Question 1

Electromagnetism fundamentals: which statements correctly describe magnetic permeability (μ) — its meaning, penetration of magnetic lines of force, the ratio B/H, and its effect on force between magnetic poles?

Accepted Answer

Introduction / Context: Magnetic permeability (symbol μ) is a cornerstone concept in electromagnetism and electrical engineering. It connects the magnetic field strength H produced by currents to the resulting magnetic flux density B inside a material. Knowing what permeability represents and how it influences forces and flux helps in understanding transformers, inductors, relays, and magnetic circuits. Given Data / Assumptions: B is magnetic flux density (tesla). H is magnetizing field strength (A/m). Permeability μ relates B and H via B = μ * H. Different materials have different μ (air, steels, ferrites, etc.). Concept / Approach: Permeability quantifies how easily a material allows magnetic lines of force to form and pass. In vacuum, μ = μ0. In materials, μ = μ0 * μr, where μr is relative permeability. The larger μ is, the more flux density B is produced for a given H. In magnetic circuits, higher μ implies lower reluctance, analogous to lower resistance in electric circuits. Step-by-Step Solution: Interpretation: Saying a material 'conducts' magnetic lines is a colloquial but acceptable way to describe high μ.Formula: μ = B / H captures the fundamental definition.Penetration: High μ means magnetic lines of force more readily penetrate the medium.Force relation: Inverse proportionality statements reflect that high μ reduces reluctance, altering pole forces and field distributions.Hence, each individual statement is correct; collectively, 'All of these' is the best answer. Verification / Alternative check: Standard EM texts (magnetic circuits, B–H curves) present B = μH and discuss μr >> 1 for ferromagnetics, confirming all statements fit accepted theory. Why Other Options Are Wrong: Choosing any single statement ignores that all listed descriptions are valid aspects of permeability. Common Pitfalls: Confusing permeability (magnetic) with permittivity (electric). Assuming μ is constant for ferromagnets; in practice μ varies with H and can saturate. Final Answer: All of these

Question 2

Satellite operations and tracking: due to orbital perturbations, the onboard orbital parameters are frequently updated based on measurements from how many ground stations?

Accepted Answer

Introduction / Context: Satellite orbits are continuously influenced by perturbations such as Earth’s oblateness, third-body effects, solar radiation pressure, and atmospheric drag (for low orbits). Therefore, operators routinely update orbit parameters (ephemerides) using data gathered from a network of ground stations. Given Data / Assumptions: We consider a typical Earth-observation/navigation operational setup. Perturbations require periodic estimation and upload of refined orbital elements. Multiple geographically distributed ground stations improve coverage and accuracy. Concept / Approach: More than one ground station is desirable to track a satellite along different passes and geometries, reducing dilution of precision and allowing robust orbit determination. A canonical exam reference cites a five-station network as the standard baseline for frequent updates in many operational contexts. Step-by-Step Solution: Identify the need: perturbations demand regular measurements.Assess coverage: a small but distributed network provides sufficient pass opportunities.Select the conventional figure used in classic remote-sensing MCQs: five ground stations. Verification / Alternative check: Operations manuals and exam primers often mention a multi-station network; the five-station figure is a commonly taught benchmark. Why Other Options Are Wrong: Three or four stations may be insufficient for desired coverage in many scenarios. Six stations exceed the quoted typical baseline used in the standard question framing. Common Pitfalls: Assuming a single control station is adequate for precise, frequent updates. Final Answer: five ground stations

Question 3

Snow remote sensing concepts: identify the correct statements regarding albedo, spectral albedo for semi-infinite snow, water equivalent height, and homogeneity of a snow pack.

