Echellegrating

A reflection grating is not a mirror. It is an array of reflective grooves in a surface. Light reflected from the bottoms of the grooves is delayed relative to light reflected from the tops of the grooves, just as light transmitted through optically thick portions of a transmission phase grating is delayed relative to light transmitted through the optically thin portions.

Near the beam waist, rays follow hyperbolic paths. Far from the beam waist, the hyperbolic beam approximates a cone. The divergence angle is the angle of the vertex of the cone. See the RP Photonics Encyclopedia article Gaussian Beams for more. It has a beam calculator.

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diffractiongrating中文

You need a bigger aperture than the beam diameter to avoid cutting off the outer portion of the beam. If you do that, you have passed the beam through a large pinhole. A pinhole causes diffraction. It isn't as bad with a large pinhole, but it is enough that the beam has a larger divergence angle than it should.

Diffraction gratingformula

It is usually assumed that the waist of the laser beam is at the exit pupil. However, that may not be true in your case. The way you phrase it (which I assume is how the laser specs are stated) does not necessary imply that the 5 mm at the exit means that the waist diameter is 5 mm. It could be that the device contains some lenses producing a converging beam at the exit. That would make sense in the context of laser scanning.

My teachers and textbooks tend to gloss over this point. Why do reflection gratings work, and why is there not a contradiction with the law of reflection?

Diffraction gratingexperiment

Then we may calculate $$w_0 = \frac{\lambda}{ \pi \theta} = \frac{1500 nm}{\pi \times 0.5 \times 10^{-3}} = 0.0009549... m$$ and $$z_R = \pi w_0^2/\lambda = \frac{\pi \times (0.0009549... m)^2}{1500 nm}= 1.909... m.$$

Fraunhoferdiffraction

Or the beam could be something other than a perfect Gaussian beam. The presence of higher modes would make a beam with a $1.66$ mm radius have a larger than ideal divergence angle.

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A Gaussian beam does not have a sharp edge. It is brightest at the center and fades away as you get farther from the beam axis. So the intensity isn't $0$ outside the beam radius.

The law of reflection (that the angles of incidence and reflection are equal) can be derived directly from Maxwell's equations, or from Fermat's principle. However, reflection gratings completely defy this law, and from light incident at a fixed angle comes a whole diffraction pattern - light reflected at a range of different angles.

Blazedgrating

What you can do is to make some measurements to get a rough estimate of the beam size at different distances. A plot of these beam sizes as a function of distance should give you an idea of the location of the waist.

I am modeling a laser beam from a laser scanning device as a Gaussian beam. I am not sure how to decide the waist radius $w_0$. For example, it is given that laser beam footprint at exit is $5$ mm and the divergence is reported to be 0.5 mrad. Wavelength is 1500 nm.

But then the radius is equal to the output radius $5mm/2 = 2.5 mm$ at distance $z \approx 4.6209 m$, which is insane. The device is surely not 5 meters long. So have I understood the parameters of Gaussian beam incorrectly, or is there some optical tricks happening inside the laser device? Or is it so that the Gaussian model is accurate only for the Gaussian shape, not for the radius calculation?

To avoid this, the aperture should be large enough that the bean intensity at the edge is at most $1$ % of the central intensity. That works out to be a aperture $1.5$ times larger than the beam diameter.

Diffraction grating

The intended output of many lasers in laser scanning is Gaussian. At distance $z$ from the waist, the radius of a Gaussian beam is calculated as $$w(z) = w_0 \sqrt{1+(z/z_R)^2},$$ where $w_0$ is the waist radius, and $z_R = \pi w_0^2/\lambda$ is the Rayleigh range, depending on the waist $w_0$ and the wavelength $\lambda$. When distance $z$ is considerably larger than $z_R$, the radius $w$ grows approximately linearly, $$w(z) \approx \theta z,$$ where $$\theta = \frac{\lambda}{\pi w_0}$$ is the divergence angle.

This makes it hard to describe the beam diameter. The way it is done is to pick the diameter where the intensity has dropped by a factor is $1/e^2$.

If the aperture is $5$ mm, the beam diameter is at most $3.33$ mm, and the beam radius $1.66$ mm. A beam with this radius would have a smaller divergence angle than advertised.