Why do daytime images in the visible band record reflected sunlight, while thermal infrared images of the same scene record the Earth's own emission?

The Sun, near 5800 K, peaks near 0.5 µm, while the Earth, near 300 K, peaks near 10 µm, so each dominates its band

The Earth emits no radiation at visible wavelengths and the Sun emits none at all in the thermal infrared region

The atmosphere absorbs every wavelength longer than 3 µm coming from the Sun, so thermal sensors can see only the Earth's own emission

The emissivity of the Earth is zero in the visible band and equal to one throughout the thermal infrared band

For Rayleigh scattering, by what factor is light of wavelength 0.45 µm scattered more strongly than light of wavelength 0.90 µm?

16 times, because scattering varies as the inverse fourth power of the wavelength

2 times, because scattering is simply inversely proportional to the wavelength itself

4 times, because scattering varies as the inverse square of the wavelength, as with many wave effects

8 times, because scattering varies as the inverse cube of the wavelength, as for a volume effect

16 times, because scattering varies as the inverse fourth power of the wavelength

Which band is best suited to mapping the temperature of the land surface from a satellite?

The 8 to 14 µm window, where the Earth's emission is large and the atmosphere is largely transparent

The 15 µm band, because the Earth's emission is strongest there and carbon dioxide does not absorb it

The 0.5 µm band, because the solar flux is strongest there and gives the clearest picture of surface heat

The 8 to 14 µm window, where the Earth's emission is large and the atmosphere is largely transparent

The 1.9 µm band, because leaf water makes the surface radiate strongly there and the air is clear

Which of the following statements about atmospheric scattering and transmission are correct?

Rayleigh scattering by particles much smaller than the wavelength is the reason for the blue colour of a clear sky

Clouds look white because their large droplets scatter all visible wavelengths about equally

Rayleigh scattering by particles much smaller than the wavelength is the reason for the blue colour of a clear sky

Clouds look white because their large droplets scatter all visible wavelengths about equally

Thin cloud blocks microwaves of centimetre wavelength, so radar cannot image the ground in cloudy weather

Rayleigh scattering increases with wavelength, so haze affects NIR images more than blue images

A pixel has reflectance 0.04 in blue, 0.09 in green, 0.05 in red, 0.45 in NIR and 0.20 in SWIR near 1.6 µm. Which surface does it most probably represent?

Healthy vegetation, with chlorophyll absorbing blue and red and a high NIR plateau

Deep clear water, which is dark in all bands and has only a slight blue-green peak in the visible

Fresh snow, which is very bright in the visible and stays bright in the NIR bands as well

Dry bare soil, whose reflectance rises slowly and smoothly with wavelength without any sharp edge

Healthy vegetation, with chlorophyll absorbing blue and red and a high NIR plateau

Which of the following statements about the reflectance of surfaces are correct?

Clear water absorbs strongly in the NIR, so water bodies appear dark in NIR images

Snow is bright in the visible but dark in SWIR near 1.6 µm, which helps to separate it from cloud

Chlorophyll in healthy leaves reflects strongly in the red and absorbs strongly in the NIR region

Clear water absorbs strongly in the NIR, so water bodies appear dark in NIR images

Wetting a dry soil with water raises its reflectance at all wavelengths from the visible to the SWIR

Snow is bright in the visible but dark in SWIR near 1.6 µm, which helps to separate it from cloud

A natural slope of dry sand has been stable for years, but it fails after prolonged heavy rain although its geometry has not changed. What is the main reason?

Rising pore pressure lowers the effective normal stress and therefore the frictional strength σ′ tan φ

The friction angle of the sand grains is permanently reduced to zero when they become wet

Rising pore pressure lowers the effective normal stress and therefore the frictional strength σ′ tan φ

The added water makes the sand lighter, so the normal force on every potential slip surface is reduced

Water adds cohesion to dry sand, and cohesion always makes a slope less stable than friction does

A 200 cm core run recovers pieces of lengths 40, 8, 25, 6, 35, 12, 9 and 20 cm. RQD classes: below 25 very poor, 25 to 50 poor, 50 to 75 fair, 75 to 90 good, above 90 excellent. What is the RQD, and its class?

66 per cent, fair, from the pieces longer than 10 cm divided by the run length

77.5 per cent, good, from all the pieces divided by the run length

85 per cent, good, from the pieces longer than 10 cm divided by the recovered length

66 per cent, fair, from the pieces longer than 10 cm divided by the run length

66 per cent, poor, from the pieces longer than 10 cm divided by the run length

Which statement correctly distinguishes the magnitude of an earthquake from its intensity?

Magnitude is one number for the source; intensity describes effects at each place and varies with distance and ground

Intensity is a single number for the whole event, whereas magnitude varies from town to town with distance and soil

Magnitude is one number for the source; intensity describes effects at each place and varies with distance and ground

Magnitude is judged from the damage to buildings, whereas intensity is computed from the amplitude on a seismogram

Magnitude and intensity are the same quantity measured on two different scales that differ only in name and units

Which of the following measures reduce the landslide hazard on an unstable slope?

