Which atmospheric window is used to record the Earth’s own thermal emission, and why?

8–14 µm, because the Earth’s emission peaks near 10 µm and the air is largely transparent there

0.4–0.7 µm, because the Earth’s own emission peaks in the visible part of the spectrum

1.3–3.0 µm, because the Earth emits most strongly in the shortwave infrared at ordinary temperatures

8–14 µm, because the Earth’s emission peaks near 10 µm and the air is largely transparent there

1 mm to 1 m, because the Earth’s emission peaks in the microwave region at ordinary temperatures

In a standard false colour composite (NIR shown in red, red in green, green in blue), how do a healthy crop field and a clear deep lake appear?

Bright red for the crop field and dark blue to black for the lake

Bright green for the crop field and bright red for the open lake water

Dark blue for the crop field and bright red for the deep lake water

White for the crop field and bright cyan for the clear deep lake water

A SWIR image of a hydrothermally altered zone shows a strong absorption feature near 2.2 µm. Which assemblage does this most directly indicate?

Al–OH-bearing minerals such as kaolinite and sericite, typical of argillic and phyllic zones

Ferric iron oxides such as hematite and goethite, whose diagnostic absorptions lie near 2.2 µm

Al–OH-bearing minerals such as kaolinite and sericite, typical of argillic and phyllic zones

Calcite-rich carbonate, because the carbonate absorption lies at the same position near 2.2 µm

Quartz-rich silicified rock, because quartz has a strong diagnostic absorption near 2.2 µm in the SWIR

Which statement correctly compares pushbroom and whiskbroom scanners?

A pushbroom scanner images a full line at once with a detector array, so it has the longer dwell time

A whiskbroom scanner uses a long linear detector array and no moving parts, so it has the longer dwell time

A pushbroom scanner builds each line with an oscillating mirror, so its dwell time per pixel is the shorter

Pushbroom and whiskbroom scanners have the same dwell time, because both are limited by orbital speed alone

A pushbroom scanner images a full line at once with a detector array, so it has the longer dwell time

Which statement correctly describes radar layover?

It occurs when the top of a slope facing the sensor is nearer in slant range than its base, so they swap

It is the complete lack of any radar signal in the zone behind steep slopes that face away from the sensor

It is the compression of a slope facing the sensor, with top and base still imaged in their true order

It occurs when the top of a slope facing the sensor is nearer in slant range than its base, so they swap

It is the very bright return from corner reflectors such as buildings and bridges, caused by double bounce

Two C-band SAR passes (wavelength 5.6 cm) show an interferogram with one complete fringe (a phase change of 2π) across a landslide. The line-of-sight displacement across that fringe is

2.8 cm, equal to half the wavelength

5.6 cm, equal to the wavelength

1.4 cm, equal to a quarter of the wavelength

11.2 cm, equal to twice the wavelength

2.8 cm, equal to half the wavelength

Which resampling method keeps the original digital numbers unchanged and is therefore preferred when the output will be classified?

Nearest neighbour, which copies the value of the closest input pixel

Bilinear interpolation, which averages the four nearest pixels

Cubic convolution, which fits a surface through sixteen neighbouring pixels

All three, since resampling never alters the values recorded by the sensor

Why is a band ratio such as TM 5/7 preferred to a single band for mapping clay-rich alteration in rugged terrain?

Slope and illumination scale both bands almost equally, so the ratio cancels shading but keeps clay contrast

The ratio raises the radiometric resolution of the sensor by adding together the bits of the two bands used

Slope and illumination scale both bands almost equally, so the ratio cancels shading but keeps clay contrast

The ratio cancels the atmospheric scattering that is the same in both bands, leaving only the clay signal

The ratio converts digital numbers directly to surface reflectance, thereby removing all atmospheric effects from the image

A single-band raster covers 12 km × 9 km with a 30 m cell size and stores 16 bits per cell. Its uncompressed size is closest to

240 kB, which counts 400 × 300 = 120 000 cells at two bytes each

120 kB, which counts only one byte per cell, as for an 8-bit raster

240 kB, which counts 400 × 300 = 120 000 cells at two bytes each

480 kB, which counts four bytes per cell, as for a 32-bit raster

1920 kB, which wrongly counts the 16 bits of each cell as if they were 16 bytes

A geologist wants the total area of the parts of a lithological polygon layer that lie within 500 m of the mapped fault lines. Which workflow is correct?

Buffer the faults by 500 m, intersect with the lithology polygons and measure areas in a projected coordinate system

Buffer the lithology polygons by 500 m, dissolve them into one feature and measure its area in latitude and longitude degrees

Intersect the fault lines directly with the lithology polygons and then add up the areas of all the resulting line features

Union the fault and lithology layers, then add up every polygon whose attributes include the word fault

Which of the following statements about hyperspectral and multispectral sensing are correct?

