Remote Sensing I: Physical Principles, Spectral Signatures and the Four Resolutions
1. The basic concept and the electromagnetic spectrum
Remote sensing acquires information about an object without contact, by recording electromagnetic energy reflected or emitted from it. The chain is: energy source → propagation through the atmosphere → interaction with the target → return through the atmosphere → recording by a sensor → processing and interpretation → application. Energy travels at c = λν (c ≈ 3 × 10⁸ m/s), and each photon carries E = hν = hc/λ, so short wavelengths carry more energy per photon — one reason thermal and microwave sensors need larger detectors or longer dwell times to collect enough signal.
| Region | Wavelength | What is sensed |
|---|---|---|
| Visible: blue, green, red | 0.4–0.5, 0.5–0.6, 0.6–0.7 µm | reflected sunlight |
| Near infrared (NIR) | 0.7–1.3 µm | reflected; vegetation vigour |
| Shortwave infrared (SWIR) | 1.3–3 µm | reflected; moisture, minerals |
| Mid-wave infrared | 3–5 µm | mixed; hot targets such as fires |
| Thermal infrared (TIR) | 8–14 µm | emitted by the Earth |
| Microwave | about 1 mm – 1 m | emitted (passive) or backscattered (radar) |
2. Physical principles: radiation laws and the atmosphere
Every body above absolute zero emits. A blackbody absorbs and emits perfectly; Planck's law gives its spectral exitance, and two consequences are examined. Stefan–Boltzmann: total exitance M = σT⁴ with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (so doubling T multiplies M by 16). Wien's displacement law: λ_max = 2898 µm·K/T. The Sun (about 6000 K) peaks near 0.48 µm, in the visible; the Earth (about 300 K) near 9.7 µm, which is why the 8–14 µm window is the thermal band. Real surfaces have emissivity ε < 1 (M = εσT⁴), and by Kirchhoff's law ε equals absorptance at each wavelength; a radiometer therefore reads a radiant temperature T_rad = ε^(1/4) T_kin, lower than the true kinetic temperature.
| Type | Particle size vs λ | Dependence | Effect |
|---|---|---|---|
| Rayleigh | much smaller (gas molecules) | ∝ λ⁻⁴ | blue sky; haze in the blue band |
| Mie | comparable (dust, smoke, aerosols) | weaker, roughly λ⁻¹ to λ⁰ | hazy, polluted days |
| Non-selective | much larger (cloud droplets) | independent of λ | clouds look white |
Blue light at 0.4 µm is scattered (0.7/0.4)⁴ = 9.4 times as strongly as red at 0.7 µm. Absorption by H₂O, CO₂ and O₃ closes much of the spectrum; the atmospheric windows used by sensors are roughly 0.3–1.3 µm, around 1.6 and 2.2 µm, 3–5 µm, 8–14 µm, and microwaves longer than about 1 cm, which pass through cloud — the reason radar is the all-weather sensor.
3. Interaction at the surface and spectral signatures
Energy incident on a surface is conserved: E_I(λ) = E_R(λ) + E_A(λ) + E_T(λ). Spectral reflectance ρ(λ) = E_R/E_I, plotted against wavelength, is the spectral signature (reflectance curve) of a material. A smooth surface reflects specularly, a rough one diffusely; a surface that scatters equally in all directions is Lambertian, the assumption most corrections make. “Rough” is relative to wavelength: a surface rough to visible light can be smooth to radar.
