Microscopy, diffraction and structure evidence

A micrograph shows selected contrast

An image can reveal grains, particles or pores when preparation and contrast are suitable. Magnification, resolution and calibration are different: a large image need not resolve small features accurately. Sampling and image interpretation determine what structural conclusions can be made.

A micrograph is an image made using a microscope. Its brightness is a measured response to the specimen and the imaging method, rather than a label attached directly to a phase. In reflected-light metallography, polishing produces a suitable section; controlled etching can develop surface relief that changes reflected-light contrast at boundaries and within regions. A boundary can become visible because the preparation made it respond differently to light.

Record the section direction, preparation and imaging conditions. A dark feature might be an opening, a differently responding constituent, or damage introduced during preparation. Comparing unetched and suitably etched observations, or using a second contrast method, can test that interpretation. Dark does not mean pore in every image, and two shades do not automatically mean two phases.

An ordinary section is a two-dimensional intersection through a specimen. A round section through a long cylinder can look like a round particle. Record where the section came from before inferring a three-dimensional shape or a whole-batch amount. The distinction between phase identity, grain regions and arrangement is developed in “Phases, grains and microstructure”.

Convert pixels into length before interpreting size

A pixel is one stored image sample. Suppose a calibrated reference establishes that 100 original-image pixels span 25.0 µm under the acquisition conditions. The horizontal scale factor is s = 25.0/100 = 0.250 µm per pixel. A horizontal feature spanning 40 pixels then has estimated projected width w = Ns = 40 × 0.250 = 10.0 µm. Here N is its pixel span, not the number of atoms across it.

The first figure is an original synthetic image with a stipulated uniform, isotropic scale, resolved feature edges and no perspective distortion. Its dark outlined region is a chosen feature, not a chemically identified phase or a real measured specimen. The scale calculation demonstrates dimensional reasoning. Actual edge location, calibration uncertainty, distortion and preparation can limit a reported width.

Enlarging the whole image by a factor of two doubles both the drawn feature width and the drawn scale-bar length. Their ratio remains 40/100 = 80/200 = 0.40, so the same labelled 25.0 µm bar still gives a 10.0 µm feature. The original 0.250 µm per pixel factor must not be applied unchanged to the resampled pixels: their corresponding factor is now 0.125 µm per pixel.

A scale bar stays meaningful when it is resized with the image and its length label is retained. A magnification label depends on the final display or print size. If horizontal and vertical resizing differ, one horizontal bar does not establish the vertical scale. Calibration also needs a justified transfer to the actual instrument settings and specimen geometry; “Calibration, uncertainty and repeatability” explains why a stable reading is not sufficient.

A small pixel is not a guarantee of a resolved feature

Magnification describes enlargement. Resolution concerns distinguishing nearby features as separate. The optics, wavelength, detector sampling, signal and specimen conditions influence what can be distinguished. Enlarging an already blurred spot makes a larger blurred spot; it does not recover two features that the original acquisition failed to separate.

Calibration converts an observed separation into units. It does not establish that two edges were distinguishable in the first place. For example, under an independently established 1.0 µm resolving limit for a stated criterion, two features separated by 0.50 µm cannot be claimed as a resolved pair merely because the image sampling is 0.10 µm per pixel. Five pixel spacings cover that separation, but the blur can still merge the signals. These are chosen teaching values, not a specification for all microscopes.

Conversely, a feature can be visible without its boundary being accurately located. Report the defined quantity: projected width at a stated contrast threshold, a boundary intercept, or another specified measure. A broad claim about “grain size” needs a suitable boundary-identification and sampling method. Merely measuring one attractive region does not supply that method.

Diffraction adds a different kind of evidence

Scattered waves interfere according to spatial arrangement. Crystalline phases can produce characteristic diffraction features. Peak positions, widths and intensities can inform different questions, but instrument response, texture and overlapping peaks must be considered. An image and a diffraction pattern complement each other rather than acting as interchangeable pictures.

For ordinary periodic crystals, think of regularly spaced parallel crystal planes. X-rays scattered by their electrons have wave character. Contributions arriving with crests aligned reinforce one another: constructive interference. An extra travel distance equal to a whole number of wavelengths restores that alignment. The plane construction explains this path difference; the planes are not literal stacks of mirrors.

