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I'm very new the topic of SPPs and have been trying to understand this particular method of exciting surface plasmons using a 1D periodic grating of grooves, with distance $a$ between each groove. If the light incident on the grating is at an angle $\theta$ from the normal and has wavevector ${\bf k}$, then apparently if this condition is met:

$\beta = k \sin\theta \pm\nu g$

where $\beta$ is corresponding SPP wavevector, $g$ is the lattice constant $2 \pi/a$ and $\nu={1,2,3,...}$ then SPP excitation is possible.

I haven't ever really had a formal course in optics, so my question is where this condition comes from. It seems like Fraunhofer diffraction, but only for the light being diffracted at a $90^\circ$ angle to the normal. Most books don't state how they get this result, they just say it's because of the grating "roughness" which really confuses me.

Any help would be greatly appreciated.

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Don't worry, I did research in surface plasmons and even then I was more than a year into it before I truly understood, on an intuitive level, how the light gets a 'kick' from the grating. You are correct that it is diffraction at a 90 degree angle to the normal, but there is an easier way to think about it.

You say you've never taken a formal course in optics so I'll talk a little bit about diffraction gratings in general. You might have come across one before and know that if a beam of light hits it, it is diffracted into several different beams. Transmissive diffraction gratings are what one usually encounters in high school physics so I'll illustrate one below:

Transmissive diffraction grating

The numbers at the end of each beam are known as the order $\nu$ of that beam. The grating equation is $d(\sin\theta_i + \sin\theta_o) = \nu\lambda$, where d is the distance between lines of the grating, $\lambda$ is the wavelength of the light, $\theta_i$ is the angle of incidence, and $\theta_o$ is the angle of the outgoing beam. In the above illustration, $\theta_i$ is zero.

Next we consider a reflective grating (for example a piece of metal with 1D periodic grooves), as in the following illustration:

Reflective grating

The same mathematics govern this situation as well. You'll notice the $\nu=+2$ order being very close to grazing the grating surface. Adjusting the angle of incidence a little bit would cause it to do so. In that case, it would have the required wave vector to launch a surface plasmon, which is the phase matching condition that you started out with. You get the $\beta = k\sin\theta\pm\nu g$ when you convert the grating formula to wave vectors (reciprocal space) which I'm too lazy to do right now.

I suppose you could technically say that the light got a momentum 'kick' from the $\nu=-2$ order being launched in the opposite direction, but thinking of it as the light getting a 'kick' is really misleading in my experience.

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  • $\begingroup$ Thank you but seriously how do you get from the grating formula you wrote down to $β=ksinθ±νg$. I thought $k<\beta$ $\endgroup$
    – Atreyu
    Commented Dec 4, 2013 at 19:50
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    $\begingroup$ Take $\theta_o = \pi/2$, so the grating equation becomes $d(\sin\theta+1)=\nu\lambda$. Then $k=2\pi/\lambda$, and $g=2\pi/d$. The difference between $k$ and $\beta$ is because $\lambda$ is different for the surface plasmon. $\endgroup$
    – ptomato
    Commented Dec 12, 2013 at 4:48
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The condition comes from "phase-matching" - or in other words that the wavevector of the SPP ($\beta$ in your example) is matched to the wavevector of the in-plane component of the incident light.

Now before the light hits the surface, this in-plane wavevector is given by $k \sin \theta$, but when it hits the grating, it receives a momentum "kick" of $\pm vg$.

The reason that a grating is required is that the SPP's wavevector $\beta$ is usually significantly larger than free-space wavevector $k$.

If the phase matching condition isn't met, then the SPP can't be excited, which is why you don't get SPPs from just shining light onto a metal-dielectric interface. You need the grating or some other mechanism (usually by forming evanescent waves) to provide the extra momentum to excite the SPP.

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  • $\begingroup$ I know all of that, I was just wondering how the light is getting "kicked" by the grating. Is this through diffraction? If so, isn't that weird since we're concerned with scattered light that is parallel to the surface (is this near field or far?) $\endgroup$
    – Atreyu
    Commented Dec 3, 2013 at 22:07
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Yes light can change propagation direction when scattering off of a grating into pre-defined directions that depend on the angle of incidence and the grating period (as well as the refractive medium the grating resides in). This is just Bragg's Law. The change in direction (or momentum) is determined by the above parameters. You can think of the change in momentum as a "kick" if you wish, though I've never heard it explained that way before. Anyway, the diffracted light can couple into a Surface Plasmon Polariton (SPP) grating mode, also called an SPP Bloch mode. I believe, as you mention, the diffracted light needs to be propagating parallel to the surface. This is similar to the TIR condition being met in prism-coupled SPP. You need to couple through a higher index material through the (thin) metal and into a lower index material to see the SPP resonance in the prism-coupling approach. This condition is relaxed when using grating coupling.

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