Structural and Compositional Characterisation of Materials

Question 1

State the 3 kinds of microscopy.

Answer 1

Projection image
Optical image
Scanning image

Question 2

How does projection image microscopy work?

Answer 2

An object is placed in from of a point source of illumination.

Question 3

How does optical image microscopy work?

Answer 3

Object is magnified using conventional lenses

Question 4

How does scanning image microscopy work?

Answer 4

Each point of an object is presented serially.

Question 5

How does a field ion microscope work?

Answer 5

High potential is applied between a very fine tip of radius r and a scene at distance R (magnification R/r)
A low pressure inert gas is maintained. Gas atoms are ionised in the very high field near the surface of the specimen.
Each ion is accelerated towards the screen which contributes a speck if light to the image (phosphor screen)
Ionisation occurs preferentially above the atoms which protrude the specimen surface.

Question 6

What are the limitations of field ion microscopes?

Answer 6

Only conducting samples can be studied (conductor or semi-conductor)
Only the surface can be studied which may not be representative of the bulk.

Question 7

Explain how time-of-flight mass spectrometry works.

Answer 7

Aim to determine the elements by mass number and isotopes.
Vaporised sample injected at low pressure. Electron gun fires and kooks an outer electron out of orbit forming positive ions.
Positive ions accelerated by electric field to a constant KE.
Positive ions with smaller m/z will take longer to move through drift area.
Ions distinguished by different flight times by generating a small current when reaching a sensor.

Question 8

Explain how positive sense atom probe works.

Answer 8

Aim to map the atom concentration within a given volume.
High potential or short laser pulse us applied to the tip, a layer of ionised atoms is stripped from the surface.
Ionised atoms will accelerate towards an imaging screen which is specifically orientated so that the ion falls over a hole in the fluorescent screen which leads to a TOF mass spectrometer.
The mass spectrometer can then identify the element within the sample.

Question 9

What are the limitations of atom probe?

Answer 9

Only conducting samples can be studied.
Material needs to be needle shaped specimen
Only a volume of 20x20x50nm can be studied at a time.

Question 10

Describe Abbe’s theory of image formation.

Answer 10

Interaction of illuminating beam and the sample leads to diffraction.
Objective lens corrects the diffraction and a pattern is formed at the back focal plane.
Waves propagate to the image plane and re-interfere to form a magnified image of the sample.
Finite size limits the maximum angle of diffraction collected which limits the smallest features to be seen.

Question 11

Explain the Rayleigh criterion.

Answer 11

Rayleigh’s criterion specifies the minimum separation between two light sources that may be resolved into distinct objects.
This is when the maximum of one diffraction pattern caused by the circular aperture coincides with the minimum of the other diffraction pattern.
See notion answers

Question 12

Explain numerical aperture.

Answer 12

In a microscope, NA is important as it relates to the resolving power if a lens.
A lens with a large NA will be able to resolve finer details.
Lenses with larger NA will also be able to collect more light and so give brighter images. Another way to describe this situation is that the larger the NA the larger the cone of light that can be brought into the lens and so the more diffraction modes will be collected.
Thus the microscope has more information to form a clear image, hence its resolving power will be brighter.

Question 13

What is the role of a condenser lens?

Answer 13

It is the lens that focuses the illuminating beam into the specimen.

Question 14

What is the role of the objective lens?

Answer 14

The lens that leads to the diffraction pattern in a real image.

Question 15

What is the role of the projection lens?

Answer 15

The lens that is used to magnify the image to give a final upright (virtual) image.

Question 16

Describe an optical lens.

Answer 16

Glad with spherical or near spherical polished surfaces, coated to minimise reflection so that light refracts from the surface.
Used in combinations to minimise spherical and chromatic aberrations.

Question 17

Describe a magnetic lens.

Answer 17

A magnetic field (solenoid) is created in which electrons are focus by having a decreasing radius to spiral about the optical lens.
Lenses always convergent
Electron lenses have both spherical and chromatic aberrations
The aberrations of the lenses scale with focal length. Objective lens focal length is just a few mm are used and sample sits in the magnetic field of the lens

Question 18

Describe how spherical aberration arises.

Answer 18

Lenses are made spherical and is not made in the ideal shape for focusing.
Arises because of different path lengths of different rays from an object point to the image point.
Rays furthest from optical axis brings rays to focus near the lens (marginal focus) than does the central portion of the lens (axial focus)
A disk of least confusion exists at the best compromise position of focus.
Check notion to see diagram.