Improving surface and subsurface drainage so that pore pressure stays low

Installing rock bolts and a retaining wall at the toe

Improving surface and subsurface drainage so that pore pressure stays low

Adding heavy fill to the crest of the slope to increase the load on it

Installing rock bolts and a retaining wall at the toe

Clearing the vegetation so that the slope carries less weight

A tsunami travels in water 4000 m deep and then reaches a shelf 100 m deep. Take g = 9.8 m/s². What happens as it enters the shallow water?

The speed falls from about 198 to about 31 m/s, the wavelength shortens and the wave height increases

The speed rises from about 31 m/s to about 198 m/s because the energy is squeezed into less water

The speed stays at about 198 m/s, since a tsunami's speed depends on the earthquake that made it and not on the depth

The speed falls from about 198 m/s to about 31 m/s and the wave height also decreases in proportion

The observed mean surface temperature of the Earth is about 288 K, whereas the effective radiating temperature is about 255 K. What is the main reason for the difference of about 33 K?

Heat flowing up from the Earth's interior supplies most of the energy that warms the surface

The stratospheric ozone layer absorbs sunlight and heats the surface directly from above

The real albedo of the Earth is close to zero, so the calculation underestimates the absorbed sunlight

For CO₂ take the forcing as ΔF = 5.35 ln(C/C₀) W/m² and a sensitivity parameter λ = 0.8 K per W/m². The equilibrium warming when CO₂ rises from 280 ppm to 420 ppm is closest to which value?

1.7 K, from a forcing of 2.17 W/m² multiplied by the sensitivity parameter

0.8 K, from using a base-10 logarithm in the forcing relation instead of the natural logarithm

1.7 K, from a forcing of 2.17 W/m² multiplied by the sensitivity parameter

2.2 K, from taking the forcing in W/m² as if it were the temperature change itself

3.0 K, from the forcing of a full doubling of CO₂ rather than the actual rise

Remote Sensing, Rock and Soil Properties, Natural Hazards and Climate Change

This chapter answers the Part A heading of the Geology and Geophysics (GG) paper that reads “Introduction to remote sensing: Energy sources and radiation principles, atmospheric absorption, interaction of energy with Earth’s surface; Engineering properties of rocks and soils; Natural hazards (landslide, volcanic, seismogenic, coastal) and mitigation; Principles of climate change”. It teaches the radiation laws behind remote sensing, atmospheric windows and scattering, spectral signatures, the phase relations, strength and rock-quality measures of rocks and soils, a simple slope factor of safety, each of the four hazards with its mitigation, and the radiative balance that underlies climate change. The numericals are Stefan–Boltzmann and Wien calculations, unit weights, RQD, a block factor of safety, magnitude-energy and P–S distance, and the effective temperature of the Earth, with all constants given in the question.

1. Energy sources and radiation principles

Remote sensing records the energy that a surface reflects or emits without touching it. The energy travels as electromagnetic waves with c = fλ = 3 × 10⁸ m/s, and a photon carries E = hf = hc/λ, so shorter wavelengths are more energetic. The bands used are visible (0.4 to 0.7 µm), near-infrared (NIR, 0.7 to about 1.3 µm), shortwave infrared (SWIR, about 1.3 to 3 µm), thermal infrared (mainly 8 to 14 µm) and microwave (about 1 mm to 1 m). Passive sensors use the Sun (reflected light, visible to SWIR) or the Earth’s own emission (thermal IR, passive microwave); active sensors such as radar and lidar send their own beam and work by day or night.

Blackbody laws. A blackbody absorbs all incident energy and emits the maximum possible at every wavelength. Its total emitted flux is M = σT⁴ (Stefan–Boltzmann, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, T in kelvin); a real surface emits M = εσT⁴ with emissivity ε < 1. The wavelength of peak emission is λmax = b/T (Wien, b = 2898 µm·K). The Sun (about 5800 K) peaks near 0.5 µm, in visible light; the Earth (about 300 K) peaks near 9.7 µm, in the thermal infrared. So visible-to-SWIR images record reflected sunlight, while thermal images record the Earth’s own emission. Doubling T raises M sixteen-fold and halves λmax. Worked example: a surface at 300 K with ε = 0.95 emits 0.95 × 5.67 × 10⁻⁸ × 300⁴ = 0.95 × 459 = 436 W/m².

⚠️ Hotter means shorter peak, and more at every wavelength

A hotter blackbody peaks at a shorter wavelength (Wien) and also emits more at every wavelength, and its total emission rises as T⁴. Use kelvin, never degrees Celsius, in σT⁴. A good absorber is a good emitter at the same wavelength (Kirchhoff), so ε equals absorptivity.