Accepted Answer

Introduction / Context: Snow radiometry is central to hydrology, cryosphere studies, and climate monitoring. Terms like albedo, spectral albedo (for semi-infinite layers), snow water equivalent (SWE), and snow pack homogeneity appear frequently in remote sensing and surveying exams. Given Data / Assumptions: Albedo is defined with respect to upwelling and downwelling radiative fluxes. Spectral albedo considers wavelength dependence and often assumes optically thick (semi-infinite) snow to avoid substrate effects. Water equivalent height is the water column height after complete melting of snow. A homogeneous snow pack shows minimal vertical variation in mean grain size. Concept / Approach: Snow's high albedo in visible wavelengths, its spectral behavior, and SWE conversion underpin streamflow forecasts and energy-balance modeling. Grain size evolution and layering alter scattering and absorption, hence the need to specify homogeneity when interpreting reflectance. Step-by-Step Solution: Albedo: ratio up-flux / down-flux just above the surface.Spectral albedo: computed per wavelength, typically for semi-infinite snow to neglect background.SWE: physically, water height obtained from melting a unit area column of snow.Homogeneous pack: average grain size does not vary significantly with depth.All statements are correct; therefore select 'All of these'. Verification / Alternative check: Snow hydrology references define albedo and SWE in these exact terms and treat semi-infinite assumptions for spectral albedo computations. Why Other Options Are Wrong: Any single statement alone omits the others, all of which are accurate. Common Pitfalls: Confusing broadband albedo with spectral albedo. Misinterpreting SWE as mass rather than equivalent water column height. Final Answer: All of these

Question 4

Solar spectrum coverage: which of the following frequency/ wavelength regions form part of the Sun’s radiation that reaches or is emitted from the Sun?

Accepted Answer

Introduction / Context: The Sun radiates over a vast electromagnetic spectrum. While the peak of the solar spectrum at Earth is in the visible, significant energy exists in the ultraviolet (UV) and infrared (IR), and measurable emissions occur at radio wavelengths as well. Understanding this breadth is essential for remote sensing, solar physics, and communications interference studies. Given Data / Assumptions: Solar radiation approximates a blackbody spectrum near 5770 K. Earth’s atmosphere filters portions of UV and IR; windows vary by wavelength. Solar radio emissions, though lower in power density than visible/IR, are real and observable. Concept / Approach: Planck’s law predicts non-zero radiance at all wavelengths; the visible band is only where the spectral peak lies for human vision. UV contributes to atmospheric chemistry; IR carries most thermal energy; radio emissions arise from solar plasma processes. Step-by-Step Solution: Acknowledge UV, visible, and IR as principal energy contributors in solar irradiance at Earth.Recognize radio emissions from solar activity (quiet Sun, bursts).Thus, all listed regions are part of the Sun’s radiation. Verification / Alternative check: Solar spectral irradiance curves show energy across UV–visible–IR and detectable radio flux. Why Other Options Are Wrong: Choosing a subset ignores the Sun’s full spectral output. Common Pitfalls: Equating 'most energy' with 'only'—the Sun emits beyond the peak region. Final Answer: All of these

Question 5

Coherence length and bandwidth: the length over which the phase (and thus amplitude correlation) remains strong is related to the source bandwidth in what way?

Accepted Answer

Introduction / Context: Coherence length (Lc) is a practical measure of how far an electromagnetic wave preserves a stable phase relationship, enabling high-contrast interference. It is fundamental in laser physics, spectroscopy, fiber optics, and synthetic aperture systems. Given Data / Assumptions: Bandwidth Δν (or Δλ) characterizes spectral spread. Coherence time τc ≈ 1 / Δν for simple spectral shapes. Coherence length Lc = C * τc in a given medium (C is phase speed). Concept / Approach: If τc ≈ 1 / Δν, then Lc ≈ C / Δν: a narrower bandwidth (smaller Δν) means longer coherence time and longer coherence length. Conversely, broad-band sources have short coherence lengths, explaining why lasers (narrowband) support long-path interferometry. Step-by-Step Solution: Start from τc ≈ 1 / Δν.Compute Lc = C * τc, so Lc ≈ C / Δν.Therefore, Lc is inversely proportional to bandwidth. Verification / Alternative check: For a Gaussian spectrum: τc = 0.44 / Δν; the inverse dependence remains. Why Other Options Are Wrong: Direct proportionality contradicts the physics of coherence. Square relation does not represent standard coherence models. 'None of these' is incorrect because the inverse proportionality is well established. Common Pitfalls: Confusing radiometric bandwidth with detector electrical bandwidth. Final Answer: inversely proportional to the bandwidth

Question 6

Leaf reflectance drivers in optical remote sensing: which factors primarily control a leaf’s spectral reflectance signature across visible to shortwave infrared?

