Question 1

Passive versus active sensors — source of energy A passive remote sensing sensor relies on which source of energy for illumination of the target?

Accepted Answer

Introduction / Context: Remote sensing systems are broadly classified as passive or active, based on how the target is illuminated. This distinction is fundamental for understanding sensor design, data characteristics, and optimal acquisition times. Given Data / Assumptions: Passive sensors measure naturally available radiation. Active sensors transmit their own radiation and measure returns. We focus on Earth observation at optical/thermal wavelengths. Concept / Approach: Passive optical sensors (e.g., Landsat, Sentinel-2) rely primarily on reflected solar radiation (visible/NIR/SWIR) and emitted thermal radiation (TIR) from the Earth–atmosphere system. Active sensors (e.g., LiDAR, SAR) supply their own energy. Step-by-Step Reasoning: 1) Identify passive: no onboard transmitter in the measured band.2) Illumination source: the Sun for reflective bands; Earth’s thermal emission at night for thermal bands.3) Hence, for typical reflective imaging, the Sun is the operative energy source. Verification / Alternative check: Night-time optical imaging in reflective bands is impractical for passive sensors due to lack of solar illumination, confirming their dependence on the Sun. Why Other Options Are Wrong: flash light as a source of energy: Not used in orbital imaging; unrealistic scale. its own source of energy: Defines active sensors (e.g., radar, LiDAR), not passive. None/All of these: Contradict the passive definition. Common Pitfalls: Confusing passive thermal emission with “own source”; the sensor measures Earth’s emission—it does not generate it. Final Answer: sun as the source of energy

Question 2

Definition of active remote sensing Name the type of remote sensing in which the system generates electromagnetic radiation of a specific wavelength to illuminate a target and then measures the scattered/backscattered radiance.

Accepted Answer

Introduction / Context: Remote sensing techniques differ in whether the sensor provides illumination. Active systems transmit energy and record the return, enabling control over wavelength, polarisation, and timing—vital for all-weather or night-time operations (e.g., radar, LiDAR). Given Data / Assumptions: System emits EMR of a chosen wavelength. Target is illuminated intentionally by the sensor. Backscatter or reflectance is measured at the sensor. Concept / Approach: By generating its own signal, an active sensor decouples imaging from solar illumination and can penetrate clouds (microwave SAR) or measure precise ranges (LiDAR). The definition in the stem exactly matches “active remote sensing”. Step-by-Step Reasoning: 1) Identify presence of onboard transmitter.2) Confirm target illumination by sensor-generated EMR.3) Conclude modality: active remote sensing. Verification / Alternative check: Examples include SAR missions (Sentinel-1, RISAT) and LiDAR altimetry (ICESat). Passive imagers lack this transmitter. Why Other Options Are Wrong: passive remote sensing: uses natural illumination (primarily Sun), not emitted by the sensor. neutral remote sensing: not a standard term. None/All of these: Do not match the precise definition. Common Pitfalls: Assuming “active” implies only radar; LiDAR is also active but in optical wavelengths. Final Answer: active remote sensing

Question 3

Infrared portion of the electromagnetic spectrum (EMR) Select the wavelength range that most appropriately covers the infrared region used in remote sensing.

Accepted Answer

Introduction / Context: The electromagnetic spectrum is segmented by wavelength. Remote sensing commonly divides infrared into near-infrared (NIR), shortwave infrared (SWIR), and thermal infrared (TIR), each with unique interactions with surface and atmosphere used for vegetation, mineral, and temperature mapping. Given Data / Assumptions: We consider the broad IR span relevant to EO sensors. Visible light is shorter than IR; microwave is longer. Concept / Approach: Remote sensing conventionally treats NIR starting near 0.7 μm, extending through SWIR bands (e.g., 1.6 μm, 2.2 μm) into thermal windows around 8–14 μm. This expansive 0.7–14 μm bracket captures widely used bands. Step-by-Step Reasoning: 1) Visible: ~0.4–0.7 μm (option a).2) Infrared used in EO spans NIR/SWIR/TIR: ~0.7 to ~14 μm (option d).3) Microwave: millimetre to metre waves (option b); too long for IR.4) 0.7–1.3 μm (option c) is only NIR subset, not full IR. Verification / Alternative check: Typical sensor bands: NIR ~0.84 μm, SWIR ~1.6 and 2.2 μm, TIR ~10–12 μm; all fall within 0.7–14 μm. Why Other Options Are Wrong: 0.4–0.7 μm: visible range, not IR. 0.5 mm to 1 m: microwave range, not IR. 0.7–1.3 μm: only a portion of IR (NIR), excludes SWIR/TIR. None of these: Incorrect since 0.7–14 μm is correct. Common Pitfalls: Equating IR only with thermal; NIR and SWIR are also IR and widely used for vegetation and geology. Final Answer: 0.7 to 14 μm