2. Atmospheric absorption and scattering

The atmosphere alters the signal on the way down and again on the way up. Absorption by gas molecules removes energy at particular wavelengths: ozone absorbs ultraviolet shorter than about 0.3 µm, water vapour absorbs in bands near 1.4 and 1.9 µm and strongly beyond about 5 µm, and carbon dioxide absorbs near 2.7, 4.3 and 15 µm. The wavelength ranges that pass well are the atmospheric windows, and sensors are built to sit in them.

Atmospheric windows and what they serve
Region	Behaviour	Typical use
Visible and NIR, 0.4 to 1.3 µm	Open window; haze affects the blue end most	Multispectral imaging, vegetation, land cover
SWIR windows near 1.6 and 2.2 µm	Water-vapour bands at 1.4 and 1.9 µm are closed	Clay and carbonate minerals, snow, moisture
Thermal IR, 8 to 14 µm	Open window with a notch from ozone near 9.6 µm	Surface temperature, thermal anomalies
Microwave, centimetre wavelengths and longer	Passes through cloud, haze and rain of light intensity	All-weather radar imaging, soil moisture

Scattering redirects light without removing its energy from the system. Rayleigh scattering, by particles much smaller than the wavelength, varies as λ⁻⁴: blue light of 0.45 µm is scattered about 4 times more than red of 0.65 µm, which makes the sky blue and sunsets red, and a doubling of wavelength cuts it sixteen-fold, which is why NIR images see through haze better than visible ones. Mie scattering, by particles comparable with the wavelength (smoke, dust, haze), depends much less on wavelength. Non-selective scattering, by particles much larger than the wavelength such as cloud droplets, scatters all visible wavelengths equally, so clouds look white. Scattered light adds a veil of path radiance that lowers contrast, and a standard correction subtracts the signal seen over a dark target such as deep clear water.

3. Interaction of energy with the Earth's surface

Incident energy is reflected, absorbed or transmitted: E_I = E_R + E_A + E_T, so reflectance + absorptance + transmittance = 1, and for an opaque surface the transmittance is zero. The curve of reflectance against wavelength is the spectral signature, and it identifies the material. Smooth surfaces such as calm water reflect specularly (in one mirror-like direction); rough surfaces reflect diffusely, and “rough” is judged against the wavelength. Absorbed energy is re-emitted as heat, which thermal sensors record, and in radar the backscatter rises with roughness and with moisture.

Typical reflectance behaviour of common surfaces
Surface	Spectral behaviour	Reason
Healthy vegetation	Low in blue and red, small peak in green, sharp rise near 0.7 µm to a high NIR plateau, dips at 1.4 and 1.9 µm	Chlorophyll absorbs blue and red; leaf cell structure scatters NIR; leaf water absorbs at 1.4 and 1.9 µm
Clear water	Low everywhere, almost zero in NIR and SWIR; turbid water is brighter in the visible	Water absorbs strongly at longer wavelengths; suspended sediment scatters visible light
Dry soil	Reflectance rises gradually with wavelength; moisture lowers it; clays absorb near 2.2 µm	Iron oxides and organic matter shape the visible; water in pores absorbs; Al–OH bonds in clay absorb near 2.2 µm
Snow and cloud	Both bright in the visible; snow turns dark in SWIR near 1.6 µm while cloud stays bright	Ice grains absorb in SWIR; cloud droplets are far smaller and absorb little

Vegetation contrast is summarised by the normalised difference vegetation index, NDVI = (NIR − Red)/(NIR + Red), which lies between −1 and +1: dense healthy vegetation is high and positive, bare soil is near 0.1 to 0.2, and water is negative. Worked example: NIR = 0.50 and red = 0.10 give (0.50 − 0.10)/(0.50 + 0.10) = 0.67.

4. Engineering properties of rocks and soils

Soil is a mixture of solids, water and air. Define the void ratio e = V_v/V_s, porosity n = V_v/V = e/(1 + e), water content w = W_w/W_s and degree of saturation S = V_w/V_v (0 dry, 1 saturated), linked by Se = wGs, where Gs is the specific gravity of solids (about 2.65 for quartz-rich soils). With γw = 9.81 kN/m³: dry unit weight γd = Gsγw/(1 + e), saturated unit weight γsat = (Gs + e)γw/(1 + e) and submerged unit weight γ′ = γsat − γw. Worked example: Gs = 2.65, e = 0.80, w = 0.20 give S = 0.20 × 2.65/0.80 = 0.66, γd = 2.65 × 9.81/1.80 = 14.4 kN/m³ and γsat = 3.45 × 9.81/1.80 = 18.8 kN/m³.