A hyperspectral sensor records hundreds of narrow contiguous bands, so each pixel yields a nearly continuous spectrum

The data volume of a hyperspectral image grows in proportion to the number of bands, so the cube is much larger than a multispectral image of the same size

A hyperspectral sensor records hundreds of narrow contiguous bands, so each pixel yields a nearly continuous spectrum

A multispectral sensor with a few broad bands resolves absorption features about 20 nm wide as well as a hyperspectral sensor

The data volume of a hyperspectral image grows in proportion to the number of bands, so the cube is much larger than a multispectral image of the same size

Hyperspectral imaging operates only in the thermal infrared window between 8 and 14 µm

Which of the following statements about GIS data and operations are correct?

A vector layer represents discrete features as points, lines and polygons and can store topology such as adjacency and connectivity

Cell-by-cell overlay of two rasters needs a common coordinate system, cell size and alignment, or prior resampling

Area measured on a layer stored in geographic coordinates in degrees is directly in square metres

A vector layer represents discrete features as points, lines and polygons and can store topology such as adjacency and connectivity

A focal raster operation assigns each output cell a value computed from that single input cell alone

Cell-by-cell overlay of two rasters needs a common coordinate system, cell size and alignment, or prior resampling

Which of the following statements about geological applications of remote sensing are correct?

Optical and thermal sensing sees only a surface skin, so the subsurface geology is inferred and must be checked in the field

Lineaments mapped on imagery or a DEM often mark fractures and faults, and their intersections are favoured targets for groundwater

Optical images taken at a high sun elevation give the best relief enhancement for mapping subtle structures

Optical and thermal sensing sees only a surface skin, so the subsurface geology is inferred and must be checked in the field

Lineaments mapped on imagery or a DEM often mark fractures and faults, and their intersections are favoured targets for groundwater

L-band radar cannot penetrate any dry soil, so it is useless for detecting buried features

Remote Sensing and GIS: EM Spectrum, Multispectral and Hyperspectral Sensing, Image Processing, GIS and Geological Applications

This chapter answers the Part B1 heading “Remote Sensing”: “EM spectrum, multispectral remote sensing in visible, infrared, thermal IR and microwave regions including hyperspectral; Digital processing of satellite images. GIS – basic concepts, raster and vector mode operations; Applications of remote sensing tools and techniques in geology.” It is the third subject of Part B1 in the Geology and Geophysics (GG) paper. It teaches radiation laws and atmospheric windows, sensors and resolution, optical, thermal, microwave and hyperspectral sensing, image processing and accuracy, GIS data models and operations, and geological uses. The numericals are ground resolution and swath, radiant temperature, radar range resolution, NDVI, kappa, raster size and weighted overlay.

1. The EM spectrum, radiation principles, atmosphere and surface interaction

Electromagnetic radiation obeys c = fλ and E = hf. The bands used are visible 0.4–0.7 µm, near-infrared (NIR) 0.7–1.3 µm, shortwave infrared (SWIR) 1.3–3 µm, mid-infrared 3–8 µm, thermal infrared (TIR) 8–14 µm and microwave 1 mm–1 m. Every body above 0 K emits: a blackbody emits M = σT⁴ (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴), with the peak at λ_max = 2898/T µm (T in kelvin, Wien’s displacement law). The Sun (about 5800 K) peaks near 0.5 µm and the Earth (about 300 K) near 9.7 µm, which is why reflected sunlight dominates the visible to SWIR and the Earth’s own emission dominates the TIR. A real surface emits a fraction ε (emissivity) of the blackbody value, and at equilibrium absorptivity equals emissivity.

The atmosphere acts through scattering (Rayleigh, by particles much smaller than λ, ∝ λ⁻⁴, hence blue haze; Mie, by particles comparable to λ such as dust and smoke; non-selective, by water droplets much larger than λ, so clouds look white) and absorption by water vapour, CO₂ and ozone. What is left are the atmospheric windows: the visible, parts of the NIR and SWIR, 3–5 µm, 8–14 µm and the microwave region, where wavelengths of a few centimetres and longer pass through cloud and haze. Sensors record at-sensor radiance, which atmospheric correction converts to surface reflectance.

At the surface, incident energy is reflected, absorbed or transmitted, so ρ + α + τ = 1 at each wavelength. The wavelength dependence is the spectral signature. Healthy vegetation reflects little in the blue and red (chlorophyll absorption), more in the green, and rises sharply near 0.7 µm (the red edge) to a high NIR plateau from leaf cell structure, with water dips near 1.4 and 1.9 µm. Water is dark and nearly black in NIR and SWIR. Dry soil brightens smoothly with wavelength; moisture and organic matter darken it. Minerals show diagnostic absorptions: ferric iron oxides in the visible and near 0.9 µm; Al–OH in clays and micas near 2.2 µm; Mg–OH (chlorite, epidote) and carbonates near 2.3–2.35 µm; and silicates, with quartz the clearest, have emissivity minima in the 8–10 µm part of the TIR.