Question 19

Explain how chromatic aberration arises.

Answer 19

Arises due to different wavelengths in photons or the different energies of electrons.
The faster an electron travel, the less it is diffracted by a magnetic lens. The higher the wavelength, the less it is refracted by an optical lens, thus different focal lengths.
A disk of least confusion exists at the compromise position of focus.

Question 20

How can chromatic aberration be removed?

Answer 20

By using monochromatic light (filters) or multiple lenses of different shapes.

Question 21

Explain how astigmatism arises.

Answer 21

Arises because focus for rays travelling in the horizontal plane are at a different position from the focus for rays travelling in the vertical plane because of different path lengths.
A disk of least confusion exists at the compromise position of focus.

Question 22

How can astigmatism be removed?

Answer 22

Using stigmators (correcting lenses) by imposing a weak electric or magnetic quadrupole field on the electron beam.

Question 23

List the different ways that photons can be produced for microscopy.

Answer 23

Heated filaments (cheap, white light, low intensity)
Arc discharge (small, white light, high intensity)
Gas discharge (monochromatic)
Laser (monochromatic, intense)

Question 24

Explain how an electron gun works.

Answer 24

Requires:
-Stable voltage supply
-Controllable filament heater
-controlled bias voltage
Steps:
-Cathode is heated which emits a stream of electrons.
-Electrons are accelerated using an electric field in a vacuum.

Question 25

How can an electron gun be improved?

Answer 25

Pointed filaments to reduce are in which electrons are emitted. Lanthanum Boron filaments which is a brighter source due to lower work function. High electric field to increase intensity.

Question 26

Explain how critical and Koehler illumination works in microscopy.

Answer 26

Light source is focused on specimens using condenser lens. Critical Illumination: -Light from the source is focused on the image plane. Koehler Illumination: -An extra field diagram (Koehler lens in diagram on notion) is placed between the source and the condenser lens. -The rays of the source are then instead focus on the aperture planes (including the eyepiece) and defocus on the sample (illumination).

Question 27

State the advantages of Koehler illumination over Critical illumination.

Answer 27

More illumination on the sample (no glare) | Removes the image of the light source on the final image.

Question 28

How are virtual images detected?

Answer 28

With an eyepiece

Question 29

How are real images detected?

Answer 29

Ground glass, fluorescent phosphor

Question 30

How are photons and electrons detected?

Answer 30

Photographic film CCD TV cameras

Question 31

State the properties of an aperture.