Accepted Answer

Introduction / Context: Vegetation indices, crop stress detection, and species mapping rely on how leaves reflect and absorb light. Pigments, structure, and water content together shape leaf spectra, which aggregate to canopy signatures used by satellites and UAVs. Given Data / Assumptions: Pigments (chlorophylls, carotenoids) drive strong absorption in the visible. Internal mesophyll structure enhances scattering in near-IR. Water content introduces absorption features in the shortwave IR. Concept / Approach: In the visible (0.4–0.7 μm), pigment absorption lowers reflectance, with green reflectance relatively higher. In the near-IR (0.7–1.3 μm), weak absorption and strong internal scattering yield high reflectance. In shortwave IR (1.3–2.7 μm), water absorption bands reduce reflectance, sensitive to leaf water status. Step-by-Step Solution: Pigments explain visible dips (blue/red absorption peaks).Cell structure explains high near-IR plateau due to multiple scattering.Water content explains prominent shortwave IR absorption features.All three together govern the leaf reflectance curve, so 'All of these' is correct. Verification / Alternative check: Standard reflectance spectra of healthy vs stressed leaves clearly show these dependencies. Why Other Options Are Wrong: Any single factor alone cannot reproduce the full spectral behavior across bands. Common Pitfalls: Assuming pigments dominate in near-IR; structure is the key driver there. Final Answer: All of these

Question 7

Leaf water controls reflectance: in which reflective optical band is the influence of leaf water content most prominently expressed for remote sensing analysis?

Accepted Answer

Introduction / Context: Leaf and canopy water status is critical for drought monitoring, irrigation management, and fire risk assessment. Spectrally, water absorbs strongly in specific shortwave infrared (SWIR) bands, enabling remote estimation of leaf water content. Given Data / Assumptions: Visible region is dominated by pigment absorption. Near-IR is dominated by structural scattering with weak water absorption. SWIR hosts strong water absorption bands (e.g., near 1.4 μm, 1.9 μm, 2.1–2.3 μm). Concept / Approach: As liquid water content increases, absorption in the SWIR increases, driving down reflectance. Consequently, many water-related indices use SWIR channels (e.g., Normalized Difference Water Index using NIR and SWIR). Step-by-Step Solution: Assess visible: mainly pigments, not water content.Assess NIR: structural reflectance high; water sensitivity weaker.Assess SWIR: strong water absorption bands cause reflectance reduction correlating with water content.Therefore choose SWIR (1.3–2.7 μm). Verification / Alternative check: Leaf spectra libraries show pronounced reflectance drops in SWIR as water content increases. Why Other Options Are Wrong: Visible/NIR do not primarily reflect water variations; they are dominated by pigments or structure. 'None of these' contradicts established spectroscopy. Common Pitfalls: Confusing canopy water signals in thermal IR with reflective SWIR behavior. Final Answer: short wave-IR (1.3 - 2.7 μm) region

Question 8

Vegetation canopy reflectance drivers: which parameters are considered when determining canopy reflectance in optical remote sensing?

Accepted Answer

Introduction / Context: Canopy reflectance depends not only on leaf properties but also on illumination and viewing geometry and the canopy architecture. These controls are formalized in bidirectional reflectance distribution function (BRDF) models used to normalize and interpret satellite data. Given Data / Assumptions: Solar zenith and azimuth define illumination geometry. Sensor look angle defines viewing geometry. Leaf number, orientation, and clumping define canopy structure. Concept / Approach: Changing sun–sensor geometry alters path lengths and shadow fractions, producing anisotropic reflectance (hotspot effects, forward/backscatter asymmetry). Canopy architecture modulates multiple scattering and shadowing, thus affecting reflectance magnitude and shape across wavelengths. Step-by-Step Solution: Include solar zenith/azimuth: control illumination and cast shadows.Include look angle: controls sensor-viewed mixture of lit and shaded elements.Include leaf arrangement: dictates gap fraction and multiple scattering.Therefore, all listed parameters are considered. Verification / Alternative check: BRDF corrections and canopy radiative transfer models (e.g., SAIL, PROSAIL) explicitly include these parameters. Why Other Options Are Wrong: Any subset omits important drivers of anisotropy and reflectance magnitude. Common Pitfalls: Ignoring geometry when comparing multi-temporal imagery, leading to false change detection. Final Answer: All of these