⚠️ Which window for which source

Do not use the thermal window to look for reflected colour, or the visible window to see the Earth’s own heat. Reflected sunlight carries the visible to SWIR signature; the 8–14 µm window carries emission, which depends on temperature and emissivity together.

2. Sensors, platforms and resolution

A satellite remote-sensing system has four kinds of resolution. Spatial is the ground size of a pixel, ground sample distance GSD = H × IFOV (H altitude, IFOV the instantaneous field of view in radians). Spectral is the number and narrowness of bands. Radiometric is the number of levels, 2ⁿ for n bits. Temporal is the revisit interval. The swath is 2H tan(FOV/2). Example: H = 800 km and IFOV = 0.03 mrad give GSD = 800 × 10³ × 0.03 × 10⁻³ = 24 m; a total FOV of 10° gives a swath of 2 × 800 × tan 5° = 140 km, about 5830 pixels across.

Scanners. A whiskbroom (across-track) scanner sweeps a rotating mirror over a few detectors (Landsat TM and ETM+). A pushbroom (along-track) scanner has a linear detector array that images a whole line at once, with no moving mirror and a longer dwell time per pixel, hence better signal-to-noise and geometry (IRS LISS, Sentinel-2, Landsat OLI).
Orbits. Earth-observation satellites are mostly sun-synchronous, near-polar orbits of 600–900 km that cross the equator at a fixed local time, giving constant illumination for comparison; weather satellites are geostationary (about 35 800 km).
Typical sensors. Landsat multispectral at 30 m (TM, ETM+, OLI; TIR at 100 m on Landsat 8); Sentinel-2 with 13 bands at 10, 20 and 60 m; IRS LISS-IV (about 5.8 m), LISS-III (about 23.5 m) and AWiFS (about 56 m); ASTER with 14 bands (VNIR 15 m, SWIR 30 m, TIR 90 m), valued in geology.

⚠️ Resolution trade-offs

Finer GSD with the same detector means a narrower swath and a longer revisit. More bits per pixel improve radiometric resolution, not spatial detail. Higher spectral resolution means narrower bands, so each band receives less energy and the signal-to-noise falls.

3. Visible, infrared and thermal-IR remote sensing

In the visible to SWIR the sensor records reflected sunlight. In a standard false colour composite NIR is shown in red, red in green and green in blue, so healthy vegetation is bright red, clear deep water is dark blue to black, and bare soil is grey to brown. Choosing bands at the diagnostic features separates materials: SWIR bands near 2.2 and 2.3 µm for clays and carbonates, visible and NIR bands for iron oxides, and NIR and red for vegetation.

In the thermal IR (8–14 µm window) the sensor records emitted radiation, day or night. The radiant temperature is lower than the true kinetic temperature when ε < 1: since εσT_kin⁴ = σT_rad⁴, T_rad = ε^(1/4) T_kin. For ε = 0.95 and T_kin = 300 K, T_rad = 296.2 K. The temperature of a surface depends also on thermal inertia P = √(kρc): high-inertia materials (rock, water, wet soil) show a small day-night range and low-inertia ones (dry sand, loose soil) a large range, so day-night pairs separate materials. Uses include lithology (through silicate emissivity), geothermal anomalies, coal-fire and volcano monitoring, and soil moisture.

🎯 Emissivity matters in TIR

Two surfaces at the same true temperature can look different in TIR if their emissivities differ, and one surface can look cooler than it is. Multi-band TIR sensors such as ASTER separate temperature from emissivity, and the emissivity spectrum then identifies silicate content.

4. Microwave and hyperspectral remote sensing

Passive microwave radiometers measure natural emission at coarse resolution. Active imaging radar illuminates the ground from the side with its own pulses, so it works by day and night and through cloud. Common bands are X (about 3 cm), C (about 5.6 cm), L (about 23 cm) and P (about 70 cm); penetration into dry soil and vegetation canopy grows with wavelength. With pulse length τ, the slant-range resolution is R_s = cτ/2 and the ground-range resolution is R_g = cτ/(2 sin θ), θ the incidence angle from the vertical, so it is finer at far range. Example: τ = 0.1 µs gives R_s = 15 m, and 30 m on the ground at θ = 30°. The along-track (azimuth) resolution of a real-aperture radar is λR/L (antenna length L, range R) and worsens with range; a synthetic-aperture radar (SAR) achieves about L/2, independent of range.