Answer 31

``` Opaque to radiation. Thin in the axial direction Zero reflectivity Adjustable in size Adjustable in position ```

Question 32

Explain how phases contrast microscopy works.

Answer 32

A condenser annulus limits the number of rays entering the illuminated specimen. Condenser focuses the illuminating light on the specimen The specimens then scatters and refracted the light causing a phase shift.The thicker the sample, the large the phase contrast. An objective lens focuses the scattered and illuminating light on to a detector. A phase plate is placed between the detector and the objective lens. The plate is thicker in the region where background light hits it, thus creating a larger phase change. The background light is dimmed by a grey filter ring. The resultant waves interfere to give rise to contrast in the final intensity in regions of the field of view that contain the specimen.

Question 33

Explain how polarising microscopy works.

Answer 33

The incident light becomes plan polarised using a polarising filter. The light is shone on a specimen which causes the polarisation to change depending on the angle of incidence (normally causes elliptical polarisation) The reflected light is then analyses by the polarising unit which is perpendicular to the polariser of the incident light. The reliant intensities are dependent on the angles of polarisation.

Question 34

Explain how interference microscopy works.

Answer 34

Light from the specimen only differs in phase and has the same amplitudes. A reference wave is added to the system which superposes with the existing waves A contrast is seen as the waves have different amplitudes.

Question 35

Explain the difference between 2-beam and multiple beam TEM.

Answer 35

2-beam: -Reciprocal lattice is rotated by double slit only (x-y coordinates) -g(hkl) satisfying the Bragg condition will have a diffraction spot just as intense as the central (000) spot -Only one reciprocal lattice vector (with small indices) is seen due to a decrease in intensity of higher integer vectors Multiple beam: -Many g(hkl) satisfies the Bragg condition and there are many diffraction spots but less intensity than the 000 spot. Many reciprocal lattice vectors are seen.

Question 36

Why do electron diffraction patterns of thin crystals have many spots?

Answer 36

Wavelength of electron is much less than the spacing between the planes. Bragg angle is very small thus the large Ewald sphere intersects many reciprocal lattice points. Bragg condition in thin samples is less rigorous as there are less scattering centres which contribute to the diffraction beam (consider double slit vs multiple slits)

Question 37

Explain what happens to the electron diffraction pattern when the sample thickness increases.

Answer 37

As sample thickness increases, the diffraction spots strait to blur out and disappear due to the stricter Bragg angle as many scattering centres contribute to the diffracted beam.

Question 38

Explain how selective area diffraction (SAD) works in electron microscopy.

Answer 38

Objective lens forms a diffraction pattern at the back focal plane. An intermediate lens is to focus on the back focal plane instead of the specimen. An aperture is place where the intermediate image forms (selected area). The diffraction pattern is projected by the projection lens.

Question 39

Explain the usual imaging mode in electron diffraction.

Answer 39

An aperture is placed at the back focal plane | The intermediate lens focuses on the specimen

Question 40

What happens in convergent beam diffraction and what is the advantage of it in electron microscopy?

Answer 40

It is where an incident beam is condensed into a small spot thus the patter comes from the whole of the illumination area. Advantages: -patterns can be obtained from very small crystallites -direct determination of space groups.

Question 41

Explain the differences in electron patterns from a single crystal, polycrystal and fine scale polycrystal.

Answer 41

Single crystal - regular array of spots. Polycrystal - superimposed spots patterns. Fine scale polycrystal - ring pattern In a polycrystal, there are more diffraction spots as the grains are orientated in different directions. In a fine scale polycrystal, there are grains orientated in almost all directions which gives many more reciprocal lattice points.

Question 42

State the structure factor for FCC, BCC and diamond structures.

Answer 42

FCC: h,k,l are either all odd or all even BCC: h+k+l = even Diamond: h,k,l are all odd or all even AND h+k+l=4n

Question 43

Explain how Kikuchi patterns arise.

Answer 43

Occurs in thicker foils Inelastic scattering followed by subsequent elastic scattering Inelastic scattering causes rays to be scattered diffusely Some of these electrons travel at the Bragg angle and may undergo Bragg diffraction. Inelastic scattering is strongly biased towards low angles, the ray that diffracted at Q, is more intense than the one diffracted at R. (Notion diagram) There is an increase in intensity to the left and a decrease to the right on the distribution of energies. Kikchi line intensity increases with sample thickness.

Question 44

What happens to Kikuchi lines and diffraction pattern if the sample is tilted?

Answer 44

Kikuchi lines move proportional to the tilt angle as inelastic scattering events still happens and the Bragg angle is satisfied. Diffraction pattern remains fixed as the Bragg angle is still satisfied at low tilts but at lower intensity.

Question 45

What causes streaking in electron diffraction patterns?

Answer 45

Platelet and rod like precipitates Planar defects Strain Compositional variations

Question 46

What causes extra spots in electron diffraction patterns?

Answer 46

Ordering of atoms | Precipitates or other phases with particular orientation relationship

Question 47

Explain how phases shifts occur in TEM as electrons go through a specimen.

Answer 47

Negligible amount of electrons are absorbed as they pass through a sample and the amplitude change is negligible. Different refractive indexes and thicknesses lead too variations in phases of the electron wave as they exit.