Strength. Soil strength depends on the effective stress σ′ = σ − u (total stress minus pore-water pressure) and follows the Mohr–Coulomb line τf = c + σ′ tan φ, with cohesion c and friction angle φ. Clean sand and gravel have c ≈ 0 and draw strength from friction; clays are cohesive and plastic, described by the Atterberg limits (liquid limit LL, plastic limit PL, plasticity index PI = LL − PL), and montmorillonite-rich clays swell and shrink with moisture and damage foundations. Water in the pores lowers σ′; when u rises to equal σ in loose saturated sand during shaking, σ′ falls to zero and the sand liquefies.

Rocks. The strength of intact rock is measured by the uniaxial compressive strength (UCS) or the point-load index, and it falls with weathering, saturation and rising porosity. A rock mass is weaker than intact rock because joints, bedding and foliation control its behaviour. The Rock Quality Designation is RQD = (sum of lengths of intact core pieces longer than 10 cm ÷ length of the core run) × 100, graded very poor (below 25), poor (25 to 50), fair (50 to 75), good (75 to 90) and excellent (above 90).

Slope stability. The factor of safety is FS = resisting force (strength) ÷ driving force, and FS > 1 means stable. For a dry cohesionless slope of angle β, FS = tan φ/tan β, so the steepest stable slope equals φ, the angle of repose. For a rigid block of weight W on a plane dipping at θ with contact area A, FS = (cA + W cos θ tan φ)/(W sin θ); water pressure on the plane lowers the normal force and so lowers FS. Worked example: a dry block with c = 0, θ = 30° and φ = 35° has FS = tan 35°/tan 30° = 0.700/0.577 = 1.21.

⚠️ Rain lowers safety without changing the slope

A rising water table or heavy rain raises the pore pressure u, lowers σ′ and therefore the frictional strength σ′ tan φ, so FS can fall below 1 although the geometry is unchanged. Total weight also rises, but the loss of effective stress is the main cause. Cohesion c and φ are strength parameters of the material, not of the slope angle.

5. Natural hazards and their mitigation

A hazard is a natural process that can cause harm; risk combines the hazard with exposure and vulnerability. The hazard itself usually cannot be stopped, so mitigation works on exposure and vulnerability through zoning, design, monitoring, warning and preparedness.

The four hazards: causes and mitigation
Hazard	Causes and main effects	Mitigation
Landslide	Steep slope, rain raising pore pressure, earthquake shaking, cutting at the toe, loading, loss of vegetation; falls, slides and flows	Drainage, regrading and benching, retaining walls, rock bolts, vegetation, hazard zonation, rainfall-threshold warning, no building on unstable slopes
Volcanic	Lava flows, pyroclastic density currents, ash fall, lahars and gases; the Volcanic Explosivity Index (0 to 8) is logarithmic in erupted volume; Barren Island (Andaman Sea) is India’s active volcano	Monitoring of earthquakes, ground swelling and gas, hazard maps, exclusion zones and timely evacuation
Seismogenic	Ground shaking, surface rupture, liquefaction, triggered landslides and tsunami	Seismic zonation (India: zones II to V, V the most severe), earthquake-resistant ductile design, base isolation, retrofitting, land-use planning
Coastal	Tsunami, cyclone storm surge, erosion, flooding and sea-level rise	Tsunami warning (the national centre is at Hyderabad), mangroves and shelter-belts, seawalls, setback and coastal-zone rules, cyclone shelters

Quantifying earthquakes. Magnitude M measures the size of the source on a logarithmic scale: the ground amplitude changes tenfold per unit and the energy by about 31.6 times, by the Gutenberg–Richter relation log₁₀E = 1.5M + 4.8 (E in joules). So an M 7 earthquake releases 10^(1.5 × 2) = 1000 times the energy of an M 5. The moment magnitude is Mw = (2/3) log₁₀M₀ − 6.07 with the seismic moment M₀ in N·m. Intensity (MMI or MSK, twelve degrees) describes the effects at a place, so one earthquake has one magnitude but many intensities. The epicentral distance from one station is d = Δt × v_P v_S/(v_P − v_S), with Δt the S−P delay: for v_P = 6.0 and v_S = 3.5 km/s, d = 8.4 km per second of delay, so Δt = 10 s means 84 km.

Tsunami. A tsunami is a long water wave raised by sudden displacement of the sea floor, most often in a subduction-zone thrust earthquake, and also by submarine landslides and volcanic collapse. In water of depth h its speed is v = √(gh): for h = 4000 m and g = 9.8 m/s², v = 198 m/s, about 710 km/h. Near the coast the depth falls, so the speed and wavelength fall and the height grows, which is why a wave that is hardly noticed at sea floods the shore.

6. Principles of climate change

Radiative balance. The Sun delivers a flux S₀ ≈ 1361 W/m² on a surface facing it at the top of the atmosphere. The Earth intercepts πR² of it but has a surface area of 4πR², so the average incoming flux is S₀/4. With a planetary albedo α ≈ 0.30 the absorbed flux is S₀(1 − α)/4 ≈ 238 W/m². At equilibrium this equals the emitted flux σTₑ⁴, so Tₑ = [S₀(1 − α)/(4σ)]^(1/4) ≈ 255 K. The observed mean surface temperature is about 288 K, and the difference of about 33 K is the natural greenhouse effect: gases such as water vapour, CO₂, CH₄, N₂O and O₃ pass incoming sunlight but absorb outgoing thermal infrared and re-emit part of it downward.