Question 4

Applications of code-based GPS receivers Code-based (C/A-code or differential code) GPS receivers are typically used for general navigation and tracking applications. Select the most comprehensive use-case grouping.

Accepted Answer

Introduction / Context: GPS receivers derive positions by measuring pseudorange using PRN codes. Code-based receivers (as opposed to carrier-phase survey-grade units) provide metre-level accuracy suitable for consumer and fleet applications that require robustness and low cost rather than centimetre precision. Given Data / Assumptions: Receiver uses code tracking, possibly with augmentation (SBAS/DGPS). Use cases include navigation and asset tracking. No requirement for geodetic-grade accuracies. Concept / Approach: Code measurements yield typical accuracies of ~3–10 m (better with augmentation). This suffices for turn-by-turn guidance, vehicle/asset tracking, and transport movement monitoring. Carrier-phase receivers are reserved for surveying, precision agriculture, and structural monitoring. Step-by-Step Reasoning: 1) Vehicle tracking: fleet logistics and anti-theft require metre-level positions.2) Land navigation: hiking, automotive navigation, mobile maps—ideal for code receivers.3) Transport movement (“Trans movement”): monitoring shipments and public transport. Verification / Alternative check: Commercial GNSS modules in smartphones and tracking devices are predominantly code-based; differential corrections enhance performance but remain in the code domain for mass market. Why Other Options Are Wrong: Choosing any single option is incomplete; all listed applications are standard. None of these: Incorrect since these are canonical code-receiver uses. Common Pitfalls: Expecting centimetre-level accuracy from standalone code receivers; that requires carrier-phase RTK/PPP solutions. Final Answer: All of these

Question 5

Remote Sensing Programmes – CAPE (1983) Crop Acreage and Production Estimation (CAPE) was funded and formally taken up in 1983 by which country/agency?

Accepted Answer

Introduction / Context: CAPE stands for Crop Acreage and Production Estimation, an early large-scale agricultural remote sensing initiative aimed at statistically robust estimates of sown area and production. In India, it was conceived to support the Ministry of Agriculture using Indian Remote Sensing satellites and allied aerial/survey inputs. Given Data / Assumptions: Year of formal launch referenced: 1983. Programme focus: agricultural acreage and production estimation. Context: public-sector, national remote sensing capacity. Concept / Approach: Identify which country’s space/agriculture establishment initiated and funded CAPE in 1983. India’s space programme (ISRO/NRSA, now NRSC) partnered with the Ministry of Agriculture to operationalise satellite-aided crop statistics, making India the correct answer. Step-by-Step Solution: Relate CAPE to national remote sensing infrastructure.Note the 1983 timeframe aligns with India’s early operational experiments in acreage estimation.Conclude the programme was funded/undertaken by India with its ministries and remote sensing centres. Verification / Alternative check: Indian exam syllabi and standard remote sensing texts consistently list CAPE as an Indian initiative under ISRO/NRSA with the Ministry of Agriculture, dating to the early 1980s for pilot-to-operational phases. Why Other Options Are Wrong: USA/EU/Russia/UK: all have remote sensing programmes, but CAPE (as termed in Indian syllabi) refers specifically to the Indian initiative. Common Pitfalls: Confusing CAPE with similarly named international projects or later Indian programmes like FASAL; the question explicitly anchors the year 1983. Final Answer: India

Question 6

Basic Radiometry & Geometry – Solid angle facts Which of the following statements about radians, solid angle, and its unit are correct?