Question 48

Explain how high resolution phases contrast imaging works.

Answer 48

Thin specimens are required as it is modelled using single scattering event The phases shift is modelled by using the contrast transfer function in which the phase shift can be approximated as contributions from defocus and spherical aberration These phase changed electrons then interfere with other electrons creating an interference pattern.

Question 49

Why are images from high resolution phase contrast microscopy hard to interpret?

Answer 49

Phases shifts are theoretically approximate to the structure and must be compared to simulations to find the right function. It is even harder for thicker specimens as multiple scattering can alter the phases.

Question 50

Explain how diffraction contrast TEM works.

Answer 50

Samples exhibit diffraction contrast, whereby the electron beam undergoes Bragg scattering which in the case of a crystalline sample, disperses electrons to discrete locations in the back focal plane by the objective lens. A small objective aperture is put into the back focal plane of the objective lens, which limits the area of the transmitted beam or the diffracted beam. The diffraction pattern is created by the focusing of the intermediate lens.

Question 51

Explain the difference between light and dark field imaging.

Answer 51

Bright field - the sample is at the optical axis and the transmitted beam is therefore bright. Beams that diffract therefore do not go through the aperture. Dark field - the sample is at an angle (tilted beam) and the diffracted beam is therefore bright. Only the beam that diffracted goes through the aperture.

Question 52

Explain the condition when diffractions caused by dislocations can be seen using bright and dark field imaging.

Answer 52

- Diffraction condition is determined by Braggs law. - At a dislocation atomic planes are either closer/further apart and will therefore either dsatisfy (dislocation causes diffraction) or not satisfy Bragg's law. - For an edge dislocation, the burgers vector is perpendicular to the atomic planes. Therefore if the diffraction vector g is perpendicular to b, the dislocation cannot be seen (parallel to the planes)

Question 53

Explain how bend contours can be seen by TEM.

Answer 53

Bend contours change the orientation wrt the beam causing diffraction (or no diffraction).

Question 54

Explain how coherent ppts can be seen by TEM.

Answer 54

Lattice planes on either side of the ppt distorted due to difference in lattice parameters. Planes passing through the centre of the precipitates are undistorted. A line of no contrast is seen as the planes are normal to the diffraction vector g in the middle of the ppt (obeying Bragg's law) while the lattice planes on either side of the ppt does not. A dark field image shows a bright line of no contrast.

Question 55

Explain how a stacking fault can be seen in TEM.

Answer 55

The region of crystal above the fault is out of register with the crystal below fault (disregistry R) For a fault not to be seen R.g = n as the disregistry. The vector R moves the reflecting planes normal to themselves by a distance equal to an integer number of spacing between planes.

Question 56

Explain how thickness contrast imaging works in TEM.

Answer 56

The objective aperture is placed around the forward scattered beam (excluding many scattered electrons). Scattering increases with mass and thickness, reducing the intensity of the image.

Question 57

What is the principle behind structure factor contrast imaging in TEM.

Answer 57

Strength of diffracted beam controlled by the structure factor which produces contrast.

Question 58

What are the uses of structure factor contrast imaging in TEM?

Answer 58

Amorphous damaging regions in irradiated samples

Question 59

Explain how diffraction conditions are set up.

Answer 59

Set up by titling specimen with reference to Kikuchi lines. In 2 beam conditions, g is strongly excited by tilting to the Bragg angle. The reciprocal lattice is rotated by appropriate double tilt (x and y) so that a particular g(hkl) satisfying the Weiss zone law is brought exactly on the Ewald sphere. In bright field conditions, sample is just tilted away from the 2 beam condition giving brighter field images. In weak beam conditions, image is formed in dark field. The optical axis goes through g, but the sample is titled so that it satisfies 3g. Intensity and then increase only from regions close to core of dislocations and high strain fields to tilt to Bragg condition giving high resolution of highly strained regions.

Question 60

Explain how an SEM works.

Answer 60

Electron gun accelerates electrons using energy between 2-40keV Condenser lenses focus the beam until it has a diameter of only 2-10nm. Fine electron beam scanned across specimens using deflector coils. Detector collects signal and intensity os plotted against scan position to create an image.

Question 61

What are secondary electrons.

Answer 61

Electrons in the sample that are ejected by ionisation that arises from impact of the primary beam in SEM. (Low energy, many electrons)

Question 62

What are backscattered electrons.

Answer 62

Primary electron. that have been multiply scattered in the sample so they are scattered back up towards the lenses in SEM. (High energy, little electrons).

Question 63

Sketch a diagram of the SEM signals generated and their sampling volumes.

Answer 63

Check notion.

Question 64

Explain how secondary electron images are detected and why tilting the sample gives larger contrast.

Answer 64

Detected by a scintillator of photomultiplier system (not ether energy of the secondary electrons are too low so the detector is coated in thin Al foil). The secondary electrons are attracted to the detector by a bias voltage on the grid (10keV) and accelerated towards a phosphor screen which emits light. Light is transmitted through a light pipe and into a photomultiplier which converts photons into pulses of electrons. Amplified and used to modulate the intensity on a Cathode ray tube. Tilting gives a larger interaction volume that is close to the surface so more secondary electrons so more contrast.

