Do Spherical Waves from Point Sources Transform into Plane Waves?

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Spherical waves from a point source do tend to approximate plane waves as they propagate outward. Plane waves experience less intensity loss with distance compared to spherical waves, as their wavefront area does not increase. The intensity of spherical waves decreases according to the inverse square law, where intensity is inversely proportional to the square of the distance from the source. The discussion clarifies that the intensity loss rate is not expected to decrease as distance increases; rather, it decreases rapidly. Overall, the relationship between spherical and plane waves is defined by their differing intensity loss characteristics as they propagate.
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Hi guys,

Let's assume I have a point source in a free field. Now, correct me if I'm wrong on any of the following points:

  • As the spherical waves spread out from the point source, they will tend towards plane waves.
  • Since planes waves lose less intensity with distance than spherical waves (due to wavefront area not increasing), the rate at which the intensity falls will decrease with distance.
  • A graph will look like the attachment. The black line is for an ideal point source where the waves stay spherical with distance, the blue line for the situation I've described.

Thanks,
Stewart
 

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TheEdge said:
Hi guys,

Let's assume I have a point source in a free field. Now, correct me if I'm wrong on any of the following points:

  • As the spherical waves spread out from the point source, they will tend towards plane waves.
  • Since planes waves lose less intensity with distance than spherical waves (due to wavefront area not increasing), the rate at which the intensity falls will decrease with distance.
  • A graph will look like the attachment. The black line is for an ideal point source where the waves stay spherical with distance, the blue line for the situation I've described.

Thanks,
Stewart
Your black line already accounts for the reduced rate of intensity loss. If there are no energy losses, the intensity (energy per unit area) will be inversely proportional to area, which is proportional to distance^2

I=K/r^2

At some reference distance say 1m, the intensity is Io

Io = K/(1m)^2

So

I/Io = (r/1m)^(-2)

Taking log of both sides

log (I/Io) = -2log(r/1m)

This is what your black line is showing

The rate of intensity loss with increasing distance is the derivative of intensity wrt r

dI/dr = K(-2)/r^3

This is a rapidly decreasing function of r. There is no reason to expect a slower rate.
 
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