For equal incident and outgoing angles θ to the planes, the extra distance for the lower of two representative rays is 2d sin θ. Here d is the perpendicular plane spacing, λ is the X-ray wavelength, sin is the sine function, and n is a positive whole number. The Bragg relation is nλ = 2d sin θ. Both d and λ must use the same length unit. A calculator in degree mode is appropriate for angles stated in degrees.

In this geometry the scattering angle between the forward incident direction and the outgoing beam is 2θ; powder patterns commonly use 2θ as their horizontal axis. A peak at 2θ = 60° therefore uses θ = 30° in the relation. For n = 1 and λ = 0.150 nm, d = 0.150/(2 × sin 30°) = 0.150 nm, since sin 30° = 0.5. This is an invented single-spacing example, not the identification of a material.

At fixed wavelength and order, increasing d requires a smaller sin θ and hence a smaller θ in the 0° to 90° range. A lower-angle position can therefore indicate a larger spacing for a correctly associated reflection. First establish angle calibration and the reflection being compared; a setup shift can also move peaks. The earlier “Crystalline, amorphous and molecular arrangements” lesson supplies the distinction between long-range order and local arrangement.

Position, width and intensity answer different questions

A powder pattern plots measured scattered intensity against angle. Many differently oriented crystalline regions can contribute to a given reflection. A set of compatible peak positions can support a proposed crystalline phase; one coincident peak is rarely a unique identity. Atomic arrangement, phase abundance, preferred orientations and measurement conditions also influence intensities. A tall peak is not a direct map of where a phase sits in the specimen.

Width describes how intensity spreads around a peak position. In the final plot, two invented isolated profiles share a centre at 2θ = 60° but have full widths at half maximum of 2° and 4°. Full width means the angular distance between the two half-height crossings. Half height is measured above the local background: with background 10 and peak intensity 90 in chosen arbitrary units, the halfway level is 10 + (90 − 10)/2 = 50.

The plotted profiles are simple Gaussian illustrations, separately scaled to the same height; their widths are chosen for visibility. They are not instrument data, a phase-abundance comparison or a fit to a real line. Their centres agree while their widths differ, demonstrating why a peak position alone does not describe its shape.

A short extent of coherent crystal order can broaden a line: fewer repeats contribute to a sharply defined interference condition. A coherent domain is a region over which the crystal arrangement contributes with the required phase relationship. It need not be an entire optically outlined grain. A grain can contain defects or subregions that shorten that coherent extent.

Variations in lattice spacing within the sampled material, often described as microstrain, can spread the angles that meet the diffraction condition and also broaden a peak. Instrument response broadens it too. Overlapping reflections can mimic a wide single line. Standards, several suitable reflections and an appropriate line-profile model are needed to separate contributions; subtracting raw widths without a model is not a general correction. This lesson does not extract a domain size from a single width.

Use a second measurement to test an explanation

Suppose a deformed metal still shows approximately the same optical grain outlines, but its diffraction peaks become broader. “Nothing changed because the grains look the same” and “the grains must all be smaller because the peaks widened” both exceed the evidence. The image tests visible outlines under its resolution and contrast conditions; line width is sensitive to a different structural description.

First compare a suitable reference measured under the relevant diffraction conditions, check specimen positioning and inspect the full profiles. If a reference also broadens, an instrumental change is a live explanation. If the reference remains comparable while several specimen lines broaden, investigate specimen contributions such as strain variation or coherent-domain subdivision. Neither comparison alone identifies a unique mechanism.

Then choose the complementary observation for the remaining question. Calibrated microscopy with suitable boundary contrast can test changes in visible grain dimensions. An orientation-sensitive map at an adequate spatial scale can test subregions; suitable multi-line diffraction analysis can assess the proposed broadening contributions. Composition-sensitive evidence may help distinguish a new constituent from a contrast change. State what each observation can and cannot discriminate.

Retain the same specimen location or justify representative sampling, and record preparation and acquisition conditions. The goal is to turn a structural claim into competing predictions and a discriminating observation, building on “From a property claim to a measurement”. Extending that result to an unseen volume or a batch needs the further reasoning in “Sampling, scale and extrapolation limits”.