Forcing and feedback. A change in the net energy flux, a radiative forcing ΔF in W/m², shifts the equilibrium temperature by ΔT = λΔF, with λ the climate sensitivity parameter. For CO₂ a simple expression is ΔF = 5.35 ln(C/C₀), so doubling gives 3.7 W/m². With an illustrative λ = 0.8 K per W/m², doubling warms the Earth by about 3.0 K. Positive feedbacks amplify a change (melting ice lowers albedo; warmer air holds more water vapour), while the Planck response, more emission from a warmer Earth, is the main stabilising feedback.

Causes and evidence. Natural drivers include orbital (Milankovitch) cycles of eccentricity (about 100,000 years), obliquity (about 41,000 years) and precession (about 23,000 years), which paced the Quaternary glacial and interglacial cycles; changes in solar output; volcanic aerosols, which cool for a few years; and, over millions of years, silicate weathering and plate tectonics, which regulate CO₂. Burning of fossil fuels and deforestation have raised CO₂ from about 280 ppm before industry to well above 400 ppm. Evidence comes from ice cores (trapped air and oxygen isotopes), tree rings, corals, sediments and instrument records. Consequences include sea-level rise by thermal expansion and melting of land ice, glacier retreat, ocean acidification (CO₂ + H₂O gives H₂CO₃, which lowers pH), and shifts in rainfall and extremes.

⚠️ The ozone hole is not the greenhouse effect

Greenhouse warming comes from gases that absorb outgoing thermal infrared. The ozone hole is the destruction of stratospheric ozone by chlorine and bromine compounds, which lets more ultraviolet reach the ground. They are separate problems, although CFCs take part in both. Also, 255 K is the temperature the Earth would have without a greenhouse atmosphere, not the actual surface temperature.

Key takeaways

Radiation: M = σT⁴ (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, kelvin only) and λmax = 2898 µm·K/T, so the Sun peaks near 0.5 µm and the Earth near 9.7 µm; visible to SWIR images use reflected sunlight, thermal images use emission.
The atmosphere absorbs (ozone in the ultraviolet, water vapour and CO₂ in the infrared) and scatters: Rayleigh scattering varies as λ⁻⁴, Mie is weakly selective, large droplets scatter non-selectively. Windows are the visible and NIR, SWIR near 1.6 and 2.2 µm, thermal 8 to 14 µm and the microwave.
Surfaces are identified by spectral signature: vegetation has a red edge near 0.7 µm and NDVI = (NIR − Red)/(NIR + Red); water is dark in NIR; moisture darkens soil; snow turns dark in SWIR while cloud does not.
Soil: Se = wGs, γd = Gsγw/(1 + e), γsat = (Gs + e)γw/(1 + e), σ′ = σ − u, τf = c + σ′ tan φ. RQD counts core pieces over 10 cm. FS = tan φ/tan β for a dry cohesionless slope; rain raises u and lowers FS.
Hazards: log₁₀E = 1.5M + 4.8 (one magnitude unit is about 31.6 times the energy), one magnitude and many intensities, d = Δt·v_P v_S/(v_P − v_S), tsunami speed √(gh). Mitigation works through zoning, design, monitoring, warning and preparedness.
Climate: Tₑ = [S₀(1 − α)/(4σ)]^(1/4) ≈ 255 K against the observed 288 K, so the greenhouse effect is about 33 K. ΔT = λΔF with ΔF = 5.35 ln(C/C₀) for CO₂; ice-albedo and water-vapour feedbacks are positive, the Planck response is stabilising.

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.

