Calculate the diffraction vector from arbitrary incident direction

Question

I need to ray trace polychromatic light through an optical system that includes reflective diffraction gratings. I got stuck here.

If I have a reflecting plane with normal $\mathbf{\hat{n}}$ and an incident wave described by wave vector $\mathbf{k_i} = (2 \pi / \lambda) \ \mathbf{\hat{k}_i}$ where $\lambda$ is the wavelength and $\mathbf{\hat{k}_i}$ is the normal of the incident ray, I can get the wave vector of the reflected wave as

$$\mathbf{k_r} = \mathbf{k_i} - 2\mathbf{\hat{n}}(\mathbf{\hat{n}} \cdot \mathbf{k_i})$$

Here $(\mathbf{\hat{n}} \cdot \mathbf{k_i})$ is the component of $\mathbf{k_i}$ perpendicular to the surface and the $-2\mathbf{\hat{n}}$ factor "reflects it back up".

If I now add a periodic 1D grating of period $d$ to the surface such that the reciprocal vector of the grating $\mathbf{g}$ is in the plane ($|g| = 2 \pi/d$), the specularly reflected wave vector $\mathbf{k_r}$ is now also the 0th order diffracted wave vector.

But now I'd like to calculate vectors for the $\pm1$ orders by somehow adding $\pm\mathbf{g}$ and I'm not sure how to do this correctly.

Since diffraction doesn't change the wavelength, $|k_{+1}| = |k_{0}| = |k_{-1}| = |k_{i}|$.

Do I take the component of $\mathbf{k_r}$ parallel to $\mathbf{\hat{g}}$ (i.e. ($\mathbf{k_r} \cdot \mathbf{\hat{g}})$) and add $\pm \mathbf{g}$ to it, then just somehow foreshorten the components perpendicular to $\mathbf{g}$ to keep $|k_{+1}|$ and $|k_{-1}|$ equal to $|k_i|$?

If that's the right idea, and I'm not sure of that; how would that be written in vector form?

update:

I've gone through Born & Wolf and Eugene Hecht's Optics (and several other optics tomes nearby on the shelf) and was not able to find anything I can apply in this case. Textbooks like to stick to the simpler case where $\mathbf{g}$ is in plane of incidence, where the solutions reduce to the simple, familiar grating equation:

$$d(\sin \theta_i - \sin \theta_m) = m \lambda$$

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Related in History of Science and Mathematics SE:

• Where does "the grating equation" come from? Does it have a another name?

Answer 1

OK got it.

Split off the parallel and perpendicular components of the incident wave vector (and then throw away the original perpendicular part).

$$\mathbf{k_{i∥}} = \mathbf{k_{i}} - \mathbf{\hat{n}} \ (\mathbf{k_{i∥}} \cdot \mathbf{\hat{n}})$$

Add the point in reciprocal space to it:

$$\mathbf{k_{∥m}} = \mathbf{k_{i∥}} + m\mathbf{g}$$

then get the new perpendicular component by giving it the direction of the outwards-pointing normal and whatever magnitude adds up to the initial $|k_i|$:

$$\mathbf{k_{⟂m}} = \mathbf{\hat{n}} \ \sqrt{|\mathbf{k_{i}}|^2 - |\mathbf{k_{∥m}}|^2}$$

and add the parallel and perpendicular components back together.

$$\mathbf{k_{m}} = \mathbf{k_{m∥}} + \mathbf{k_{m⟂}} $$

Adjust the sign convention for the direction of $\mathbf{g}$ to taste.

When the expression under the square root goes negative, you've got an evanescent situation, and for free-space ray-tracing purposes on a macroscopic scale, you terminate the ray at this point.

The intensities of all the propagating (non-evanescent) orders will sum to the incident intensity.