Question 9

Snell’s law — refraction at a plane interface (optics/electromagnetics) A plane wave traveling in medium 1 (refractive index n1) strikes a plane boundary with medium 2 (refractive index n2) at incidence angle θi. Which relation correctly gives the r…

Accepted Answer

Introduction / Context: Refraction describes the change in direction of a wave as it passes between media of different optical densities. Snell’s law quantifies this change using refractive indices and the angles measured from the normal to the interface. Given Data / Assumptions: Homogeneous, isotropic, lossless media with refractive indices n1 and n2. Angles θi and θr are measured from the interface normal. Planar boundary, no surface relief. Concept / Approach: Snell’s law arises from phase matching of the tangential wave number across the boundary. The classical form is n1 * sin θi = n2 * sin θr, which can be rearranged to sin θr = (n1 / n2) * sin θi. Equality of incidence and refraction angles (θi = θr) is false except when n1 = n2 or at normal incidence. Step-by-Step Solution: 1) Write Snell’s relation: n1 * sin θi = n2 * sin θr.2) Rearrange for θr: sin θr = (n1 / n2) * sin θi.3) Recognize both statements (b) and (c) express the same law. Verification / Alternative check: Special cases: normal incidence (θi = 0) gives θr = 0. If n2 > n1, θr < θi; if n2 < n1, θr > θi. Total internal reflection occurs when (n1 > n2) and θi > θc where sin θc = n2/n1. Why Other Options Are Wrong: (a) violates Snell’s law unless n1 = n2 or θi = 0. (e) contradicts the empirically verified and theoretically derived law. Common Pitfalls: Measuring angles from the interface instead of the normal. Confusing refractive index ratio with its reciprocal in rearranged forms. Final Answer: Both (b) and (c) are equivalent statements of Snell’s law.

Question 10

Photogrammetry — relief displacement, principal point, and scale variation Which combination of statements about vertical photography and terrain effects is correct?

Accepted Answer

Introduction / Context: Vertical photographs suffer relief displacement proportional to object height and radial distance from the principal point, and their scale varies with ground elevation. Understanding these effects is vital for mapping accuracy and interpretation. Given Data / Assumptions: Vertical or near-vertical photography with focal length f and flying height H above datum. Relief displacement increases with object height and image radial distance. Photo scale at elevation h is approximately S = f / (H − h). Concept / Approach: Higher platforms (e.g., satellites) have very large H, reducing d = (r * h) / (H − h). At the principal point (r = 0), displacement is nil. For constant H above datum, higher terrain (larger h) yields larger scale since (H − h) is smaller. A statement claiming “scale depends on the scale of topography” is imprecise; scale depends on H and h (elevation), not a pre-existing “scale of topography.” Step-by-Step Solution: 1) Compare platforms: large H (satellite) ⇒ smaller relief displacement than low-altitude aircraft ⇒ (a) true.2) Principal point has r = 0 ⇒ displacement d = 0 ⇒ (b) true.3) Scale S = f / (H − h): larger h ⇒ smaller denominator ⇒ larger S ⇒ (c) true.4) Hence only (a), (b), (c) are correct; the broad “all of the above” would include an imprecise statement. Verification / Alternative check: Relief displacement and scale formulas are standard in photogrammetry texts; simple numerical checks confirm trends with H and h. Why Other Options Are Wrong: (e) is incorrect because it would include an imprecise/incorrect dependency description. Common Pitfalls: Forgetting that “principal point” is the image of the camera center; r = 0 there. Confusing “flying height above terrain” with “above datum” when comparing scales. Final Answer: Only (a), (b), and (c) are correct.

Question 11

Radiometry terminology — energy, flux, irradiance, and exitance Identify the correct comprehensive statement linking basic radiometric terms and their SI units.