Accepted Answer

Introduction / Context: Remote sensing and geodesy rely on angular measures. Plane angles (radians) and solid angles (steradians) appear in sensor instantaneous field of view (IFOV), radiance definitions, and calibration geometry. Understanding these basics prevents unit and interpretation errors. Given Data / Assumptions: Definition of radian in plane geometry. Definition of solid angle at the centre of a sphere. Mathematical form of a solid angle Ω. SI units for angles. Concept / Approach: Check each statement against standard definitions: π rad = 180°; Ω is the conical spread from an area A on a sphere of radius r; Ω = A / r^2; SI unit is steradian, symbol sr. Step-by-Step Solution: Confirm plane-angle relation: π rad = 180°.Interpret “cone subtended by area” as the geometric definition of solid angle Ω.Use Ω = A / r^2 to connect area and radius.Recall SI unit: steradian (sr). Verification / Alternative check: On a unit sphere (r = 1), Ω = A; thus steradians numerically equal area on the unit sphere, ensuring consistency with Ω's definition. Why Other Options Are Wrong: a–d are all correct, so the combined correct choice is “All of these”. Common Pitfalls: Using degrees for calculations requiring radians, or confusing planar angle (rad) with solid angle (sr), leads to scaling mistakes. Final Answer: All of these

Question 7

Wave Optics at Interfaces – Reflection and refraction basics For electromagnetic radiation incident on a smooth interface, which relation is always true?

Accepted Answer

Introduction / Context: When an electromagnetic wave meets a planar boundary, part of it may reflect and part refract. The geometric relationships between the incident, reflected, and refracted rays govern sensor viewing geometry and radar/lidar returns. Given Data / Assumptions: Smooth, plane interface between two homogeneous media. Angles measured from the normal to the interface. Standard laws of reflection and refraction. Concept / Approach: The law of reflection states θ_i = θ_r (incidence equals reflection). Refraction follows Snell’s law n1 * sin θ_i = n2 * sin θ_t, which does not imply θ_i = θ_t except for equal indices. Step-by-Step Solution: Apply reflection law: θ_incidence = θ_reflection ⇒ true.Apply refraction law: θ_refraction generally differs from θ_incidence unless n1 = n2.Option (c) is ill-posed; angles do not add that way. Verification / Alternative check: Derivations from Fermat’s principle or boundary conditions on EM fields yield the same reflection law; experiments confirm mirror-like reflection symmetry. Why Other Options Are Wrong: (a) Violates Snell’s law in general. (c) Not a recognized geometric relationship. (d) Cannot be correct because (a) and (c) are false. Common Pitfalls: Measuring angles from the surface instead of the normal or assuming equal refractive indices causes mistakes. Final Answer: Angle of incidence = angle of reflection

Question 8

SAR Coherence Length – Using Doppler bandwidth For a C-band Synthetic Aperture Radar (SAR) with Doppler bandwidth Δf_D = 1300 Hz, estimate the coherence length l_coh (use the relation l_coh ≈ c / Δf for coherence length based on spectral width).

Accepted Answer

Introduction / Context: Coherence length characterizes the path-length over which a wave maintains a predictable phase relationship. In radar/optics, it is inversely related to spectral width. When bandwidth is given in hertz, a useful first-order estimate is l_coh ≈ c / Δf, where c is the speed of light and Δf is the effective spectral width. Given Data / Assumptions: Doppler bandwidth Δf_D = 1300 Hz. Speed of light c ≈ 3.0 * 10^8 m/s. Using the simple coherence-length estimate l_coh ≈ c / Δf (no factor of 2 for this definition). Concept / Approach: For a finite spectral width, coherence time τ_coh ≈ 1 / Δf. Coherence length is l_coh ≈ c * τ_coh ≈ c / Δf. Substitute Δf = 1300 Hz to compute l_coh and convert to kilometres. Step-by-Step Solution: τ_coh = 1 / Δf = 1 / 1300 s ≈ 7.6923 * 10^-4 sl_coh = c * τ_coh ≈ (3.0 * 10^8 m/s) * (7.6923 * 10^-4 s)l_coh ≈ 2.3077 * 10^5 mConvert to km: 2.3077 * 10^5 m ≈ 230.8 kmRounded to the nearest option ≈ 230 km Verification / Alternative check: Using the compact form l_coh = c / Δf directly: 3 * 10^8 / 1300 ≈ 2.31 * 10^5 m, confirming ≈ 230 km. Why Other Options Are Wrong: 130 km and 250 km: off by large rounding; not supported by the calculation. 500 km: roughly double the computed value. 65 km: about one-fourth of the correct magnitude. Common Pitfalls: Confusing range bandwidth (MHz) with Doppler bandwidth (Hz) or inserting an unnecessary factor of 2 (which would correspond to certain coherence definitions) will change the numerical result. Final Answer: 230 km

