)(550 10 9 m) 5 . 1 m = 1 . 1 10 4 m = 11 km ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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