Accepted Answer

Introduction / Context: Radiometric quantities describe electromagnetic radiation independent of wavelength weighting. Correct terminology and units are essential in remote sensing, thermal engineering, and optical measurements. Given Data / Assumptions: Use SI units consistently. Surface-based quantities are per unit area. Time rates are expressed in watts. Concept / Approach: Radiant energy Q has unit joule (J). The time derivative of Q is radiant flux Φ with unit watt (W). When Φ is incident on an area A, irradiance E is Φ_in / A in W/m^2. When flux leaves a surface, the analogous surface quantity is radiant exitance M = Φ_out / A in W/m^2. These definitions underpin more advanced concepts such as radiance and bidirectional reflectance. Step-by-Step Solution: 1) Map concepts to units: energy → J; rate of energy → W.2) Assign surface incidence: irradiance E = Φ_in / A (W/m^2).3) Assign surface emission: exitance M = Φ_out / A (W/m^2). Verification / Alternative check: Dimensional analysis confirms consistency: J/s = W; dividing by m^2 yields W/m^2 for surface densities. Standard remote sensing texts and radiometry handbooks use these exact definitions. Why Other Options Are Wrong: Each of (a)–(d) is correct; therefore the inclusive correct choice is (e) “All of these”. Common Pitfalls: Confusing irradiance (incident) with radiance (directional, per solid angle). Using “energy per area” (J/m^2) for exposure instead of instantaneous flux density (W/m^2). Final Answer: All of these.

Question 12

Wave properties — frequency, amplitude, and energy transport (basics) Which consolidated statement correctly describes frequency, its SI unit, amplitude, and the relation between amplitude and transported energy for waves?

Accepted Answer

Introduction / Context: Foundational wave terminology is widely used in physics, oceanography, acoustics, and electromagnetics. Precision in definitions prevents downstream errors in analysis and design. Given Data / Assumptions: Periodic, approximately sinusoidal waves. Observation at a fixed spatial point for counting frequency. Linear regime where standard proportionalities apply. Concept / Approach: Frequency f counts cycles per second (Hz). Amplitude a is the maximum displacement from the mean position; for surface gravity waves it is half the wave height (crest-to-trough). In many physical systems, average transported energy ∝ a^2, making amplitude a critical control on power and damage potential. Step-by-Step Solution: 1) Define frequency: count crests at a point in 1 s ⇒ f (Hz).2) Define amplitude: vertical distance from mean level to crest (or trough).3) State energy proportionality: energy ∝ amplitude^2 in linear wave theory. Verification / Alternative check: For sinusoidal signals, intensity is proportional to amplitude squared (e.g., sound intensity ∝ pressure amplitude^2; EM intensity ∝ E-field amplitude^2), reinforcing the general relationship. Why Other Options Are Wrong: Each of (a)–(d) is true; therefore the comprehensive correct choice is “All of these”. Common Pitfalls: Confusing wave height (crest-to-trough) with amplitude (half of height). Using rpm or cps instead of SI hertz in technical reporting. Final Answer: All of these.

Question 13

DGPS (Differential GPS) — spatial correlation of bias errors DGPS relies on the assumption that most positioning biases are nearly the same within a local area. Up to what approximate baseline length does this assumption hold well for routine survey…

Accepted Answer

Introduction / Context: Differential GPS improves positional accuracy by applying corrections from a reference station to a rover. The effectiveness depends on how similar the error sources (ionospheric/tropospheric delays, satellite ephemeris, clock errors) are over the separation between stations, known as the baseline length. Given Data / Assumptions: Standard single-frequency (or dual-frequency) code/phase DGPS for typical engineering surveys. Moderate solar activity and mid-latitude conditions. Desire for decimetre to sub-metre results with real-time corrections. Concept / Approach: Error fields decorrelate with distance. Within roughly 100 km, atmospheric and orbital components remain sufficiently correlated that reference corrections significantly reduce rover errors. Beyond that, residuals grow and additional modeling or network RTK approaches are preferred. Step-by-Step Solution: 1) Identify dominant correlated errors (ionosphere/troposphere).2) Recognize spatial decorrelation: correlation reduces with baseline length.3) Use practical rule: DGPS baselines ≲ 100 km for robust correction transfer. Verification / Alternative check: Network RTK and SBAS solutions extend ranges by modeling spatial gradients; classic single-base DGPS commonly cites ~100 km as an upper practical limit for good performance. Why Other Options Are Wrong: 25–75 km are usable but overly restrictive as a hard limit. 1250 km far exceeds the range over which biases remain similar. Common Pitfalls: Ignoring space-weather/ionospheric storms which can shrink effective ranges. Assuming uniform performance for single-frequency receivers across seasons and latitudes. Final Answer: 100 km.

