I want to excite surface plasmon in Kretschmann configuration (nanometer gold layer sputtered on a prism).
I used a 532nm diode laser as the light source, and a polarizing beam splitter cube to separate TE and TM polarizations of the laser beam. But I am not sure if the output beam is TM polarized or not.
How can I become sure of the polarization of output light?
If you mean that you want TM polarized surface wave to be excited (because there is no TE mode in SPP) than you should just use p-polarization, that is E field in the plane of incidence. Use any polarizer with known axis. If you have no - check the specification of your PBS, usually it reflects the wave with E field in the beamsplitting (tilted) plane. Or use Brewster's effect to find p-polarization.
TM or TE have meaning only in regard to the incidence plane. What you have after the polarizing cube beam splitter is just two orthogonally polarised beams. Beams may be denoted as TE or TM but this has only meaning in regard to the splitting element in the cube. This is not what you should be interested in. Actually any of beams could be used as TM in Kretschmann configuration because whether it is TE or TM will be defined by how you position the prism within the beam path. So knowing the polarisation state of beam in free space is useless, what counts is its orientation in regard to the gold coated prism surface.
By the way - with 532 nm and gold surface the SPR curve will be hardly recognisable. Better use red laser diode if possible.
If your initial laser polarization is linear, you can put a polarizer at the output. In the first case, you make a polarizer rotation of 90° (parallel to the Y-axis) according to the laboratory axis. If you remark a light beam after the polarizer, you will have a TE polarization (or vertical). In the second case, you take a 0° rotation of the polarizer and you see if a light beam appears. If it happens you still have a TM mode (horizontal linear polarization)
Use a linear polariser after the laser source (providing that your laser is actually linearly polarised) and have a photodiode or detector after the polariser. Rotate the polariser until you get a maximum signal. This way, you know the angle of the polarisation of the laser and thus can rotate your prism accordingly (or the laser itself if it's easier) to choose TE or TM modes.
Thanks for your kind and useful answers.
I found the polarization of light by using the concept of Brewster angle. I used a glass slab after my polarizing beam splitter, and rotated it (in the same plane as prism rotation) to see if reflection spot disappears or not. And as expected for p-polarized light, it disappeared at Brewster angle.
Dear Dr. Nizamov
I think the wavelength of laser light influences mostly the angle at which the SP excitation occurs. for example, acc. to figures attached, SP excitation occurs at approximately 64 degrees of incident angle for 980nm light, but it occurs at 77 degrees for 594nm light.
What other important effects does the wavelength have?
Why is it hard to recognize SP excitation with green laser?
Its hard to recognise for the Green laser as the K vector no longer matches as good (you go to the right of the dispersion curve) and thus further from the crossing point between the light line and the SPR dispersion curve.
Thanks Mitchell,
Yes, for green laser K vector increases and that's why we don't have excitation at the same incident angle. but if we increase the incident angle(or use a higher refractive index prism), then we will have a new crossing point and excitation will occur.
Hi, Mohsen,
sorry, my estimations were flawed. You are actually right - it is possible to observe quite a nice SPR at 532 nm with gold. There may be couple of technological problems (the gold layer should be almost twice thinner then usually and high RI substrate is important). I doubt its usefulness after all, but nevertheless it seems to be quite doable.
'What other important effects does the wavelength have?'
You should consider the imaginary parts of the permittivities as well. Then, you can find that the wavelength affects not only excitation conditions but also absorption. The other thing that changes is the field localization factor.
i have a question to ask from Shavakat
how do calculate the or determine the parameters which effects the excitation .
i am new to this field and is tryng to do the same experiment as the original poster of this question .
it would be grateful if you could help mw with the caluculation method . thank you
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