Radar images record backscatter, which depends on roughness (a surface is smooth if its relief h < λ/(8 cos θ), so calm water reflects away and looks dark), on the dielectric constant (wet soil and metal return more), on incidence angle and polarisation (HH, VV, HV), and on local slope. Because the geometry is side-looking, slopes facing the sensor are foreshortened; if a slope is steeper than the incidence angle the top is closer in slant range than the base and the order reverses (layover); slopes facing away may receive no pulse (shadow). InSAR compares the phase of two passes: Δφ = 4πd/λ, so one fringe (2π) is a line-of-sight displacement of λ/2, 2.8 cm for C-band, which maps earthquake, landslide, subsidence and volcano deformation.

Hyperspectral (imaging spectrometry) records hundreds of narrow, contiguous bands (about 10 nm wide) across 0.4–2.5 µm, for example AVIRIS with 224 bands and Hyperion with 220 bands at 30 m. Each pixel yields a nearly continuous spectrum, so narrow diagnostic absorptions (position, depth, shape) identify minerals directly, using spectral matching such as the spectral angle mapper, continuum removal and unmixing. A multispectral sensor with a few broad bands (Landsat bands are roughly 60–270 nm wide) averages such features away. The price is data volume (rows × columns × bands × bytes), lower signal per band and strong band-to-band redundancy, handled by dimensionality reduction (PCA, MNF).

⚠️ Layover is not shadow

Foreshortening compresses a slope facing the sensor, layover reverses top and base when that slope is steeper than the incidence angle, and shadow appears behind steep slopes facing away. Radar slant range is also not ground distance: it must be converted using the incidence angle.

5. Digital processing of satellite images

A digital image is a matrix of digital numbers (DN) per band. Pre-processing has two parts. Radiometric: sensor calibration, repair of striping and drop-outs, haze removal (for example dark-object subtraction) and atmospheric correction. Geometric: removing distortion by choosing ground control points, fitting a polynomial, then resampling: nearest neighbour keeps the original DN values (preferred before classification), bilinear and cubic convolution interpolate and smooth. Orthorectification with a DEM removes relief displacement. The fit is judged by the RMSE of the control points, ideally below one pixel.

Enhancement includes contrast stretching (linear or histogram equalisation) and spatial filtering: low-pass for smoothing, high-pass for edges and directional filters for lineaments. Spectral transforms create new bands. A band ratio divides one band by another; because shading from slope and illumination affects both bands almost equally, it suppresses topography and isolates spectral slope. For Landsat TM, 5/7 is high for clays (absorption in band 7, 2.2 µm) and 3/1 for iron oxides; for Landsat 8 OLI the equivalents are 6/7 and 4/2. NDVI = (NIR − Red)/(NIR + Red) ranges from −1 to +1: water is negative, bare soil near 0.1 to 0.2 and dense healthy vegetation about 0.6 to 0.9. Example: NIR = 0.50 and red = 0.08 give 0.42/0.58 = 0.72. PCA rotates correlated bands into uncorrelated components, with PC1 carrying the largest variance, mainly albedo and topography.

Classification assigns each pixel to a class. Unsupervised methods (K-means, ISODATA) group pixels by spectral clustering and the analyst labels the clusters; supervised methods (minimum distance, parallelepiped, maximum likelihood) use training areas of known cover. The result is checked against reference samples in an error matrix (map classes in rows, reference in columns). Overall accuracy = diagonal total ÷ N; user’s accuracy = correct ÷ row total (reliability of a map class); producer’s accuracy = correct ÷ column total (how much of the reference was found). Kappa κ = (p_o − p_e)/(1 − p_e), with p_o the overall accuracy and p_e = Σ(row_i × col_i)/N², corrects for chance agreement.

Example error matrix, N = 100 (rows: map, columns: reference)
Map class	Ref. A	Ref. B	Ref. C	Row total
A	30	6	4	40
B	3	20	2	25
C	2	4	29	35
Column total	35	30	35	100

Reading the matrix: overall accuracy = (30 + 20 + 29)/100 = 79 per cent. User’s accuracy of A is 30/40 = 75 per cent, while producer’s accuracy of B is 20/30 = 66.7 per cent. With p_e = (40 × 35 + 25 × 30 + 35 × 35)/100² = 0.3375, κ = (0.79 − 0.3375)/(1 − 0.3375) = 0.68.

6. GIS: basic concepts, raster and vector operations

A Geographic Information System captures, stores, analyses and displays spatial data tied to the Earth. Its parts are data, software, hardware, people and methods. Every feature has a geometry (where) and attributes (what). Vector data store discrete features as points (wells, samples), lines (faults, streams) and polygons (lithological units), with exact boundaries and optional topology (connectivity, adjacency, containment). Raster data store a grid of cells, each with one value, which suits continuous surfaces (DEM, satellite images, rainfall); a finer cell size gives more detail and a larger file, and boundaries become stair-stepped. Raster size = rows × columns × bytes per cell.