Question 9

GPS Space Segment – Number of NAVSTAR satellites The GPS space segment (NAVSTAR constellation) nominally consists of how many operational satellites?

Accepted Answer

Introduction / Context: GPS (NAVSTAR) provides global coverage using a medium-Earth-orbit satellite constellation. Understanding the nominal constellation size is fundamental to dilution of precision (DOP), satellite visibility, and mission planning for surveying and navigation. Given Data / Assumptions: Nominal, textbook constellation (legacy design). Focus on operational satellites required for full global coverage. Concept / Approach: The classic NAVSTAR design specifies 24 satellites arranged in 6 orbital planes (4 per plane), with appropriate RAAN spacing and 55° inclination to ensure at least 4 satellites are visible worldwide at most times. Step-by-Step Solution: Recall the standard: 6 planes, 4 satellites each.Total = 6 * 4 = 24 operational satellites (not counting on-orbit spares). Verification / Alternative check: Although modern GPS often operates with more than 24 spacecraft (including spares), the nominal constellation in textbooks/exams remains 24. Why Other Options Are Wrong: 8, 12, 16: too few for continuous global 3D positioning. 30: can be seen in practice but exceeds the nominal design. Common Pitfalls: Confusing nominal constellation size with the actual number currently on orbit; exams typically ask for the nominal 24. Final Answer: 24

Question 10

Total Internal Reflection – Identifying the critical angle condition Critical angle for electromagnetic radiation at an interface occurs under which angular condition?

Accepted Answer

Introduction / Context: Critical angle marks the onset of total internal reflection when light (or EM waves) travels from a denser to a rarer medium. Beyond this, refraction ceases and all energy reflects along with an evanescent field at the boundary. Given Data / Assumptions: Two homogeneous media with refractive indices n1 > n2. Angles measured from the normal. Snell’s law governs refraction. Concept / Approach: Snell’s law: n1 * sin θ_i = n2 * sin θ_t. At the critical angle θ_c, the refracted ray grazes the interface, i.e., θ_t = 90°. Therefore sin θ_c = n2 / n1 (valid only when n1 > n2). Step-by-Step Solution: Set θ_t = 90° at critical condition.Apply Snell’s law: n1 * sin θ_c = n2 * sin 90° = n2.Thus sin θ_c = n2 / n1 (requires n1 > n2). Verification / Alternative check: For glass (n ≈ 1.5) to air (n ≈ 1.0), sin θ_c ≈ 1/1.5 ≈ 0.667, so θ_c ≈ 41.8°, consistent with experience. Why Other Options Are Wrong: (a) Reflection equality is unrelated to refraction. (b) True for many incidences but does not define critical angle. (c) Incidence at 90° is not the critical condition. (e) Reflection angle is never 90° for typical incidence on a plane surface. Common Pitfalls: Measuring angles from the surface instead of the normal, or applying the formula when n1 ≤ n2, where no critical angle exists. Final Answer: Angle of refraction equals 90°

Question 11

Directional Signatures – Variations of reflectivity/emissivity The changes in reflectivity or emissivity with time are termed as what kind of variation?