Why do daytime images in the visible band record reflected sunlight, while thermal infrared images of the same scene record the Earth's own emission?
1. The Sun, near 5800 K, peaks near 0.5 µm, while the Earth, near 300 K, peaks near 10 µm, so each dominates its band
2. The Earth emits no radiation at visible wavelengths and the Sun emits none at all in the thermal infrared region
3. The atmosphere absorbs every wavelength longer than 3 µm coming from the Sun, so thermal sensors can see only the Earth's own emission
4. The emissivity of the Earth is zero in the visible band and equal to one throughout the thermal infrared band
Show answer
Answer: A — The Sun, near 5800 K, peaks near 0.5 µm, while the Earth, near 300 K, peaks near 10 µm, so each dominates its band
By Wien's law λmax = 2898/T µm·K, a 5800 K Sun peaks near 0.5 µm and a 300 K Earth near 9.7 µm. In the visible band reflected sunlight is overwhelmingly larger than the Earth's weak emission, and in the 8 to 14 µm band the Earth's emission is far larger than the reflected solar energy. The Sun does emit a little thermal infrared and the Earth emits a negligible amount in the visible, so statements of absolute zero are wrong, and the atmosphere has windows rather than absorbing everything beyond 3 µm.
A land surface at 320 K has an emissivity of 0.90. Take the Stefan–Boltzmann constant as 5.67 × 10⁻⁸ W m⁻² K⁻⁴. The flux it emits, in W/m² to the nearest integer, is ____.
Numerical answer — type the value.
Show answer
Answer: 535
M = εσT⁴ = 0.90 × 5.67 × 10⁻⁸ × 320⁴. Now 320⁴ = 1.0486 × 10¹⁰, so σT⁴ = 594.5 W/m² and M = 0.90 × 594.5 = 535 W/m². Dropping the emissivity would give 594.5 (blackbody), and using 320 on the Celsius scale (593 K) would be a larger error.
For Rayleigh scattering, by what factor is light of wavelength 0.45 µm scattered more strongly than light of wavelength 0.90 µm?
1. 2 times, because scattering is simply inversely proportional to the wavelength itself
2. 4 times, because scattering varies as the inverse square of the wavelength, as with many wave effects
3. 8 times, because scattering varies as the inverse cube of the wavelength, as for a volume effect
4. 16 times, because scattering varies as the inverse fourth power of the wavelength
Show answer
Answer: D — 16 times, because scattering varies as the inverse fourth power of the wavelength
Rayleigh scattering varies as λ⁻⁴, so the ratio is (0.90/0.45)⁴ = 2⁴ = 16. The smaller exponents in the other options belong to scattering regimes (such as Mie scattering) in which the dependence on wavelength is weaker, not to Rayleigh scattering by particles much smaller than the wavelength.
Which band is best suited to mapping the temperature of the land surface from a satellite?
1. The 15 µm band, because the Earth's emission is strongest there and carbon dioxide does not absorb it
2. The 0.5 µm band, because the solar flux is strongest there and gives the clearest picture of surface heat
3. The 8 to 14 µm window, where the Earth's emission is large and the atmosphere is largely transparent
4. The 1.9 µm band, because leaf water makes the surface radiate strongly there and the air is clear
Show answer
Answer: C — The 8 to 14 µm window, where the Earth's emission is large and the atmosphere is largely transparent
The Earth's emission peaks near 10 µm and the 8 to 14 µm region is an atmospheric window, so the sensor sees the surface. At 15 µm carbon dioxide absorbs strongly and the surface signal is blocked, which is the tempting choice because the emission there is not small; 0.5 µm records reflected sunlight, and 1.9 µm is a water-vapour absorption band that is closed to the surface.
Which of the following statements about atmospheric scattering and transmission are correct?
1. Rayleigh scattering by particles much smaller than the wavelength is the reason for the blue colour of a clear sky
2. Clouds look white because their large droplets scatter all visible wavelengths about equally
3. Thin cloud blocks microwaves of centimetre wavelength, so radar cannot image the ground in cloudy weather
4. Rayleigh scattering increases with wavelength, so haze affects NIR images more than blue images
Show answer
Answer: A — Rayleigh scattering by particles much smaller than the wavelength is the reason for the blue colour of a clear sky; B — Clouds look white because their large droplets scatter all visible wavelengths about equally
Rayleigh scattering by tiny particles gives the blue sky, and droplets much larger than the wavelength scatter non-selectively, which makes cloud white. Centimetre-wavelength microwaves pass through thin cloud and light rain, which is why radar is an all-weather tool. Rayleigh scattering falls as λ⁻⁴, so haze affects blue light far more than NIR.
A pixel has reflectance 0.04 in blue, 0.09 in green, 0.05 in red, 0.45 in NIR and 0.20 in SWIR near 1.6 µm. Which surface does it most probably represent?
1. Deep clear water, which is dark in all bands and has only a slight blue-green peak in the visible
2. Fresh snow, which is very bright in the visible and stays bright in the NIR bands as well
3. Dry bare soil, whose reflectance rises slowly and smoothly with wavelength without any sharp edge
4. Healthy vegetation, with chlorophyll absorbing blue and red and a high NIR plateau
Show answer
Answer: D — Healthy vegetation, with chlorophyll absorbing blue and red and a high NIR plateau