Vector operations. Buffer (zone of fixed distance), overlay (intersect keeps the common part, union keeps all, clip cuts to a boundary, erase removes), dissolve (merge by attribute), spatial join and proximity or network analysis. Lines and points have no area, so only polygons can be intersected to give an area.
Raster operations (map algebra). Local (cell by cell: reclassify, arithmetic, weighted overlay), focal (neighbourhood: slope, smoothing filters), zonal (statistics per zone) and global (distance, cost surface, viewshed, flow accumulation). Layers combined must share coordinate system, cell size and alignment, or be resampled.
Coordinate systems. A geographic system uses latitude and longitude in degrees on a datum such as WGS84. A projected system (for example UTM, in 6° zones) uses metres. Distance and area must be computed in a suitable projected system, never in degrees.

Examples. A 12 km × 9 km area at 30 m cells has 400 × 300 = 120 000 cells, so 240 000 bytes at 16 bits; halving the resolution to 60 m cuts it to a quarter. From a DEM, slope = arctan(rise/run): two neighbouring cells 30 m apart with elevations 15 m apart give arctan(0.5) = 26.6°. A weighted overlay adds rating × weight: lithology (weight 0.4, rating 8), lineament density (0.3, rating 6) and slope (0.3, rating 4) give 3.2 + 1.8 + 1.2 = 6.2.

⚠️ Degrees are not metres

Area or buffer distance measured in a layer stored in latitude and longitude is meaningless (a degree of longitude shrinks towards the poles). Reproject first. Likewise raster overlay with unequal cell sizes gives misaligned cells unless one layer is resampled.

7. Applications of remote sensing in geology

Lithological mapping: spectral contrast and texture, with SWIR bands for carbonates and clays and TIR bands for silica-rich and silica-poor rocks.
Structural mapping: lineaments (often fractures and faults), folds from outcrop patterns, drainage anomalies, offsets. A low sun angle enhances relief; DEMs and radar add terrain information.
Mineral exploration: hydrothermal alteration zones such as argillic and phyllic (Al–OH near 2.2 µm), propylitic (Mg–OH near 2.3 µm) and iron-oxide gossans, mapped by band ratios, PCA and, in detail, hyperspectral matching.
Groundwater: potential zones from weighted overlay of lithology, lineament density, geomorphology, drainage density, slope and land use.
Hazards and energy: landslide inventory and susceptibility, InSAR deformation, volcano and coal-fire thermal monitoring, and SAR detection of oil slicks and surface structure for petroleum.

⚠️ The image shows the surface

Optical and thermal sensing sees only a surface skin, often hidden by soil, weathering, vegetation and cloud; L-band radar penetrates some metres of very dry sand but not wet or vegetated ground. Remote sensing gives a hypothesis for the geology, which must be checked by ground truth and, for depth, by geophysics and drilling.