Accepted Answer

Introduction / Context: Remote sensing signals depend on wavelength (spectrum), location (space), viewing/illumination geometry (angle), and time. Properly naming each variation helps interpret multitemporal imagery for change detection and monitoring. Given Data / Assumptions: Quantity of interest: reflectivity/emissivity of a target. Axis of change under consideration: time. Concept / Approach: Terminology: spectral = across wavelengths; spatial = across locations; temporal = across time; angular/bidirectional = across viewing/illumination angles (BRDF/BTDF). Time-based changes, such as seasonal vegetation dynamics or diurnal thermal cycles, are therefore “temporal variation”. Step-by-Step Solution: Identify the dimension of change: time.Map to term: temporal variation.Exclude spectral, spatial, and angular categories. Verification / Alternative check: Time-series analyses (e.g., NDVI curves, nighttime land surface temperature cycles) are canonical examples of temporal variation in remote sensing signals. Why Other Options Are Wrong: Spectral: wavelength dependence, not time. Spatial: geographic dependence. Angular: geometry-dependent BRDF effects. None: incorrect because a standard term exists. Common Pitfalls: Confusing temporal trends with spectral signatures or BRDF, leading to misinterpretation of multitemporal datasets. Final Answer: Temporal variation

Question 12

GPS Orbits – Typical altitude of NAVSTAR satellites What is the normal (nominal) altitude of a GPS (NAVSTAR) satellite above Earth's surface?

Accepted Answer

Introduction / Context: GPS satellites occupy medium Earth orbits (MEO), chosen to balance global coverage, signal strength, and orbital perturbations. Knowing their altitude supports link budget estimates and timing/geometry calculations in surveying. Given Data / Assumptions: GPS (NAVSTAR) system. Nominal design orbit, not geostationary or LEO. Concept / Approach: GPS satellites orbit at a semi-major axis of ~26,560 km from Earth’s centre, corresponding to a nominal altitude of ~20,200 km above mean sea level, with 55° inclination and ~12-hour orbital period (sidereal). Step-by-Step Solution: Recall standard GPS altitude ≈ 20,200 km.Reject LEO (~400–1200 km) and GEO (~35,786 km) values. Verification / Alternative check: Period ~11 h 58 min, consistent with MEO altitude and constellation geometry. Why Other Options Are Wrong: 16,200/24,400 km: not standard GPS altitudes. 36,100 km: close to GEO; GPS is not geostationary. 10,400 km: too low for NAVSTAR. Common Pitfalls: Confusing GLONASS, Galileo, or geostationary altitudes with GPS values. Final Answer: 20,200 km

Question 13

EM Wave Propagation – Field orientations in homogeneous, isotropic media As an electromagnetic wave propagates through a homogeneous, isotropic, lossless medium, which statement is correct?

Accepted Answer

Introduction / Context: In free space or any homogeneous, isotropic, lossless medium, plane electromagnetic waves are transverse. This geometric relationship underlies polarization, antenna patterns, and radiative transfer models used in remote sensing. Given Data / Assumptions: Medium: homogeneous, isotropic, and lossless. Plane-wave approximation. Concept / Approach: For a plane wave, the electric field E, magnetic field H, and the direction of propagation k form a mutually orthogonal triad. Hence E ⟂ H, and both are ⟂ to k (the Poynting vector S is along k). Step-by-Step Solution: Identify plane-wave property: transversality.E ⟂ H ⇒ option (a) true.E ⟂ k and H ⟂ k ⇒ option (b) true.Therefore combined statement (c) is correct. Verification / Alternative check: Maxwell’s equations in homogeneous media lead to wave equations with ∇·E = 0 and ∇·H = 0 for plane waves, enforcing transversality. Why Other Options Are Wrong: (d) contradicts wave transversality. (e) applies to longitudinal waves, not EM plane waves in such media. Common Pitfalls: Assuming longitudinal components exist in free space; they appear only in waveguides/plasmas under special conditions. Final Answer: Both (a) and (b)

Question 14

Ocean Colour & Bio-optics – Effect of phytoplankton on spectral response Which statements about phytoplankton and water-leaving radiance in the visible spectrum are correct?