Low blue and red, a small green peak and a very high NIR value (a red edge) are the signature of healthy vegetation, with NDVI = (0.45 − 0.05)/(0.45 + 0.05) = 0.8. Water would be nearly zero in NIR, snow would be bright in the visible, and dry soil would show only a modest NIR to red difference.
Which of the following statements about the reflectance of surfaces are correct?
1. Chlorophyll in healthy leaves reflects strongly in the red and absorbs strongly in the NIR region
2. Clear water absorbs strongly in the NIR, so water bodies appear dark in NIR images
3. Wetting a dry soil with water raises its reflectance at all wavelengths from the visible to the SWIR
4. Snow is bright in the visible but dark in SWIR near 1.6 µm, which helps to separate it from cloud
Show answer
Answer: B — Clear water absorbs strongly in the NIR, so water bodies appear dark in NIR images; D — Snow is bright in the visible but dark in SWIR near 1.6 µm, which helps to separate it from cloud
Water absorbs strongly in NIR and SWIR, so it is dark there, and ice grains absorb near 1.6 µm while the small droplets of cloud do not, which separates snow from cloud. Chlorophyll does the reverse of what is claimed about it: it absorbs red, and the leaf structure reflects NIR. Moisture fills the pores and darkens a soil, it does not brighten it.
A saturated clay has a water content of 25 per cent and the specific gravity of its solids is 2.70. Take γw = 9.81 kN/m³. Its dry unit weight, in kN/m³ to one decimal place, is ____.
Numerical answer — type the value.
Show answer
Answer: 15.8
For a saturated soil S = 1, so e = wGs = 0.25 × 2.70 = 0.675. Then γd = Gsγw/(1 + e) = 2.70 × 9.81/1.675 = 26.487/1.675 = 15.8 kN/m³. Using the saturated unit weight (Gs + e)γw/(1 + e) = 19.8 kN/m³ instead is the usual slip.
A rigid rock block of weight 2000 kN rests on a plane dipping at 30°. The contact area is 20 m², the cohesion is 10 kPa, the friction angle is 35°, and there is no water pressure. Taking tan 35° = 0.700 and cos 30° = 0.866, the factor of safety against sliding, to two decimal places, is ____.
Numerical answer — type the value.
Show answer
Answer: 1.41
Resisting force = cA + W cos θ tan φ = 10 × 20 + 2000 × 0.866 × 0.700 = 200 + 1212.4 = 1412.4 kN. Driving force = W sin θ = 2000 × 0.5 = 1000 kN. FS = 1412.4/1000 = 1.41. Omitting the cohesion term gives 1.21, and using sin θ in the friction term is a common error.
A natural slope of dry sand has been stable for years, but it fails after prolonged heavy rain although its geometry has not changed. What is the main reason?
1. The friction angle of the sand grains is permanently reduced to zero when they become wet
2. Rising pore pressure lowers the effective normal stress and therefore the frictional strength σ′ tan φ
3. The added water makes the sand lighter, so the normal force on every potential slip surface is reduced
4. Water adds cohesion to dry sand, and cohesion always makes a slope less stable than friction does
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Answer: B — Rising pore pressure lowers the effective normal stress and therefore the frictional strength σ′ tan φ
Rain raises the pore-water pressure u, so the effective stress σ′ = σ − u falls and the strength σ′ tan φ falls with it, while the weight and the slope angle stay almost the same. The friction angle of the grains is a material property and does not go to zero; added water makes the mass heavier, not lighter; and cohesion, where present, adds strength.
A 200 cm core run recovers pieces of lengths 40, 8, 25, 6, 35, 12, 9 and 20 cm. RQD classes: below 25 very poor, 25 to 50 poor, 50 to 75 fair, 75 to 90 good, above 90 excellent. What is the RQD, and its class?
1. 77.5 per cent, good, from all the pieces divided by the run length
2. 85 per cent, good, from the pieces longer than 10 cm divided by the recovered length
3. 66 per cent, fair, from the pieces longer than 10 cm divided by the run length
4. 66 per cent, poor, from the pieces longer than 10 cm divided by the run length
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Answer: C — 66 per cent, fair, from the pieces longer than 10 cm divided by the run length
Only pieces longer than 10 cm count: 40 + 25 + 35 + 12 + 20 = 132 cm. RQD = 132/200 × 100 = 66 per cent, which lies in the 50 to 75 band, fair. Counting all pieces gives 77.5, and dividing by the recovered length (155 cm) instead of the run length gives 85; both overstate the quality. Labelling 66 per cent as poor misreads the class limits.
Which statement correctly distinguishes the magnitude of an earthquake from its intensity?
1. Intensity is a single number for the whole event, whereas magnitude varies from town to town with distance and soil
2. Magnitude is one number for the source; intensity describes effects at each place and varies with distance and ground
3. Magnitude is judged from the damage to buildings, whereas intensity is computed from the amplitude on a seismogram
4. Magnitude and intensity are the same quantity measured on two different scales that differ only in name and units
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Answer: B — Magnitude is one number for the source; intensity describes effects at each place and varies with distance and ground
