When I studied physics, we learned how to write down planar waves and spherical waves. But, when I turn on my flashlight, I see a cone of light. How can I see that there is a solution to the wave equation which describes a wave in a conical region, dropping off sharply outside that cone? And am I right that the wave equation cannot describe a cylindrical beam?
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asked Oct 28, 2009 at 3:37
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2 Answers
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At human scales, the wavelength of visible light is so tiny (or equivalently, the frequency is so high) that the wave equation can be modeled by geometric optics (this is the high frequency limit of the wave equation). The cone you see from the flashlight is then nothing more than the shadow that the casing of the light casts from the light bulb.
At larger wavelengths, diffractive effects become more pronounced (when was the last time you saw a water wave propagating along a cone?). The asymptotic behaviour of any localised wave disturbance (in odd dimensions) is then an outgoing spherical wave, modulated by a scattering amplitude that depends only on the direction of propagation (and which is basically a kind of Radon transform of the initial data); this can be seen from asymptotics of the fundamental solution. (It's slightly more complicated in even dimensions due to the failure of the sharp Huygens principle.)
answered Oct 28, 2009 at 16:49
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I can answer your last question: you're right, due to diffractive effects. See Huygens-Fresnel.
answered Oct 28, 2009 at 4:30