Key takeaways

c = fλ; blackbody M = σT⁴ and λ_max = 2898/T µm, so the Sun peaks near 0.5 µm and the Earth near 9.7 µm. Windows: visible, parts of NIR and SWIR, 3–5 µm, 8–14 µm, microwave. ρ + α + τ = 1.
GSD = H × IFOV, swath = 2H tan(FOV/2). Pushbroom arrays have a longer dwell time than whiskbroom mirrors. Sun-synchronous orbits give constant illumination; finer pixels mean narrower swaths.
Vegetation has the red edge near 0.7 µm; Al–OH clays absorb near 2.2 µm, carbonates and Mg–OH near 2.3 µm. In TIR T_rad = ε^(1/4) T_kin and thermal inertia separates materials by day-night range.
Radar: R_s = cτ/2, R_g = cτ/(2 sin θ); SAR azimuth resolution is about L/2; layover, foreshortening and shadow come from side-looking geometry; InSAR fringe = λ/2 of line-of-sight motion. Hyperspectral gives hundreds of narrow bands and a continuous spectrum.
Processing: correct, resample (nearest neighbour keeps DN), enhance, transform (NDVI = (NIR − Red)/(NIR + Red); ratios suppress shading; PCA), classify, then check with an error matrix: overall, user’s and producer’s accuracy and κ = (p_o − p_e)/(1 − p_e).
GIS: vector (points, lines, polygons, topology) against raster (cells; size = rows × columns × bytes). Buffer, overlay and dissolve are vector operations; local, focal, zonal and global are raster ones. Measure area in a projected system, and apply remote sensing to geology as a hypothesis checked on the ground.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Which atmospheric window is used to record the Earth’s own thermal emission, and why?
1. 0.4–0.7 µm, because the Earth’s own emission peaks in the visible part of the spectrum
2. 1.3–3.0 µm, because the Earth emits most strongly in the shortwave infrared at ordinary temperatures
3. 8–14 µm, because the Earth’s emission peaks near 10 µm and the air is largely transparent there
4. 1 mm to 1 m, because the Earth’s emission peaks in the microwave region at ordinary temperatures
Show answer
Answer: C — 8–14 µm, because the Earth’s emission peaks near 10 µm and the air is largely transparent there
By Wien’s law a body at about 300 K peaks at 2898/300 ≈ 9.7 µm, inside the 8–14 µm window where the atmosphere transmits well. The visible and SWIR contain reflected sunlight, not the Earth’s own emission, which is negligible there, and the microwave emission of the Earth is weak and is used only by coarse passive radiometers.
In a standard false colour composite (NIR shown in red, red in green, green in blue), how do a healthy crop field and a clear deep lake appear?
1. Bright red for the crop field and dark blue to black for the lake
2. Bright green for the crop field and bright red for the open lake water
3. Dark blue for the crop field and bright red for the deep lake water
4. White for the crop field and bright cyan for the clear deep lake water
Show answer
Answer: A — Bright red for the crop field and dark blue to black for the lake
Healthy vegetation has a very high NIR reflectance, which drives the red gun, and low red and blue values, so it appears bright red. Clear deep water absorbs almost all NIR and SWIR and reflects little in the visible, so it is dark blue to black. The other pairings reverse these properties or ignore that water is nearly black in NIR.
A SWIR image of a hydrothermally altered zone shows a strong absorption feature near 2.2 µm. Which assemblage does this most directly indicate?
1. Ferric iron oxides such as hematite and goethite, whose diagnostic absorptions lie near 2.2 µm
2. Al–OH-bearing minerals such as kaolinite and sericite, typical of argillic and phyllic zones
3. Calcite-rich carbonate, because the carbonate absorption lies at the same position near 2.2 µm
4. Quartz-rich silicified rock, because quartz has a strong diagnostic absorption near 2.2 µm in the SWIR
Show answer
Answer: B — Al–OH-bearing minerals such as kaolinite and sericite, typical of argillic and phyllic zones
The Al–OH vibrational absorption of clays and micas lies near 2.2 µm, which is why a SWIR band at 2.2 µm is used for argillic and phyllic zones. Carbonates and Mg–OH minerals absorb near 2.3–2.35 µm, a close but distinct position. Ferric oxides absorb in the visible and near 0.9 µm, and quartz has no diagnostic feature in the SWIR, only in the TIR.
Which statement correctly compares pushbroom and whiskbroom scanners?
1. A whiskbroom scanner uses a long linear detector array and no moving parts, so it has the longer dwell time
2. A pushbroom scanner builds each line with an oscillating mirror, so its dwell time per pixel is the shorter
3. Pushbroom and whiskbroom scanners have the same dwell time, because both are limited by orbital speed alone
4. A pushbroom scanner images a full line at once with a detector array, so it has the longer dwell time
Show answer
Answer: D — A pushbroom scanner images a full line at once with a detector array, so it has the longer dwell time
A pushbroom sensor has a linear array that records an entire cross-track line simultaneously, so each detector dwells on a pixel for the whole line time, which improves signal-to-noise and geometric stability and removes the moving mirror (IRS LISS, Sentinel-2, Landsat OLI). A whiskbroom scanner sweeps a mirror across the track over a few detectors, so its dwell time is shorter, and the two do not have equal dwell times.
Which statement correctly describes radar layover?
1. It is the complete lack of any radar signal in the zone behind steep slopes that face away from the sensor
2. It is the compression of a slope facing the sensor, with top and base still imaged in their true order
3. It occurs when the top of a slope facing the sensor is nearer in slant range than its base, so they swap
4. It is the very bright return from corner reflectors such as buildings and bridges, caused by double bounce
Show answer
Answer: C — It occurs when the top of a slope facing the sensor is nearer in slant range than its base, so they swap