Question 65

Explain the information that secondary electron image yield.

Answer 65

Tilted surfaces result in string topographical contrast. Improves yield of secondary electrons recorded on detector if surface faces the detector. Potential biases prevent secondary electrons to escape the detector. Voltage contrast as there is a work function associated with secondary electron production.

Question 66

Explain how backscattered electrons are detected.

Answer 66

Using a solid state detector "Doughnut" type arrangement, concentric with the electron beam, maximising the solid angle of collection BSE detectors are usually either of scintillator or semiconductor types Fraction of incident electrons that are backscattered increases with atomic number (strong composition contrast).

Question 67

Estimate the ultimate resolution of SEM.

Answer 67

Limited by spherical aberration and Rayleigh's criterion | Approx 2nm.

Question 68

Explain why working resolution is not na high as ultimate resolution.

Answer 68

Working resolution (pixel size) -The pixel size is determined by the minimum size that may be obtained on the cathode ray tube monitors (typically 100nm). This limits the specimens pixel size by considering the magnification: P(spec) = P(image)/M Ultimate resolution (sampling volume) -depends on probe size. The sampling volume determines the volume in which the electrons interact with the sample and can escape the surface. -if electron probe is larger than the specimen pixel then signals from adjacent pixels are merged. -if the electron probe is smaller than the specimen pixel then the signal will be weak (small beam current) with a lot of noise. -For secondary electrons this is about the same as the pixel size (good resolution) -for backscattered electrons and x-rays this is bigger than the pixel size so poor resolution.

Question 69

Define depth of field.

Answer 69

The range of positions for the object for which our eye can detect no change in sharpness of the image. Small convergence angle increases depth of field considerably

Question 70

Show that the typical depth of field for an SEM image is larger than that for an optical microscope.

Answer 70

Check notion | Depth of focus h = s/α

Question 71

How does STEM incorporate the advantages of SEM and TEM?

Answer 71

TEM quality lenses and forcing electron beam down to 0.1nm gets atomic resolution scaled image This sample required X-rays and scattered electrons can also be detected.

Question 72

Sketch a typical X-ray spectrum and explain its key features.

Answer 72

Check notion

Question 73

How are characteristic X-rays produced in EDX?

Answer 73

Produced by electrons knocking the inner electron and causing an electron at a higher energy orbital to fall to the lower one emitting an electron.

Question 74

How does Brehmstrahlung arise in EDX?

Answer 74

It arises as electrons get close to the nucleus and change direction producing a photon.

Question 75

How are the highest energy X-rays formed in EDX?

Answer 75

When the nucleus completely stops the electron all the kinetic energy is released as an x-ray,

Question 76

Explain the mechanism of Auger emission and why it tends to dominate.

Answer 76

Auger emission is where an inner electron is knocked out An electron is emitted from the outer shell An electron at a lower energy level lowers its energy to an inner shell

Question 77

Sketch a diagram showing the sampling volume of x-ray microanalysis.

Answer 77

Check notion.

Question 78

Explain the advantages and disadvantages of thick samples in x-ray microanalysis.

Answer 78

Ads: Sampling volume diameter of x-rays a lot larger than the resolution for secondary and backscattered electrons. Disads: Bad resolution as this is a lot larger than pixel size and distinct images in specimen cannot be resolved.

Question 79

Explain the advantages and disadvantages of thin samples in x-ray microanalysis.

Answer 79

Ads: Sampling volume diameter of c-rays is a lot less due to the decreasing depth of penetration into the sample. This causes less lateral spread of the electron beam. Disads: The number of x-rays emitted is smaller than for a bulk specimen. Noise may be a problem.

Question 80

Explain how EDX works.

Answer 80

Detector is collimated so backscattered or secondary electrons cannot reach the detector. Electrons focus on sample and characteristic x-rays produced. Thin beryllium window to separate the two vacuum systems (microscope and detector) The x-rays cannot be deflected. The detector is placed near the sample in the line of sight of the sample. The incoming x-ray excites a number of electrons in the condition band of silicon reverse biased diode, leaving holes in the valence band. Lithium is included to compensate for native impurities present that cause extra holes. As the creation of electron hole pairs is typically 3.8eV, there are several electron hole pairs generated which is proportional tot he energy of the x-rat. This leads to increased conductivity. EDX is operated at low T to reduce residual conductivity from thermal excitation. A current is then produced as the electrons are now free to move in the conduction band when a potential bias is applied. It is also amplified by a pre-amp. Electron trap to prevent backscattered and secondary electrons from entering the detector. Dead zone where holes are present and recombination can occur.

Question 81

Explain how special artefacts of EDX arise.

Answer 81

- Microscope may have electromagnetic coils that change the energy of the X-ray created. This is why the detector is placed as close to the sample as possible. - Escape peaks that occur when the Si x-ray escapes the detector leading to artefact peak equal to the energy of the parent ming the energy off the Si x-ray that is generated. - Some peaks occur when two x-rays arrive at the detector nearly simultaneously so that they are not discriminated by the pulse pileup rejection electronic giving a false peak as the sum of two x-rays.

Question 82

Explain the problems in the use of EDX.

Answer 82

