Problem #0085 Quantum Physics Problem #0001 Chapters Chapters

787.

Problem 52.35 (RHK)

The use of lasers for defence against ballistic missiles is being studied. A laser beam of intensity would probably burn into and destroy a hardened (nonspinning) missile in about 1 s. (a) Assuming that the laser has a power output of 5.30 MW, a wavelength of , and a beam diameter of 3.72 m (a very powerful laser indeed), we have to answer whether it would destroy a missile at a distance of 3000 km. (b) If the wavelength could be changed, we have to find the wavelength that would work. (c) If the wavelength of the laser could not be changed, we have to find the destructive range of the laser in (a). We may use the equation for the central diffraction disk and take the focal length to be the distance to the target.

Solution:             Click For PDF Version

(a)

We know that the radius of the central diffraction disk is given by

We also know that the central disk contains about 84% of the incident power. In the equation given above for the radius of the central disk, we will use for the focal length f the distance of the target from the laser lens. The wavelength of the laser is and the beam diameter is 3.72 m. Therefore,

The output power of the laser is 5.30 MW. About 84% of the incident power is contained in the central disk. Therefore, the power flux of the laser at the target will be

It is much less than required for burning the target.

Let us find the radius of the central diffraction disk at the target, , for which the power flux at the target will be .

(b)

We will calculate next the wavelength of the laser beam that will produce a central diffraction disk of at the target distance of 3000 km.

(c)

If the wavelength of the laser beam could not be changed, the distance to the target at which the power flux will be can be calculated by the requirement that