Radar sorts targets by slant range, so when a slope facing the sensor is steeper than the incidence angle the summit is nearer the antenna than the base and is imaged first: the order reverses (layover). Lack of signal behind slopes facing away is shadow, compression with the true order kept is foreshortening, and double-bounce brightness from corner reflectors is a separate effect.
Two C-band SAR passes (wavelength 5.6 cm) show an interferogram with one complete fringe (a phase change of 2π) across a landslide. The line-of-sight displacement across that fringe is
1. 5.6 cm, equal to the wavelength
2. 1.4 cm, equal to a quarter of the wavelength
3. 11.2 cm, equal to twice the wavelength
4. 2.8 cm, equal to half the wavelength
Show answer
Answer: D — 2.8 cm, equal to half the wavelength
The two-way path gives Δφ = 4πd/λ, so Δφ = 2π corresponds to d = λ/2 = 2.8 cm along the line of sight. Equating one fringe with a full wavelength forgets that the signal travels to the ground and back, which doubles the path change for each unit of displacement.
Which resampling method keeps the original digital numbers unchanged and is therefore preferred when the output will be classified?
1. Nearest neighbour, which copies the value of the closest input pixel
2. Bilinear interpolation, which averages the four nearest pixels
3. Cubic convolution, which fits a surface through sixteen neighbouring pixels
4. All three, since resampling never alters the values recorded by the sensor
Show answer
Answer: A — Nearest neighbour, which copies the value of the closest input pixel
Nearest neighbour copies an existing DN, so no new spectral values are created and class signatures are preserved, at the cost of a blocky look and up to half-pixel displacement. Bilinear and cubic convolution compute weighted averages, which smooth the image and create DN values that were never recorded. The statement that all methods keep the values is false for the interpolating methods.
Why is a band ratio such as TM 5/7 preferred to a single band for mapping clay-rich alteration in rugged terrain?
1. The ratio raises the radiometric resolution of the sensor by adding together the bits of the two bands used
2. Slope and illumination scale both bands almost equally, so the ratio cancels shading but keeps clay contrast
3. The ratio cancels the atmospheric scattering that is the same in both bands, leaving only the clay signal
4. The ratio converts digital numbers directly to surface reflectance, thereby removing all atmospheric effects from the image
Show answer
Answer: B — Slope and illumination scale both bands almost equally, so the ratio cancels shading but keeps clay contrast
Topographic shading multiplies the radiance of both bands by almost the same factor, so it cancels in the ratio, whereas clays darken band 7 (2.2 µm) relative to band 5, so 5/7 stays high over clays on both sunlit and shaded slopes. A ratio adds no radiometric bits and is not a calibration to reflectance. Atmospheric path radiance is additive and differs between bands, so a ratio does not remove it.
A single-band raster covers 12 km × 9 km with a 30 m cell size and stores 16 bits per cell. Its uncompressed size is closest to
1. 120 kB, which counts only one byte per cell, as for an 8-bit raster
2. 240 kB, which counts 400 × 300 = 120 000 cells at two bytes each
3. 480 kB, which counts four bytes per cell, as for a 32-bit raster
4. 1920 kB, which wrongly counts the 16 bits of each cell as if they were 16 bytes
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Answer: B — 240 kB, which counts 400 × 300 = 120 000 cells at two bytes each
The grid has 12 000/30 = 400 columns and 9000/30 = 300 rows, so 120 000 cells. At 16 bits = 2 bytes per cell the size is 240 000 bytes, about 240 kB (taking 1 kB = 1000 bytes). One byte per cell would be an 8-bit raster, four bytes a 32-bit one, and treating bits as bytes overstates the size eightfold.
A geologist wants the total area of the parts of a lithological polygon layer that lie within 500 m of the mapped fault lines. Which workflow is correct?
1. Buffer the faults by 500 m, intersect with the lithology polygons and measure areas in a projected coordinate system
2. Buffer the lithology polygons by 500 m, dissolve them into one feature and measure its area in latitude and longitude degrees
3. Intersect the fault lines directly with the lithology polygons and then add up the areas of all the resulting line features
4. Union the fault and lithology layers, then add up every polygon whose attributes include the word fault
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Answer: A — Buffer the faults by 500 m, intersect with the lithology polygons and measure areas in a projected coordinate system
A line has no area, so the faults must first be buffered into 500 m wide polygons; the buffer is then intersected with the lithology polygons, and the areas of the pieces are measured in a projected system in metres. Intersecting lines with polygons yields lines with no area, buffering the lithology answers a different question, and areas in degrees are meaningless. A union with a text criterion does not apply the 500 m distance.
Which of the following statements about hyperspectral and multispectral sensing are correct?
1. A hyperspectral sensor records hundreds of narrow contiguous bands, so each pixel yields a nearly continuous spectrum
2. A multispectral sensor with a few broad bands resolves absorption features about 20 nm wide as well as a hyperspectral sensor
3. The data volume of a hyperspectral image grows in proportion to the number of bands, so the cube is much larger than a multispectral image of the same size
4. Hyperspectral imaging operates only in the thermal infrared window between 8 and 14 µm
