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DiKsa [7]
3 years ago
7

What are some drawbacks of electron microscopes? 3. If an object being viewed under the phase-contrast microscope has the same r

efractive index as the background material, how would it appear? +
Physics
1 answer:
gayaneshka [121]3 years ago
7 0

1. <u>Some drawbacks of electron microscopes:</u>

  • Price, size, repair, researcher learning and image artifacts arising from specimen preparation are the major drawbacks.
  • This form of magnification is a massive, burdensome, costly piece of equipment, highly sensitive to vibration and exterior magnetic field.
  • It must be held in an environment that is big enough to contain the microscope, as well as to secure and prevent any unwanted effect on the electrons.
  • Upkeep includes ensuring balanced voltage supplies, electromagnetic coil / lens currents and cool water circulation so that the specimens are not destroyed or damaged by the heat released during the electrons energization process.

2. If an object being viewed under the phase-contrast microscope has the same refractive index as the background material than "it would be difficult to see because the phase contrast microscope amplifies differences in the refractive index".

<u>Explanation:</u>

In order to improve the comparison of transparent and colorless specimens with the light microscopy pictures, thus phase contrast is used. This allows the visualization of cells and cell elements which would be hard to see using a standard light magnification. The phase comparison does not involve the destruction, fixation or staining of cells.

Due to diffraction and scattering phenomena which exist at the edges of these objects, large, extended specimens are also quickly visualized with phase contrast optical. As light transits through one medium to another, the velocity is changed in proportion to the variations in the refractive index between the two media. Therefore, the wave is either increased or decreased in velocity whenever a coherent light wave produced by the oriented microscope filament progresses via a phase specimen with a particular thickness and refractive index.

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the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

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You can write the KVL equations:

Top left loop:

  I2(4) +(I2 +I3)(2) +I1(1) = 10

Bottom left loop:

  (I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10

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In matrix form, the equations are ...

  \left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]

These equations have the solution ...

  \left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]

This means the branch currents are as follows:

  top left: I2 = 0.625 A

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_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

  (I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

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