Worked examples

Try the reasoning

1. Does increasing digital zoom improve resolving power? An acquisition merges two nearby features into one broad spot; enlarging its pixels fourfold shows a large spot. Can its two constituent widths now be measured separately?

   Show the answer and reasoning

   No. Enlargement alone has not resolved the pair or supplied their separate widths.

   Digital zoom enlarges the recorded intensity information. A scale factor can assign a length to the broad spot, but resolving the two contributors requires adequate acquisition and an appropriate analysis. Distinguish a measured combined profile from two independently located boundaries.

2. Why use diffraction standards? After an instrument change, both a suitable reference and a specimen show wider lines. A report assigns the specimen change entirely to smaller grains. What is the missing comparison?

   Show the answer and reasoning

   The changed instrument response must be characterised; the common widening prevents that unique specimen interpretation.

   Reference line profiles help evaluate the instrument under the relevant conditions. They do not automatically make every specimen correction valid or determine grain size. Recheck setup and profile analysis, then examine what specimen contribution remains under a justified model.

3. An original image has equal horizontal and vertical scale factors of 0.250 µm per pixel. A rectangular feature spans 40 pixels horizontally and 24 vertically. An editor doubles only the image width, making those spans 80 and 24 pixels. Calculate the two scale factors and feature dimensions after editing. Can the new horizontal factor be used vertically?

   Show the answer and reasoning

   The new factors are 0.125 µm per horizontal pixel and 0.250 µm per vertical pixel. The feature remains 10.0 µm wide and 6.00 µm high; one factor cannot serve both directions.

   The edited horizontal pixels cover half the old physical distance; vertical sampling is unchanged. Thus 80 × 0.125 = 10.0 and 24 × 0.250 = 6.00. Applying 0.125 vertically would incorrectly halve the height to 3.00 µm. Preserve both calibration relationships and avoid interpreting the distorted displayed aspect ratio as specimen shape.

4. A second invented first-order reflection is at 2θ = 90° with the same wavelength 0.150 nm. Using sin 45° = 0.7071, estimate its plane spacing. Is this larger or smaller than the 0.150 nm spacing at 2θ = 60°? Does the single line identify a chemical composition?

   Show the answer and reasoning

   d ≈ 0.1061 nm, which is smaller. The single line does not identify a chemical composition.

   Use half the plotted angle: θ = 45°. Then d = 0.150/(2 × 0.7071) ≈ 0.1061 nm. The greater angle corresponds to smaller spacing for fixed order and wavelength. Reflection assignment and other compatible evidence are still required before identifying a phase or composition.

5. A chosen isolated peak has background intensity 20 and maximum 100 in arbitrary units. At which intensity should its half-maximum width above background be read? If its crossings are at 39.7° and 40.3° in 2θ, find the full width.

   Show the answer and reasoning

   Read the crossings at intensity 60, giving a full width of 0.6° in 2θ.

   The peak rises 100 − 20 = 80 above background; halfway up is 20 + 80/2 = 60. The angular difference is 40.3 − 39.7 = 0.6°. Using intensity 50 ignores the background and gives a different width. Baseline choice, overlap and uncertainty need assessment in an actual profile.

6. Dark round features occupy part of one polished section. A report calls them pores throughout the whole material because the diffraction pattern matches the expected crystalline phase. Give two remaining uncertainties and a useful next observation.

   Show the answer and reasoning

   The dark-feature identity and their three-dimensional representativeness remain uncertain. Check preparation/contrast and examine suitable additional sections or another appropriate spatial measurement.

   Matching crystalline-phase evidence does not locate every dark region or prove that a region is empty. Preparation can create misleading features, and a single section does not establish three-dimensional shape or batch abundance. Repeating appropriate preparation and contrasting the same area can test an artifact explanation; planned additional spatial sampling tests representativeness. Choose the observation for the remaining claim rather than adding more zoom to the same image.

What to carry forward

• An image needs a contrast interpretation, a calibrated scale and adequate resolution; these are separate requirements.
• Resizing the image and scale bar together preserves their ratio, while pixel calibration must follow the resizing.
• Bragg reasoning connects a defined plane spacing to angle and wavelength; use θ when the pattern axis is 2θ.
• Peak position, width and intensity contain different information. Coherent domains need not coincide with visible grain outlines.
• Combine observations that test competing explanations, retaining preparation, instrument response and sampling limits.