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.
<u>Answer
</u>
4.8 A
<u>Explanation
</u>
From the ohm's law,
V = IR
where V is voltage, I is current and R is resistance.
I = V/R
= 12.0/2.5
= 4.8 A
Answer:
Explanation:
As given point p is equidistant from both the charges
It must be in the middle of both the charges
Assuming all 3 points lie on the same line
Electric Field due a charge q at a point ,distance r away
Where
- q is the charge
- r is the distance
-
is the permittivity of medium
Let electric field due to charge q be F1 and -q be F2
I is the distance of P from q and also from charge -q
⇒
F1
F2
⇒
F1+F2=
Answer:
that is equal to = 17 × 10 = <u>1</u><u>7</u><u>0</u><u> </u>= 28.33kg
6
