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Download Homework Set #6 Due: 3-28-14
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Homework Set #6 Due: 3-28-14 (1) This problem introduces a vital component in the laboratory, the waveplate. Waveplates allow easy modification of the polarization of light without significant modification to the beam path. “Half-wave” plates rotate the polarization direction and “quarter-wave” plates can transform linear polarization to elliptical polarization to circular polarization and back. Ordinary axis. Extraordinary axis. To summarize what we’re learned recently: Anisotropic media will in general have an index of refraction that depends on the polarization direction. For a given crystal and direction of the light, two polarization modes exist for which there is a definite index of refraction. These are referred to as the extraordinary and ordinary modes. When light is incident on an anisotropic medium at non-normal incidence it will in general refract into two beams, each at a different angle, inside the medium. This is because, so long as the input polarization of the light is a mixture of the two modes, each mode will refract according to its own index. This is called double refraction. Suppose we have a thin plate made of an anisotropic crystal (quartz is usually employed) oriented so that, for linearly polarized light at normal incidence, vertical polarization is ordinary (n = no) and horizontal polarization is extraordinary (n = ne) with no ≠ ne (for quartz, ne < no). If the polarization is not initially parallel to either axis, then the polarization of the light will not be preserved as the light propagates through the crystal. We can take advantage of this effect. (Note that in Type I SHG we avoided it and the polarization of all fields was well-defined and preserved.) To proceed, we can always break the polarization vector of the input light into components parallel to either axis. Thus, we take the electric field at the input face of the crystal to be: E E o (cos eˆ sin oˆ ) where ê and ô are unit vectors pointing along the extraordinary and ordinary axes (to the right and up, respectively, in the figure), and is the angle the polarization vector makes with the extraordinary axis. In practice, the waveplate is in a rotary mount so this angle can be varied. The following problems can be solved as is or you can switch to a complex representation for the electric field (I prefer to use the complex representation). (a) Half-wave plate. Suppose we have monochromatic light, of wavelength , as given above. At the output of the crystal, the ordinary and extraordinary components will have a phase difference between them because they traveled through different optical pathlengths. Suppose the thickness of the crystal is selected such that, after propagating through it, this phase difference is (N + ½) * 2 where N is a positive integer. For ease of construction and for strength the plates are usually ~1 mm thick, so N is large. (i) Find an expression for the length of the crystal taking no and ne to be given. (ii) Show that the effect of the crystal is to rotate the polarization vector and determine how the output polarization angle is related to . (iii) How does the output polarization depend on N? (b) Qualitatively describe what will happen if a different wavelength, ’ ≠ , is used. How does this answer depend on N? (c) Quarter-wave plate. Now let the phase shift be (N+¼ ) * . Show that the effect of the crystal is to produce elliptical polarization and that for = 0o, 45o, 90o, 135o the resultant polarization is linear, circular, linear, circular (with opposite handedness) respectively. Discussion. Waveplates are made from flats that have sides that are flat and parallel to high precision. This helps avoid wavefront distortion. Unfortunately, this means waveplates will also act as Fabry-Perot etalons (with R 5%) and thus have a wavelength dependent transmission. To avoid this, as well as to avoid loss of energy and stray beams, waveplates are always anti-reflection (AR) coated on both sides. Note that a beam sent through a waveplate and telescope, for example, will lose ~27% of its power if all the surfaces are uncoated. Another difficulty is that the index of refraction seen by the light will vary if the light is not normally incident upon the waveplate. The ordinary and extradinary polarization directions and the extraordinary wave index of refraction depend on input angle. If normal incidence is not used, a half-wave plate will produce elliptically polarized light: the larger the deviation, the worse the effect. If the beam is diverging or converging when it travels through the waveplate, then there is no well-defined angle of incidence; a range of angles is present. The output polarization will now vary spatially. For example, for a half-wave plate, the center of the beam would be linearly polarized, but off-center light would be slightly elliptically polarized. Waveplates require careful alignment. In practice, they usually degrade the polarization of the beam somewhat. For short pulse work, we must also take into account the consequence of a range of wavelengths being present at the same time, as you can see from your answer to part (b). All of the above effects are minimized if N = 0! Such a waveplate is called a “zero-order waveplate”. They generally aren’t used for monochromatic light, but they are mandatory for short pulse work. The question is, how do you make one? The most direct approach, using a very thin plate, is impractical. You should convince yourself that the flat would have to be a few 10’s of wavelengths long. (This can be done. The optic will be expensive and fragile, however.) One clever way to do this is to sandwich two flats together, the extraordinary axis of one aligned to the ordinary axis of the other. If the two plates were of identical thickness, there would be no polarization effect at all. By making one of the plates slightly thicker, an effective zero-order half- or quarter-waveplate is obtained. This is usually what is done. (2) In this problem we consider the mismatch, k = k p k o , for FWDM (four-wave degenerate mixing) in BK7 glass using 514.5 nm light for the geometry shown. As in the class notes, kp is the non-linear polarization k-vector and ko is the k-vector associated with the output wave generated by the polarization. The output wave has the same frequency and direction as the “polarization-wave”. There are three input beams (#1-3), all the same frequency, and we will focus on the output beam that is not parallel to any of the input beams. Beams #1 and #2 make a 10o angle with respect to the normal and beam #3 makes a 5o angle. z L x #1 #3 Note: The angles used in the figure are exaggerated for clarity. As discussed in class, there will be a delay “sweep” because of the noncollinear geometry and the beams’ finite widths that will change the effective path length. You do not need to take this into account. (a) Find k p . Use the specified coordinate system. Remember that k p must be specified inside the medium, but the input rays are given outside the medium, so take refraction into account. (b) Find k. (c) Suppose that we wish to use this process in an experiment. We can limit the phase error due to the k-vector mismatch by keeping the sample thin. What sample thickness, L, corresponds to a phase error of ? As you can see from your answer, we’ll probably use a thicker sample and depend on focusing to effectively limit the sample interaction length. (3) What is the phase matching angle for collinear sum frequency generation of 355 nm light from 1064 nm and 532 nm light in KD*P? Specify which of the fields are to be ordinary and which are to be extraordinary. (The resulting equation cannot be solved explicitly. A graphical solution will work fine or you can use a numerical equation solver such as that provided in Mathematica.) This calculation is similar to that for SHG in the notes. You do not need to write full equations of motion based on the wave equation. Just find the phase matching condition and then solve for KD*P and the specific wavelengths for this problem. KD*P information is available from the public class web site in the companies section. #2