p41_067

# p41_067 - approximate both sin and tan by , in radians....

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67. (a) The intensity at the target is given by I = P/A ,where P is the power output of the source and A is the area of the beam at the target. We want to compute I and compare the result with 10 8 W / m 2 . The beam spreads because diFraction occurs at the aperture of the laser. Consider the part of the beam that is within the central diFraction maximum. The angular position of the edge is given by sin θ =1 . 22 λ/d ,where λ is the wavelength and d is the diameter of the aperture (see Exercise 61). At the target, a distance D away, the radius of the beam is r = D tan θ .S in c
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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.

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