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Unformatted text preview: approximate both sin and tan by , in radians. Then, r = D = 1 . 22 D/d and I = P r 2 = Pd 2 (1 . 22 D ) 2 = (5 . 10 6 W)(4 . 0 m) 2 [1 . 22(3000 10 3 m)(3 . 10 6 m)] 2 = 2 . 1 10 5 W / m 2 , not great enough to destroy the missile. (b) We solve for the wavelength in terms of the intensity and substitute I = 1 . 10 8 W / m 2 : = d 1 . 22 D r P I = 4 . 0 m 1 . 22(3000 10 3 m) s 5 . 10 6 W (1 . 10 8 W / m 2 ) = 1 . 4 10 7 m = 140 nm ....
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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.
 Fall '08
 SPRUNGER
 Physics, Power

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