NetNotes
As Tina says, welcome—you are now a full, initiated member
of the club. My worst happened when I was on vacation and a tech decided that the filament was blown. Getting an enterprising group together, including several academics, including the Unit Director. Well, suffice to say the Wehnelt was so badly cross threaded when they were done that we had to replace it. While I did not laugh at the time, it proved a very valuable teaching lesson—to me and for me to use. Since then I prepared a laminated checklist titled “What to do if the screen is black.” It is well used, including by me. Paul R. Hazelton
paul_hazelton@umanitoba.ca Tu Apr 28 Yes indeed, we all do experience such events. When I first
joined the faculty of the University of Michigan the Department had just acquired an RCA Model EML TEM. Te person who had been responsible for its purchase had leſt, and I was put in charge of it. One day I walked into the lab and tried to turn it on, but despite all my best efforts I couldn’t get the thing running. Finally, I called the local RCA Service Engineer, Tony Boulet, for help. He showed up, pushed the main power switch a couple of times, then walked around behind the instrument and said, “It helps if you plug it in.” It turned out that the darned thing wasn’t hard wired into the electrical system, but simply had a power line that plugged into a wall outlet. Somehow or other the plug had come out of the socket and was laying peacefully on the floor. It was so embarrassing that it was funny. I ended up taking Tony out for a steak dinner, and I enjoyed his expert assistance and his friendship for many years thereaſter. Wilbur C. Bigelow
bigelow@umich.edu Tu Apr 28 Tere’s also those of us (who will go unnamed) who forget to
remove the digital camera from the side port of the microscope when finished using the microscope. When we fire up the scope the next day at low magnification and don’t see the beam, it may take a while for the “light” to go on and remember that the camera’s still blocking the beam! Ed Haller
ehaller@health.usf.edu Fri Apr 29
LM:
phase contrast and refractive index I was rereading the section of MicroscopyU on phase contrast
microscopy, and one aspect of it confused me (again). I usually explain the contrast as a measure of refractive index differences between parts of the cell and the surrounding medium. Te refractive index differences lead to differences in optical path length which leads to waves being out of phase. But changes in refractive index also deviates the light. MicrosopyU calls the deviated wave in phase the diffracted wave, but then proceeds to lay out the math in terms of refractive index. So what is the correct way to say this? Is it that the differences in the refractive index of the medium through which the diffracted light passes that changes the optical path length? Does deviation by refraction play any role? Dave Knecht
david.knecht@
uconn.edu Mon Apr 11 I’m not sure about the official or definitive answer but diffraction
from an aperture edge would happen in the absence of a refractive medium (such as in a vacuum). Diffraction is where a secondary wave (Huygens’ secondary source) is created by scattering rather than a modification to a primary wave, which is what I think is called refraction. Diffracted waves have a stepwise difference in phase from the primary wave—oſten 90-degrees advanced. If the refraction of the cell and the medium differ (because of differences in refractive index and possibly thickness) then there will be optical path differences that translate directly into phase contrast, resulting from the recombination of the differently refracted waves. Diffraction is better known from the X-ray and electron study of crystals; and in the electron microscope combining diffracted with non-diffracted beams can deliver atomic resolution (lattice) images. Rob Keyse
rok210@lehigh.edu Mon Apr 11
62
F. Zernike, Z. Tech. Phys., 16 (1935), 848; Physica, 9(1942),
686, 97F4. (references from Born and Wolfe, 7th ed, Principles of Optics) F. Zernike - Nobel Prize Talk:
http://nobelprize.org/nobel_ prizes/physics/laureates/1953/
zernike-lecture.html. Fred Monson
fmonson@wcupa.edu Mon Apr 11 Te two descriptions—refractive index and diffraction—are
equivalent, so it does not matter which description you use. Some people find the refractive index easier to understand, and others are more comfortable with the diffraction picture. Regarding your question about the role of deviation, remember that a lens transfers the light (or electrons in the EM) from one point in the specimen to one point in the image, regardless of the direction that the light leaves the specimen (as long as it is within the solid angle of the lens, of course). Te total path length will depend on the angle of deviation, and this will affect the phase. In TEM the angles of deviation are very small, so the phase differences are correspondingly small, and the small-angle approximation is used, and the differences are ignored. In LM with a lens with high NA, some of the angles can be large, so some of the light may be sufficiently out of phase that the contrast is lowered. Bill Tivol
william.f.tivol@aero.org Mon Apr 11 My question might be a bit basic for this list, but since we’re on the
subject. I’ve always had difficulty understanding how phase-contrast operates in terms of creating the two phases of light. (Tat way the two work to create contrast through interference I get.) But how does one light path for the background, and another sample/refracted light path off by 1/4 wavelength get created? I’ve never figured out how the condenser ring and the objective ring interact to do this. Peter Werner
germpore@sonic.net Mon Apr 11 Here is the theory as I see it (correct if necessary): Te phase shiſt is
not only due to the refraction index but also to the specimen thickness. Te phase shiſt is much more influenced by the refraction index and specimen thickness than by the difference in traveled distance due to different diffraction angles. As to the roles of the condenser annulus and phase ring: Te condenser annulus produces a ring of light, such as the uninfluenced (background) light will arrive exactly on the phase ring (this must be precisely adjusted). Te phase ring is represented by a special coating (or the glass is carved) in such a way that the background light is by 1/4 phase-shiſted (whether positively or negatively). Because the diffracted light is naturally shiſted by about 1/4 too (this is due to the refraction index and thickness of cells), the phase interference can be destructive (dark) or additive (bright) up to 180°. But even with the phase interference occurring that way, the background light would be too intense to obtain really a good contrast of small details, so the trick is to reduce the background light by up to 90° (again in the phase ring). Tis is an artificial (but effective way) to increase the proportion of interference light in the image creation process. In other words, without a phase ring the phase interference would still occur (although up to 90° and not 180°) but would be so small in comparison to the intense background light that it would not give a good contrast. Because the diffracted light does not follow a ring pattern like the background light but rather diffract in all directions, it will not be influenced by the phase ring and will join the background light (which passed through the phase ring) in the image plane. In other words the phase ring (along with the condenser annulus of course) allows the physical sorting of background and diffracted light. It is possible that I didn’t use the very accurate terms but I think (and hope) that it makes the explanations easier to understand. Stephane Nizets
nizets2@yahoo.com Wed Apr 13 Your explanation sounds much like many other explanations I
have read, and has the same problem that led to my posting the initial question. You start out talking about refractive index and then switch to talking about diffracted light. Diffraction and refraction are quite
www.microscopy-today.com • 2011 July
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