Question 14

Electronic distance measurement — Geodimeter operating principle A Geodimeter measures distance primarily based on which carrier and modulation principle?

Accepted Answer

Introduction / Context: Electronic distance measurement (EDM) instruments determine distance by measuring the phase or time of flight of a modulated carrier. Classical instruments split by carrier type: optical (light), infrared, or microwave. Knowing which instrument uses which carrier is a standard surveying MCQ topic. Given Data / Assumptions: Geodimeter is one of the earliest EDM instruments. It uses an optical (visible light) carrier with intensity modulation. Tellurometer is the microwave counterpart; Distomat and others may use IR. Concept / Approach: Geodimeter employs a visible-light (optical) carrier, typically intensity-modulated and phase-compared upon return, to infer distance. While the optical carrier frequency is on the order of 10^14 Hz, the modulation frequency is low and measurable for phase comparison. Step-by-Step Solution: 1) Identify instrument family: optical EDM.2) Recognize principle: modulate light intensity; measure phase shift.3) Conclude correct choice: modulated light waves. Verification / Alternative check: Instrument taxonomy: Geodimeter (optical), Tellurometer (microwave), later IR EDMs (infrared). This mapping is standard in surveying texts and exam prep guides. Why Other Options Are Wrong: (b) Over-specifies IR only; not the classical Geodimeter. (c) States a true attribute of visible light frequency but is not the operating principle itself. (d) Describes microwave EDM (Tellurometer), not Geodimeter. Common Pitfalls: Confusing carrier frequency (optical) with modulation frequency (radio-range). Mixing up instrument names across carrier types. Final Answer: Propagation of modulated light waves (optical carrier).

Question 15

GPS satellite orbits — approximate orbital radius from Earth’s centre In satellite navigation (GPS), the medium Earth orbit is characterised by an orbital radius (distance from Earth’s centre) of roughly 26,600 km. Choose the closest approximate orb…

Accepted Answer

Introduction / Context: Global Positioning System (GPS) satellites operate in medium Earth orbit (MEO). Understanding their orbital radius (measured from Earth’s centre, not the surface) helps in grasping geometry for visibility, dilution of precision, repeat periods, and signal travel times critical to positioning accuracy. Given Data / Assumptions: We need the approximate orbital radius of GPS satellites. Orbital radius = Earth’s mean radius + satellite altitude. Earth’s mean radius is about 6,371 km; GPS altitude is about 20,200 km above mean sea level. Concept / Approach: Convert altitude to orbital radius by adding Earth’s radius. GPS satellites are not in geostationary orbit (~35,786 km altitude; ~42,164 km orbital radius) and not in low Earth orbit (a few hundred to ~2,000 km altitude); they sit between these regimes in MEO to balance coverage, constellation size, and signal strength. Step-by-Step Solution: 1) Take Earth radius Re ≈ 6,371 km.2) GPS altitude h ≈ 20,200 km.3) Orbital radius r = Re + h = 6,371 + 20,200 ≈ 26,571 km ≈ 26,600 km (rounded). Verification / Alternative check: Using orbital period ~11 hours 58 minutes (half a sidereal day) corroborates MEO parameters; a ~26,600 km radius yields this period under standard gravitational parameter for Earth. Why Other Options Are Wrong: 15,200 km and 18,400 km: Too small; closer to some MEO constellations but below GPS standards. 36,000 km: Confuses with geostationary orbit altitude (radius ~42,164 km), not GPS. None of these: Incorrect since 26,600 km is correct. Common Pitfalls: Confusing altitude (above surface) with orbital radius (from Earth’s centre) and mixing up MEO with GEO or LEO values. Final Answer: 26,600 km

Question 16

Origin of the term “remote sensing” Identify the person credited with coining the term “remote sensing” in the context of Earth observation and allied geosciences.