``` Peak broadening (typically 150eV) and many elements overlap. Difficult or impossible to analyse light elements due to the beryllium window. Light elements get filtered. ```

Question 83

State ways that light elements can be detected in EDX.

Answer 83

Use a windowless or ultra thin window detector which is good for elements heavier than C. Use electron energy loss spectroscopy in which inelastic scattering causes ionisation and the characteristic excitation energy if the shell. The energy loss can be measured using a magnetic spectrometer.

Question 84

Explain how WDX works.

Answer 84

X-rays from specimen are filtered so only a chosen wavelength is detected. X-rays are collimated by two sets of slits so that the x-rays are focused onto the crystal. X-rays leaving the specimen are at a certain angle to the crystal which satisfies Braggs law and thus reach the detector. A further slit is used to focus x-rays on detector. The detector then counts x-ray photons arriving of the specific wavelength. The sample, crystal and detector must lie on the Rowland circle so that the x-rays can be focused

Question 85

Compare the advantages and disadvantages of WDX to EDX.

Answer 85

Ads: -Resolving power is excellent. peaks of less than 5eV in width. -Gas proportional counters are capable of very high count rates, possible to collect data from single element in short time. -Better for light element analysis. -WDX more sensitive and can detect low concs. Disads: -EDX is cheaper and simpler to operate. -Whole spectrum is acquired simultaneously using EDX whereas WDX is one wavelength at a time. -EDX better for thin foil analysis and are good at detecting relatively small quantities of material present at relatively high conc in a local area.

Question 86

State the Cliff-Lorimer equation and explain its use.

Answer 86

C1/C2 = k12 l1/l2 A first approximation of relative weight fractions of the two elements. The Cliff-Lorimer factor is dependent on two elements, operating conditions and detector response.

Question 87

Explain how a scanning probe microscope works.

Answer 87

A probe is put very close to a surface and moved by using a piezoelectric crystal that contracts and expands by a small voltage. A voltage is applied between the specimens and the probe top. This causes electron distribution to overlap between a protuberant atom and the probe. As tunnelling only occurs at short distances, the magnitude of the tunnelling current depends on the spacing of atoms in which the probe current is the strongest on top of atoms. The surface can then map to a resolution of about 0.1nm Individual atoms can also be identified according to the current generated at each point.

Question 88

Explain how AFM works.

Answer 88

A fine tip is mounted on a cantilever beam The tip is moved and scans the sample using a piezoelectric crystal. The cantilever beam (deflection remains constant) will rise and fall depending on the surface topography. A laser is shone onto the cantilever and collected by a photodiode, The angle at which the laser is reflected depends on the vertical displacement of the tip as it moves along the surface.

Question 89

Compare the use of scanning probe microscopy and AFM.

Answer 89

AFM can be conducted in most environments while STM can only be conducted in a vacuum. The resolution of AFM is worse than STM as the force between sample and the tip is determined by several neighbouring atoms as well in AFM. AFM requires computer processing and interpretation of images while STMs can directly image specimens. STMs can only work for metals (conductors) as tunnelling is harder due to electrons being less "free", while AFMs can be used for both conductors and insulators as it depends on the forces between tip and atoms.

Question 90

State an equation for contrast.

Answer 90

Check page 5

Question 91

Give the expression for Rayleigh resolution limit.

Answer 91

Check page 10

Question 92

Give the expression for size of spherical aberration.

Answer 92

Check page 15

Question 93

Give an expression for size of chromatic aberration

Answer 93

Check page 16

Question 94

Derive an expression for minimum resolution in the presence of diffraction and spherical aberrations.

Answer 94

Check page 16

Question 95

Give expressions for depth of focus, h.

Answer 95

Check notion.

Question 96

State an expression for pixel size.

Answer 96

Check notion.

Question 97

State the condition for diffraction and no diffraction for dislocation.

Answer 97

If g.b = 0 then no (or weak) contrast will be seen