Accepted Answer

Introduction / Context: Ocean-colour remote sensing interprets water-leaving radiance to infer chlorophyll concentration and particle content. Phytoplankton pigments and associated particulates modify absorption and scattering, especially in the blue–green region used by sensors like SeaWiFS, MODIS, and OCM. Given Data / Assumptions: Case-1 waters where phytoplankton and covarying particles dominate optical properties. Visible bands spanning blue (~440–490 nm) and green (~550–560 nm). Qualitative relationships between pigment concentration, absorption, and scattering. Concept / Approach: Chlorophyll-a strongly absorbs blue light while particulate matter (including phytoplankton and detritus) contributes to scattering, often enhancing green reflectance relative to blue as concentration rises. Therefore statements about pigment presence, increased green backscattering, and increased blue absorption are correct; the claim that green backscattering decreases is contrary to typical behaviour in Case-1 waters. Step-by-Step Solution: Recognize pigment: chlorophyll-a in phytoplankton ⇒ (a) true.With more phytoplankton/particles, backscattering and thus green-region reflectance tend to rise ⇒ (b) true.Chlorophyll absorption peaks in blue, reducing blue reflectance as concentration increases ⇒ (c) true.Statement (d) claims decreased green backscatter with more phytoplankton ⇒ generally false in Case-1 waters. Verification / Alternative check: Empirical ocean-colour algorithms (e.g., blue/green band ratios) exploit the decrease in blue reflectance and relatively higher green reflectance with increasing chlorophyll, consistent with (a)–(c). Why Other Options Are Wrong: (d) contradicts observed trends; phytoplankton and associated particles tend to raise scattering, not reduce it, in green bands. Common Pitfalls: Confusing absorption (which lowers reflectance) with backscattering (which raises reflectance), or applying Case-2 coastal water logic indiscriminately to open ocean scenarios. Final Answer: Only (a), (b), and (c) are correct

Question 15

Remote Sensing Signatures – Which statements are correct? Consider the nature of spectral and polarimetric signatures used for Earth observation. Identify the correct statements about how signatures are represented, how polarization behaves, and how…

Accepted Answer

Introduction / Context: Remote sensing analysts use “signatures” to distinguish land covers and materials. Signatures can be spectral (variation of reflectance/emittance with wavelength), temporal/phenologic (how those responses change with growth stage or season), angular (BRDF behaviour), and polarimetric (state and degree of polarization). Understanding what a signature encodes and its limitations is essential for reliable classification and change detection. Given Data / Assumptions: Passive optical and active microwave observations are considered. Field and image signatures are affected by illumination, atmosphere, geometry, and sensor noise. Targets are natural/agricultural surfaces that vary through time. Concept / Approach: A spectral reflectance curve of a crop, traced over successive dates, exhibits a phenologic pattern related to chlorophyll, water content, and structure. Polarization variation denotes how the polarization state (for example, degree and angle) of reflected or emitted radiation changes due to surface roughness, moisture, and geometry. Because the observed scene is influenced by many factors, signatures are best treated as statistical rather than fixed constants; hence libraries store ranges and variances, not single lines only. Step-by-Step Solution: Evaluate Option A: Time-series spectral curves across growth stages indeed form crop phenologic signatures.Evaluate Option B: Changes in polarization state of radiation from a target are referred to as polarization variation and are a central concept in polarimetry.Evaluate Option C: Degree of polarization depends on the target and illumination-view geometry, assisting target discrimination.Evaluate Option D: Field signatures vary; noise and environmental factors make them statistical rather than strictly deterministic.Therefore, all statements are correct. Verification / Alternative check: Standard texts on spectral libraries and SAR/optical polarimetry emphasize temporal profiles for crops and the diagnostic value of polarization while cautioning users about variability and the need for probabilistic classifiers. Why Other Options Are Wrong: No single statement is wrong; choosing any one would discard other equally valid properties, so the combined option is appropriate. Common Pitfalls: Expecting a unique “fingerprint” under all conditions; ignoring viewing geometry, soil background, and atmosphere; overlooking that signature statistics shift with phenology and moisture. Final Answer: All of these.

Question 16

Satellite Orbits for Monitoring Dynamic Changes For interpolating repeat satellite observations to monitor dynamic surface changes with consistent lighting conditions, which orbit type is most suitable?