Magnitude measures the size of the source (energy or moment) and is one value for an earthquake. Intensity (MMI or MSK) grades the shaking and damage at a place, so it falls with distance and differs with soil conditions. The roles in the other statements are reversed or merged.
At one station the S wave arrives 15 s after the P wave. Assume a uniform medium with vP = 7.0 km/s and vS = 4.0 km/s. The distance of the source from the station, in km, is ____.
Numerical answer — type the value.
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Answer: 140
The travel times are d/vP and d/vS, so Δt = d(1/vS − 1/vP) = d(1/4.0 − 1/7.0) = d × 3/28. Then d = 15 × 28/3 = 140 km. Equivalently d = Δt × vP vS/(vP − vS) = 15 × 28/3 = 140 km. Using the sum of the velocities or their mean instead of this relation is the common slip.
Which of the following measures reduce the landslide hazard on an unstable slope?
1. Improving surface and subsurface drainage so that pore pressure stays low
2. Adding heavy fill to the crest of the slope to increase the load on it
3. Installing rock bolts and a retaining wall at the toe
4. Clearing the vegetation so that the slope carries less weight
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Answer: A — Improving surface and subsurface drainage so that pore pressure stays low; C — Installing rock bolts and a retaining wall at the toe
Drainage lowers u and raises effective stress, and bolts and a toe wall add resisting force. Fill at the crest adds driving load, and clearing vegetation removes the root reinforcement and the interception of rain, so both make the slope less stable.
A tsunami travels in water 4000 m deep and then reaches a shelf 100 m deep. Take g = 9.8 m/s². What happens as it enters the shallow water?
1. The speed falls from about 198 to about 31 m/s, the wavelength shortens and the wave height increases
2. The speed rises from about 31 m/s to about 198 m/s because the energy is squeezed into less water
3. The speed stays at about 198 m/s, since a tsunami's speed depends on the earthquake that made it and not on the depth
4. The speed falls from about 198 m/s to about 31 m/s and the wave height also decreases in proportion
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Answer: A — The speed falls from about 198 to about 31 m/s, the wavelength shortens and the wave height increases
The long-wave speed is v = √(gh): √(9.8 × 4000) = 198 m/s offshore and √(9.8 × 100) = 31 m/s on the shelf. The front slows while the rear is still fast, so the wavelength shortens and, since the energy is conserved, the height grows (shoaling). The alternative that also lowers the height gets the speed right but the height wrong, which is the tempting error; and the speed does depend on depth.
Take the solar constant S₀ = 1370 W/m², a planetary albedo of 0.30 and σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. Treating the Earth as a blackbody in radiative balance with no atmosphere, its effective temperature, in K to the nearest integer, is ____.
Numerical answer — type the value.
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Answer: 255
Absorbed flux = S₀(1 − α)/4 = 1370 × 0.70/4 = 239.75 W/m². Setting σT⁴ = 239.75 gives T⁴ = 239.75/(5.67 × 10⁻⁸) = 4.228 × 10⁹ and T = 255.0 K. Forgetting the factor of 4 (the ratio of surface area to intercepted area) would give about 361 K, and forgetting the albedo about 279 K.
The observed mean surface temperature of the Earth is about 288 K, whereas the effective radiating temperature is about 255 K. What is the main reason for the difference of about 33 K?
1. Heat flowing up from the Earth's interior supplies most of the energy that warms the surface
2. The stratospheric ozone layer absorbs sunlight and heats the surface directly from above
3. The real albedo of the Earth is close to zero, so the calculation underestimates the absorbed sunlight
4. Greenhouse gases absorb outgoing thermal infrared and re-emit part of it downward toward the surface
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Answer: D — Greenhouse gases absorb outgoing thermal infrared and re-emit part of it downward toward the surface
The natural greenhouse effect warms the surface because the atmosphere is transparent to incoming sunlight but absorbs outgoing thermal infrared and radiates part of it back down. Interior heat flow is only about 0.09 W/m², a tiny fraction of the roughly 240 W/m² absorbed from the Sun. Ozone absorbs ultraviolet high in the atmosphere but does not warm the surface, and an albedo near zero would be contradicted by the bright clouds and ice that give the observed 0.30.
For CO₂ take the forcing as ΔF = 5.35 ln(C/C₀) W/m² and a sensitivity parameter λ = 0.8 K per W/m². The equilibrium warming when CO₂ rises from 280 ppm to 420 ppm is closest to which value?
1. 0.8 K, from using a base-10 logarithm in the forcing relation instead of the natural logarithm
2. 1.7 K, from a forcing of 2.17 W/m² multiplied by the sensitivity parameter
3. 2.2 K, from taking the forcing in W/m² as if it were the temperature change itself
4. 3.0 K, from the forcing of a full doubling of CO₂ rather than the actual rise
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Answer: B — 1.7 K, from a forcing of 2.17 W/m² multiplied by the sensitivity parameter
ΔF = 5.35 ln(420/280) = 5.35 × 0.405 = 2.17 W/m², and ΔT = λΔF = 0.8 × 2.17 = 1.7 K. A base-10 logarithm would give 0.8 K, forgetting λ would give 2.2, and the 3.0 K figure belongs to a full doubling (280 to 560 ppm), which has not occurred here.