| Material | Visible | NIR | SWIR |
|---|---|---|---|
| Healthy green vegetation | low; chlorophyll absorbs blue (~0.45 µm) and red (~0.67 µm); small green peak (~0.55 µm) | high (40–50 % or more) from leaf mesophyll structure; the sharp rise near 0.7 µm is the red edge | falls, with leaf-water absorption dips near 1.4, 1.9 and 2.7 µm |
| Dry bare soil | moderate, rising gently with λ | moderate | moderate; lowered by moisture and organic matter |
| Clear water | low, highest in blue-green; turbidity raises it | almost zero — water absorbs NIR | almost zero |
4. The four resolutions and the ground sampling distance
| Resolution | Meaning | Set by |
|---|---|---|
| Spatial | the smallest ground area resolved: pixel size, GSD, IFOV projected on the ground | altitude, focal length, detector size |
| Spectral | the number, width and position of the bands | filters, dispersing optics |
| Radiometric | the smallest difference in radiance recorded: n bits give 2ⁿ levels, 0 to 2ⁿ − 1 | quantisation and signal-to-noise ratio |
| Temporal | the revisit interval over the same place | orbit, swath, off-nadir pointing, number of satellites |
The instantaneous field of view (IFOV) β is the angle one detector sees; at nadir from altitude H the ground cell is GSD = H·β (β in radians). For a detector of pitch p behind a lens of focal length f, β = p/f, so GSD = p·H/f. A 7 µm detector, f = 0.7 m, H = 500 km: GSD = 7 × 10⁻⁶ × 500 000/0.7 = 5 m. A drone camera with 4 µm pixels, f = 8 mm, flown at 120 m: GSD = 4 × 10⁻⁶ × 120/0.008 = 0.06 m. Data volume grows with every resolution: a 6000 × 6000 scene in 4 bands at 12 bits is 36 × 10⁶ × 4 × 12/8 = 216 × 10⁶ bytes.
Key takeaways
- c = λν and E = hc/λ; the reflective region is 0.4–3 µm, the thermal window 8–14 µm, and microwaves pass through cloud.
- Wien: λ_max = 2898/T µm (Sun ≈ 0.48 µm, Earth ≈ 9.7 µm); Stefan–Boltzmann: M = εσT⁴; radiant temperature = ε^(1/4) × kinetic temperature.
- Rayleigh scattering ∝ λ⁻⁴ (molecules), Mie for aerosols of size ≈ λ, non-selective for cloud droplets.
- Vegetation: low red, high NIR, red edge near 0.7 µm; water: near zero in NIR; soil: moderate and rising.
- Spatial, spectral, radiometric (2ⁿ levels) and temporal resolution; GSD = H·IFOV = pH/f.
Practice questions (14)
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.
Using Wien's constant 2898 µm·K, the wavelength of peak emission of the Earth's surface at 300 K, in µm (to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 9.66
λ_max = 2898/T = 2898/300 = 9.66 µm, inside the 8–14 µm thermal window. For the Sun at 6000 K the same law gives 0.483 µm, in the visible.Assuming pure Rayleigh scattering, the ratio of the scattering of blue light at 0.4 µm to that of red light at 0.7 µm (to one decimal place) is ____.
Numerical answer — type the value.
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Answer: 9.4
Rayleigh scattering ∝ λ⁻⁴, so the ratio is (0.7/0.4)⁴ = 1.75⁴ = 3.0625² = 9.38, i.e. 9.4. Squaring gives 3.1, the λ⁻² slip; the ratio 1.75 alone ignores the power entirely.A surface at a kinetic temperature of 300 K has an emissivity of 0.90 in the thermal band. The radiant temperature a radiometer records, in kelvin (to one decimal place), is ____.
Numerical answer — type the value.
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Answer: 292.2
Equating exitances, σT_rad⁴ = εσT_kin⁴, so T_rad = ε^(1/4) T_kin = 0.90^0.25 × 300 = 0.97400 × 300 = 292.2 K. Multiplying by ε itself (270 K) forgets the fourth root.Clouds appear white in visible imagery because cloud droplets cause
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Answer: A — non-selective scattering
Droplets are much larger than visible wavelengths, so they scatter all visible wavelengths about equally — non-selective scattering — and equal amounts of all colours look white. Rayleigh scattering by molecules favours blue and gives the blue sky instead.The high reflectance of healthy green vegetation in the near-infrared is due mainly to
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Answer: A — the internal cell structure (spongy mesophyll) of the leaf
NIR is scattered at the many cell-wall and air-space interfaces of the mesophyll, so a healthy leaf reflects strongly; stress that collapses those cells lowers NIR first. Chlorophyll controls the visible (it absorbs blue and red), and leaf water controls the SWIR absorption dips.In a near-infrared band, which of the following would normally appear darkest?