Accepted Answer

Introduction / Context: Monitoring dynamic changes (for example, crop growth, snow cover, urban expansion) requires repeated observations that are comparable over time. The illumination geometry should be as consistent as possible between passes to minimize radiometric differences caused by changing solar angles, thereby enabling reliable temporal interpolation or time-series analysis. Given Data / Assumptions: Need repeat coverage at near-constant local solar time. Global to regional monitoring of surfaces. Preference for consistent sun angle to simplify radiometric normalization. Concept / Approach: A sun-synchronous orbit is a special near-polar, low Earth orbit that precesses so that the satellite crosses any latitude at nearly the same local solar time on each pass. This keeps illumination geometry similar across dates, which is ideal for multi-temporal analysis and interpolation of surface parameters. Step-by-Step Solution: Identify requirement: consistent illumination for temporal comparability.Check orbit candidates: equatorial or Molniya orbits do not provide global, uniform local-time revisits suited to land monitoring.Sun-synchronous, near-polar orbits do, by design, maintain constant local solar time.Therefore, choose sun-synchronous orbit. Verification / Alternative check: Most land-observing missions (for example, Landsat, Sentinel-2, Resourcesat) employ sun-synchronous orbits precisely for these reasons. Why Other Options Are Wrong: Circular equatorial orbits restrict coverage; non-sun-synchronous polar orbits vary local time; Molniya orbits target high-latitude dwell but not consistent local-time land imaging; “None” is incorrect since a standard solution exists. Common Pitfalls: Confusing “polar” with “sun-synchronous”—the latter is a designed subset ensuring constant local time. Final Answer: Sun-synchronous orbit

Question 17

Radiometry – Foundational physical quantities Radiometry quantifies electromagnetic radiation for remote sensing. Which set of quantities underpins radiometric measurement and analysis?

Accepted Answer

Introduction / Context: Radiometry provides the physical framework for measuring and modelling electromagnetic power detected by sensors. Its quantities define how energy is counted, how fast it is delivered, how it is distributed in angle, and how it is emitted or reflected per unit area and solid angle. Mastery of these terms is essential for calibration, atmospheric correction, and inversion of surface properties. Given Data / Assumptions: Passive or active sensing scenarios. Standard SI-based radiometric definitions. Idealized surfaces and far-field sensor geometry for clarity. Concept / Approach: The fundamental set comprises: radiant energy Q (total energy), radiant flux phi (time rate of energy), radiant intensity I (flux per unit solid angle), irradiance E and exitance M (flux per unit area incident or leaving), and radiance L (flux per unit projected area per unit solid angle). Together they describe what a detector measures and how surface-sensor geometry maps power into signals. Step-by-Step Solution: Option A: Q is the energy budget relevant to exposures and integration.Option B: phi is power, the time derivative of Q.Option C: I distributes flux directionally per unit solid angle.Option D: L is the most directly sensor-relevant quantity for imaging radiometers, combining area and angle.Hence all listed quantities are foundational in radiometry. Verification / Alternative check: Sensor calibration documents and textbooks define Level-1 products in terms of radiance and convertance, with flux and energy central to instrument design and exposure control. Why Other Options Are Wrong: No single quantity alone is sufficient; the framework needs the complete set to map sources, surfaces, and sensors. Common Pitfalls: Confusing radiance L with irradiance E; mixing photometric (luminous) and radiometric terms; ignoring solid angle in intensity and radiance definitions. Final Answer: All of these

Question 18

Information Systems – What do they do and how are they used? Which statement correctly describes the role and nature of an information system and its outputs for decision-making?