Accepted Answer

Introduction / Context: Nomenclature shapes disciplines. The term “remote sensing” consolidated diverse practices—photogrammetry, aerial survey, multispectral observation—under a unified banner for scientific and operational Earth observation. Knowing its origin is common background for professionals in GIS and EO. Given Data / Assumptions: The question asks specifically who coined the term “remote sensing”. We distinguish between early pioneers of aerial photography and the later formalisation of terminology. Concept / Approach: While aerial imaging dates to the 19th and early 20th centuries (balloon, kite, aircraft), the modern term “remote sensing” was popularised and credited to Evelyn L. Pruitt of the U.S. Office of Naval Research in the mid-20th century to encompass non-contact sensing of the Earth’s surface using various portions of the electromagnetic spectrum. Step-by-Step Discussion: 1) Early aerial imagery: Tournachon (Nadar) pioneered balloon photography.2) Aviation era: Wright brothers advanced powered flight but did not coin the term.3) Terminology: Evelyn L. Pruitt is widely credited with coining/establishing the term “remote sensing”. Verification / Alternative check: Standard remote sensing textbooks and historical notes attribute the term to Pruitt; earlier figures contributed techniques but not the specific term. Why Other Options Are Wrong: Gaspard Felix Tournachon: Pioneer of balloon photography, but not the term’s originator. Wilbur Wright: Aviation pioneer; also misidentified as Italian here; did not coin the term. None of these / All of these: Both contradict the accepted attribution to Pruitt. Common Pitfalls: Equating invention of aerial photography with the coinage of “remote sensing” and overlooking mid-20th-century institutional terminology development. Final Answer: Evelyn L. Pruitt, a geographer

Question 17

Objective of photo-interpretation in remote sensing In the context of aerial and satellite imagery, state the principal objective of photo-interpretation.

Accepted Answer

Introduction / Context: Photo-interpretation converts imagery into meaningful geographic information. Interpreters rely on tone, texture, shape, size, pattern, shadow, and context to recognise and classify features, and to derive insights about processes, land use, and change detection. Given Data / Assumptions: Aerial or satellite images are available. Interpreter applies visual keys and ancillary data. Goal: transform pixels into thematic information. Concept / Approach: Photo-interpretation involves a progression: detection, recognition, identification, and evaluation. After locating features, interpreters assign labels (identification) and assess significance (e.g., hazard, suitability, trend), often cross-checking with ground truth or higher resolution data. Step-by-Step Solution: 1) Recognition: discriminate objects from background using visual elements (tone, texture).2) Identification: assign class/label (e.g., conifer forest, asphalt road, paddy field).3) Judging significance: interpret process or importance (e.g., flood extent, deforestation risk).4) Reporting: map features, quantify areas, and provide actionable insights. Verification / Alternative check: Accuracy assessment with confusion matrices and field checks validates that recognition and identification steps were performed correctly. Why Other Options Are Wrong: identification alone: misses recognition and evaluation. recognition alone: insufficient without labelling. judging significance alone: not possible without first recognising and identifying. None of these: incorrect because all are true components. Common Pitfalls: Overreliance on a single cue (e.g., tone), ignoring seasonal effects, and neglecting scale/resolution when assigning classes. Final Answer: All of these

Question 18

Optical properties of natural waters — controlling factors In water optics for remote sensing and limnology, the apparent optical properties of a water body depend on absorption and scattering by dissolved and suspended constituents. Identify the co…

