Nuclear-pumped x-ray lasers are seen as a possible weapon to destroy ICBM booste

Question

Nuclear-pumped x-ray lasers are seen as a possible weapon to destroy ICBM booster rockets at ranges up to 2250 km. One limitation on such a device is the spreading of the beam due to diffraction, with resulting dilution of beam intensity. Consider such a laser operating at a wavelength of 1.40 nm. The element that emits light is the end of a wire with diameter 0.150 mm. By what factor is the beam intensity reduced in transit to the target? (The laser is fired from space, so that atmospheric absorption can be ignored.)

Explanation / Answer

sin theta = 1.22 lamda/d = 1.22*(1.4 x 10-9m)/(0.150 x 10-3m) = 1.13867E-05 theta is the same (in radians) 0.150 x 10^-3*1.13867E-05*2250000m = 3.843 mm diameter after all of this way. Intensity goes as area, which goes as diameter squared. Ratio of the two diameters squared is (.15mm/3.843mm)^2 = 0.0015 So the intensity is 0.15% as much as it was It is reduced by a factor of 1/.0015 = 656 This is amazing! Who would have thought that the beam would still be so small after 2,250 km?

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