Accepted Answer

Introduction / Context: From enterprise IT to geographic information systems, information systems integrate data, processes, tools, and outputs to support decisions. The ultimate purpose is not data storage alone but actionable insight for planning, monitoring, and control. Given Data / Assumptions: Generic information system encompassing acquisition-to-decision pipeline. Geospatial products (for example, maps) considered as structured outputs. Users require improved decision quality and timeliness. Concept / Approach: An information system carries data through a chain: capture, processing, analysis, storage, retrieval, visualization, and dissemination. Products like maps or dashboards summarize analysed data and become decision artefacts for users across domains such as urban planning, agriculture, or logistics. Therefore, statements about improved decision ability, chain of operations, and maps as stored, analysed information are consistent and complementary. Step-by-Step Solution: Option A: Decision support is the primary objective.Option B: The system is operationally a pipeline from input to output.Option C: Maps encode analysed information suitable for decisions.Thus, “All the above” is correct. Verification / Alternative check: Systems analysis frameworks (for example, input–process–output) and GIS literature both emphasise these roles. Why Other Options Are Wrong: “None of the above” contradicts established definitions. Common Pitfalls: Equating information systems with raw databases; undervaluing analysis and visualization components; ignoring user needs. Final Answer: All the above

Question 19

Scan Geometry Effects in Satellite Imaging Which statements correctly describe geometric and resolution effects as the sensor scans away from nadir?

Accepted Answer

Introduction / Context: Across-track (whiskbroom) and pushbroom imagers both experience geometric effects away from nadir. Understanding how instantaneous field of view (IFOV) projects on the ground and how perspective alters feature shape is crucial for interpreting imagery and for pre-processing steps such as orthorectification. Given Data / Assumptions: Nadir is the point directly beneath the sensor; theta is the off-nadir scan angle. Flat Earth assumption locally for intuition; true processing uses Earth curvature and ephemeris. IFOV is small enough that first-order projections apply. Concept / Approach: As the look angle increases, the ground-projected size of a fixed angular IFOV grows due to obliquity. Along one dimension this scales roughly as sec(theta), and the projected area scales approximately as sec^2(theta). Simultaneously, perspective causes panoramic distortion: features toward the scan edge are geometrically compressed in one direction while stretched in others, and relief displacement increases. Step-by-Step Solution: Option A: Resolution worsening off-nadir is a standard property of scanning imagers.Option B: Projected ground distance or area rises with sec or sec^2(theta); the statement about sec^2(theta) captures the area growth trend.Option C: Edge-of-scan compression (panoramic distortion) is well documented and corrected by geometric pre-processing.Therefore, all three are correct. Verification / Alternative check: Sensor geometry models in orthorectification software implement these relations and remove compressions through resampling using accurate ephemeris and DEMs. Why Other Options Are Wrong: “None” contradicts observed imagery and sensor design notes. Common Pitfalls: Assuming constant GSD across the swath; ignoring angular dependence when comparing edge and centre radiometry; confusing linear dimension scaling with area scaling. Final Answer: All of these

Question 20

Geostationary Orbit – Approximate altitude above Earth A geostationary satellite remains fixed relative to a point on the equator. What is its approximate altitude above the Earth’s surface?

Accepted Answer

Introduction / Context: Geostationary satellites orbit with the same angular velocity as Earth’s rotation, appearing stationary over a longitude. This requires a specific orbital radius and therefore a characteristic altitude used by communication, meteorology, and broadcast satellites. Given Data / Assumptions: Geostationary orbit is circular, equatorial, and prograde. Earth’s sidereal rotation period determines the orbital period (about 23 h 56 min). Standard gravitational parameter of Earth is assumed. Concept / Approach: Setting the orbital period equal to Earth’s sidereal day gives an orbital radius of roughly 42,164 km from Earth’s centre. Subtracting the mean Earth radius (about 6,378 km) yields an altitude near 35,786 km, commonly rounded to 36,000 km. Step-by-Step Solution: Orbital radius r ≈ 42,164 km from Earth’s centre.Earth mean radius R ≈ 6,378 km.Altitude h = r − R ≈ 35,786 km ≈ 36,000 km. Verification / Alternative check: Public ephemerides for GEO satellites and standard references agree on 35,786 km altitude, often rounded for instructional purposes. Why Other Options Are Wrong: 26,000 and 30,000 km are too low to maintain geostationarity; 44,000 km is too high; 18,000 km is MEO territory. Common Pitfalls: Confusing orbital radius with altitude; mixing solar day with sidereal day; forgetting that geosynchronous is not necessarily geostationary unless the orbit is equatorial and zero-eccentricity. Final Answer: 36,000 km

Elements of Remote Sensing Questions