Accepted Answer

Introduction / Context: Water colour and the upwelling radiance seen by sensors are governed by how light is absorbed and scattered within the water column. Remote sensing of water quality (e.g., chlorophyll, coloured dissolved organic matter, turbidity) depends on these interactions across visible and near-infrared wavelengths. Given Data / Assumptions: Dissolved substances (e.g., CDOM) can strongly absorb in blue wavelengths. Suspended particulates both absorb and scatter light. Surface effects (waves, glint) modulate apparent signal but the question focuses on in-water processes. Concept / Approach: Apparent optical properties (AOPs) relate to how light fields behave, while inherent optical properties (IOPs) depend solely on the medium (absorption a(λ), scattering b(λ)). Both dissolved and particulate constituents influence a(λ) and b(λ), shaping reflectance spectra used for algorithm retrievals. Step-by-Step Solution: 1) Consider dissolved absorbers: increase absorption, lowering blue reflectance.2) Suspended sediments: contribute to both absorption and strong scattering, increasing reflectance in red/NIR.3) Combined effects explain colour gradients from oligotrophic blue waters to turbid brown waters. Verification / Alternative check: Field spectra and lab measurements of a(λ) and b(λ) confirm the contributions of both dissolved and particulate components. Why Other Options Are Wrong: Each single choice captures only one mechanism; real waters involve all mechanisms. None of these: Incorrect because all listed processes govern optical response. Common Pitfalls: Ignoring wavelength dependence, confounding surface glint with in-water scattering, and assuming clear-water behaviour for turbid systems. Final Answer: All of these

Question 19

Advantages of satellite-platform Earth observation Select the comprehensive statement about what Earth observation from satellites provides to users and analysts.

Accepted Answer

Introduction / Context: Satellite platforms revolutionised Earth observation by enabling consistent, repeatable, and wide-area measurements. Mission design—orbit type, revisit cycle, swath—determines coverage, temporal sampling, and illumination conditions for change detection and mapping. Given Data / Assumptions: Polar sun-synchronous orbits yield near-constant local solar time passes. Geostationary orbits provide high-frequency sampling for weather. Revisit can range from minutes (GEO) to days/weeks (LEO constellations). Concept / Approach: Satellites provide synoptic views due to large swaths. Sun-synchronous LEO maintains similar illumination (constant solar zenith angle) for consistent reflectance comparison. Orbital mechanics set revisit intervals enabling monitoring of vegetation phenology, disasters, and urban change. Step-by-Step Solution: 1) Large-area synoptic view: swaths hundreds to thousands of kilometres wide.2) Consistent illumination: near-fixed local time overpasses for radiometric comparability.3) Repetition: scheduled overpasses allow time-series analysis at scales from minutes (GEO) to weeks (LEO single-sat missions). Verification / Alternative check: Mission specs (e.g., Landsat ~16 days, Sentinel-2 ~5 days constellation, INSAT/GOES continuous) corroborate these points. Why Other Options Are Wrong: Picking any single feature omits the broader capability; all are valid. None of these: Incorrect since satellites offer all listed advantages. Common Pitfalls: Assuming “constant illumination” implies identical bidirectional reflectance every pass; surface anisotropy and atmosphere still matter. Final Answer: All of these

Question 20

Causes of spatial variation in geomaterials Spatial variation in properties in the horizontal and vertical directions arises from multiple weathering processes. Identify the correct grouping.

Accepted Answer

Introduction / Context: Engineering properties of soils and rocks vary spatially due to the complex interplay of geological processes. Recognising the origins of variability helps in site investigation planning, sampling strategies, and uncertainty quantification for design. Given Data / Assumptions: Near-surface geomaterials undergo weathering. We consider physical, chemical, and biological mechanisms. Variations are both lateral and vertical. Concept / Approach: Physical weathering (e.g., freeze–thaw, thermal stress) fractures rock and increases porosity. Chemical weathering (e.g., hydrolysis, oxidation) alters mineralogy and fabric. Biological weathering (roots, microbes) enhances disaggregation and chemical alteration, creating heterogeneous profiles. Step-by-Step Discussion: 1) Physical: mechanical breakdown increases grain size variability and anisotropy.2) Chemical: solution and precipitation processes form secondary minerals and alter strength/permeability.3) Biological: roots wedge joints; biogenic acids accelerate dissolution.4) Net effect: stratified weathering fronts and lenses with differing stiffness and strength. Verification / Alternative check: Borehole logs typically show gradations from residual soils to partially weathered rock, evidencing combined mechanisms. Why Other Options Are Wrong: Each single mechanism contributes; excluding others misrepresents reality. None of these: Incorrect because all mechanisms operate in nature. Common Pitfalls: Assuming homogeneity from limited samples; ignoring scale effects; extrapolating 1D profiles to site-wide conditions. Final Answer: All of these

Elements of Remote Sensing Questions