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Answer: A — A hyperspectral sensor records hundreds of narrow contiguous bands, so each pixel yields a nearly continuous spectrum; C — The data volume of a hyperspectral image grows in proportion to the number of bands, so the cube is much larger than a multispectral image of the same size
Narrow contiguous bands give a continuous spectrum per pixel, and the cube holds rows × columns × bands values, so its volume scales with the band count. A few broad bands average over a 20 nm feature and cannot resolve it, and imaging spectrometers such as AVIRIS and Hyperion work mainly over 0.4–2.5 µm, not only in the thermal window.
Which of the following statements about GIS data and operations are correct?
1. Area measured on a layer stored in geographic coordinates in degrees is directly in square metres
2. A vector layer represents discrete features as points, lines and polygons and can store topology such as adjacency and connectivity
3. A focal raster operation assigns each output cell a value computed from that single input cell alone
4. Cell-by-cell overlay of two rasters needs a common coordinate system, cell size and alignment, or prior resampling
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Answer: B — A vector layer represents discrete features as points, lines and polygons and can store topology such as adjacency and connectivity; D — Cell-by-cell overlay of two rasters needs a common coordinate system, cell size and alignment, or prior resampling
Vector layers hold points, lines and polygons with optional topology, and raster overlay needs matching grids or resampling. Degrees are angular units, so areas in a geographic system are not in square metres. A focal operation uses a neighbourhood around each cell; using only the cell itself is a local operation.
Which of the following statements about geological applications of remote sensing are correct?
1. Optical images taken at a high sun elevation give the best relief enhancement for mapping subtle structures
2. Optical and thermal sensing sees only a surface skin, so the subsurface geology is inferred and must be checked in the field
3. Lineaments mapped on imagery or a DEM often mark fractures and faults, and their intersections are favoured targets for groundwater
4. L-band radar cannot penetrate any dry soil, so it is useless for detecting buried features
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Answer: B — Optical and thermal sensing sees only a surface skin, so the subsurface geology is inferred and must be checked in the field; C — Lineaments mapped on imagery or a DEM often mark fractures and faults, and their intersections are favoured targets for groundwater
Remote sensing records the surface layer and gives hypotheses that need ground truth, and lineaments often follow fractures, with intersections concentrating weathering and recharge. A low, not high, sun angle casts shadows that enhance relief, and L-band radar does penetrate some metres of very dry, fine sand, so the statement that it cannot penetrate any dry soil is false.
A scanner flies at an altitude of 800 km with an IFOV of 0.03 mrad and a total field of view of 10°. Treat the ground as flat, take the pixel size as the nadir value for all pixels, and use swath = 2H tan(FOV/2). The number of pixels across one scan line, to the nearest whole number, is ____.
Numerical answer — type the value.
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Answer: 5833
GSD = H × IFOV = 800 000 m × 0.03 × 10⁻³ = 24 m. Swath = 2 × 800 × tan 5° = 2 × 800 × 0.08749 = 139.98 km. Pixels = 139 982/24 = 5832.6, which is 5833 to the nearest whole number. Using the full angle 10° in place of the half-angle gives a swath about twice too large.
A surface of emissivity 0.95 has a true kinetic temperature of 300 K. Using T_rad = ε^(1/4) T_kin, the radiant temperature sensed in the thermal infrared, to one decimal place, is ____ K.
Numerical answer — type the value.
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Answer: 296.2
Since εσT_kin⁴ = σT_rad⁴, T_rad = ε^(1/4) T_kin. 0.95^0.25 = 0.98726, so T_rad = 300 × 0.98726 = 296.18 K, i.e. 296.2 K. The sensor reads about 3.8 K lower than the true temperature, because the surface emits less than a blackbody.
An imaging radar transmits pulses of duration 0.2 µs. Take the speed of light as 3 × 10⁸ m/s and flat terrain, with an incidence angle of 40° from the vertical. The ground-range resolution R_g = cτ/(2 sin θ), to one decimal place, is ____ m.
Numerical answer — type the value.
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Answer: 46.7
Slant-range resolution = cτ/2 = 3 × 10⁸ × 0.2 × 10⁻⁶/2 = 30 m. Ground-range resolution = 30/sin 40° = 30/0.6428 = 46.7 m. Using cos 40° in place of sin 40° would give 39.2 m, which is the result for a depression angle, not an incidence angle.
The mean reflectance of a vegetated pixel is 0.46 in the near-infrared band and 0.10 in the red band. Its NDVI = (NIR − Red)/(NIR + Red), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.64
NDVI = (0.46 − 0.10)/(0.46 + 0.10) = 0.36/0.56 = 0.643, i.e. 0.64, in the range of dense healthy vegetation. Inverting the ratio (sum over difference) would give 1.56, which lies outside the possible range of −1 to +1.
A two-class map is checked against 200 reference samples. The error matrix has map classes in rows and reference classes in columns: map class 1 has 70 samples reference 1 and 10 samples reference 2; map class 2 has 20 samples reference 1 and 100 samples reference 2. Kappa κ = (p_o − p_e)/(1 − p_e), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.69
p_o = (70 + 100)/200 = 0.85. Row totals are 80 and 120, column totals 90 and 110, so p_e = (80 × 90 + 120 × 110)/200² = (7200 + 13 200)/40 000 = 0.51. κ = (0.85 − 0.51)/(1 − 0.51) = 0.34/0.49 = 0.694, i.e. 0.69. The overall accuracy of 85 per cent overstates the agreement because 51 per cent would arise by chance.