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In my understanding, light works as follows: every point in space where there is light, this light works as a point source. When we progress in time, the light spreads out from there in all directions (spherically). The new locations where the light is now, act as new point sources.

From this it follows that the light from a laser beam would go in all directions, not just in the direction of a laser. However, we see in practice that (most of) the light only goes in the direction of the laser. How is this possible?

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  • $\begingroup$ Because a laser is not a point source, but the propagating result of a cavity mode. Even an LED is not purely isotropic. $\endgroup$
    – Jon Custer
    Commented Aug 8, 2023 at 12:43

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A spatially coherent source, such as laser or an antenna surface, of characteristic length $D$, say diameter, operating at wavelength $\lambda$ has a beamwidth of $k_0\frac{\lambda}{D}$, where $k_0$ is factor of order $\tfrac{1}{2} \text{to} \tfrac{3}{2}$, i.e., $\mathcal O(k_0)=1$, and it depends on the shape of the emitting surface, its illumination by the source and how the beam width itself is defined. In this context, by coherence of the emitting surface, for laaer it is output lens, I mean that every point is of the same phase, in other words the emitting surface is an equiphase surface coinciding with a phase front. The geometric factor $k_0$ depends on the amplitude of each emitting point source, each point on the aperture plane, this amplitude can be varying from point to point but no its phase. If the area is large then the each point that act as a Huygens (hemi)-spherical interfere destructively in most directions except the one orthogonal to the surface. In other directions away from the central one the Huygens wavelets interfere partially leading to "sidelobes" that are much smaller than the central "main" lobe in whose direction all points are in phase. Antenna engineers, by convention, usually specify the beamwidth as the half power direction, physicists may prefer first nulls around the peak in whose direction totally destructive interference occur. In either the case, the beamspread in steradians may also be written as $k_1\frac{4\pi \lambda^2}{A_{eff}}$ where now $\mathcal O(k_1)=1$ and $A_{eff}$ is the effective aperture area.

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  • $\begingroup$ So most light outside of the visible laser beam cancels because of destructive interference $\endgroup$
    – Riemann
    Commented Aug 8, 2023 at 13:04
  • $\begingroup$ Yes, that is correct. Another way to see all this is to think of an ideal plane wave that has rays going in only one direction. Then restrict the plane wave with a variable diaphragm, as the one placed behind the lens of a photographic camera. As you reduce the size of the diaphragm the spread of the wave increases while the intensity of the "middle" will be proportional to the area of the diaphragm, in other directions it will decrease and spread out with fluctuating intensity. That decrease and spreading is caused by the destructive interference; analytically it is a Fourier transform. $\endgroup$
    – hyportnex
    Commented Aug 8, 2023 at 13:26

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