Question

# Let a plane wave be incident on a wall with a hole in it. To turn the plane wave into a laser beam, let's make

the hole a "Gaussian hole," which has a transmission, *t*(*x*) = exp[-(1/2)(*x*/*w*)^{2}], and, to simplify the problem, we'll just consider the *x*-direction. Compute the shape of the beam intensity, *I*(*x*'), a large distance *z* away (neglect any factors that don't depend on *x*' and need only use the "proportional to," (∝) symbol, rather than "equals" when calculating the electric field).

If we call *w* in the above formula the "spot size" of the beam at the wall, use the same definition of spot size to determine the size of the beam (call it *w*') a distance, *z*, away as a function of *z*, *w*, and λ (the wavelength of the beam). In other words, set the resulting exponential of the intensity far away equal to exp[-(*x*'/*w*')^{2}] and solve for *w*'.

Notice that, as long as the Fraunhofer approximation is valid, the beam far away is always bigger than that nearby. That is, beams always broaden far away from their sources. Finally, which yields a smaller spot far away, a small or large spot nearby?

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