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Answer: A — A clear, deep lake
Water absorbs almost all incident NIR within a short depth, so clear deep water is nearly black in an NIR band. Vegetation is bright in NIR and dry sand is moderately bright, which is why NIR is the band used to delineate water bodies.A pushbroom sensor has detectors of 7 µm pitch behind optics of focal length 0.7 m and orbits at 500 km. Its ground sampling distance at nadir, in metres, is ____.
Numerical answer — type the value.
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Answer: 5
IFOV = p/f = 7 × 10⁻⁶/0.7 = 10⁻⁵ rad. GSD = H × IFOV = 500 000 × 10⁻⁵ = 5 m. Equivalently GSD = pH/f = 7 × 10⁻⁶ × 500 000/0.7 = 5 m. Forgetting to convert µm to m gives an absurd 5 × 10⁶ m.A drone camera with a pixel pitch of 4 µm and a focal length of 8 mm is flown at 120 m above flat ground. The ground sampling distance, in centimetres, is ____.
Numerical answer — type the value.
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Answer: 6
GSD = pH/f = (4 × 10⁻⁶ m × 120 m)/(8 × 10⁻³ m) = 4.8 × 10⁻⁴/8 × 10⁻³ = 0.06 m = 6 cm. The photo scale is f/H = 1:15 000, and 4 µm × 15 000 = 6 cm is the same result.A sensor quantises its signal to 11 bits. The number of distinct grey levels it can record is ____.
Numerical answer — type the value.
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Answer: 2048
n bits give 2ⁿ levels: 2¹¹ = 2048, numbered 0 to 2047. The maximum DN, 2047, is often given by mistake for the number of levels.A scene of 6000 × 6000 pixels is recorded in 4 spectral bands at 12 bits per pixel per band, with no compression. Its size, in megabytes (1 MB = 10⁶ bytes), is ____.
Numerical answer — type the value.
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Answer: 216
Bits = 6000 × 6000 × 4 × 12 = 1.728 × 10⁹; bytes = 1.728 × 10⁹/8 = 2.16 × 10⁸ = 216 MB. Storing each 12-bit value in a 16-bit word, as many formats do, would make it 288 MB, but the question asks for the packed size.A satellite revisits the same ground point every 16 days. This describes its
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Answer: A — temporal resolution
Temporal resolution is the repeat interval, set by the orbit, the swath, pointing ability and the number of satellites. Radiometric resolution is about grey levels, spectral about bands, and spatial about the ground size of a pixel.Which of the following regions are atmospheric windows commonly used for Earth observation?
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Answer: A — 0.4–0.7 µm; B — 8–14 µm; C — Microwaves of a few centimetres
The visible, the 8–14 µm thermal window and centimetre microwaves are all transmitted well; microwaves even pass cloud. Around 6.3 µm water vapour absorbs strongly — that region is used by meteorological water-vapour channels precisely because the surface cannot be seen through it.Which of the following statements about spectral signatures are correct?
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Answer: A — Increasing soil moisture lowers soil reflectance across the reflective spectrum; B — Vegetation shows absorption dips near 1.4 and 1.9 µm due to leaf water; C — Turbid water reflects more visible light than clear water
Water in the pores absorbs, so wet soil is darker at all reflective wavelengths; the 1.4 and 1.9 µm dips are liquid-water absorption in leaves; and suspended sediment back-scatters visible light, brightening turbid water. Stress damages the mesophyll, so NIR reflectance falls, not rises.Which of the following changes, taken alone, make the ground sampling distance of a nadir-looking camera smaller (finer)?
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Answer: A — Using a longer focal length; B — Flying lower; C — Using smaller detector elements
GSD = pH/f, so it falls as f increases, as H decreases and as p decreases. The number of bits sets radiometric resolution — how finely brightness is divided — and has no effect on the ground size of a pixel.