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ITC Level II practice questions

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Question 1Thermal physics and heat transfer fundamentals

Convert 27 °C to absolute temperature in kelvin.

  1. 327 K
  2. 246 K
  3. 273 K
  4. 300 K Correct answer.

Explanation

Kelvin temperature is obtained by adding 273.15 to the Celsius value. Adding 273 to 27 gives 300 K. The offset is added, not subtracted or multiplied.

Question 2Thermal physics and heat transfer fundamentals

Which quantity is defined as the energy required to raise the temperature of 1 kg of a substance by 1 K?

  1. Specific heat capacity Correct answer.
  2. Thermal conductivity
  3. Thermal diffusivity
  4. Latent heat

Explanation

Specific heat capacity is energy per unit mass per unit temperature rise, expressed in J/(kg·K). Latent heat instead describes energy absorbed during a phase change at constant temperature. Thermal conductivity and diffusivity describe how heat spreads, not how much energy is stored per kelvin.

Question 3Thermal physics and heat transfer fundamentals

Which mode of heat transfer can proceed through a perfect vacuum?

  1. Conduction
  2. Convection
  3. Diffusion
  4. Radiation Correct answer.

Explanation

Radiation carries energy as electromagnetic waves, which need no material medium. Conduction and convection both require matter to transport energy. Diffusion is mass transport rather than a distinct heat transfer mode.

Question 4Thermal physics and heat transfer fundamentals

The latent heat of fusion of a substance is the energy needed to change 1 kg of it from

  1. liquid to solid with a temperature rise
  2. liquid to vapour at constant temperature
  3. solid to vapour directly
  4. solid to liquid at constant temperature Correct answer.

Explanation

Fusion means melting, so the latent heat of fusion converts solid to liquid at the melting temperature. The liquid-to-vapour change is vaporisation, and the direct solid-to-vapour change is sublimation. Latent heat changes phase without changing temperature.

Question 5Thermal physics and heat transfer fundamentals

Two bodies are in thermal equilibrium when they

  1. are both at standard pressure
  2. exchange no net heat and share the same temperature Correct answer.
  3. have the same mass and volume
  4. have the same heat capacity

Explanation

Thermal equilibrium means no net heat flows between the bodies, which requires equal temperatures. Mass, volume, heat capacity and pressure do not by themselves prevent heat flow. Equilibrium is defined by temperature equality, not by equal energy content.

Question 6Thermal physics and heat transfer fundamentals

The SI unit of thermal conductivity is

  1. W/(m·K) Correct answer.
  2. J/(kg·K)
  3. W/(m²·K)
  4. W/K

Explanation

Fourier's law gives heat flux as conductivity times temperature gradient, so the unit is W/(m·K). Dividing W/m² by K/m yields W/(m·K). The other units describe convection coefficients, specific heat or thermal conductance.

Question 7Thermal physics and heat transfer fundamentals

For a black body at absolute temperature T, the total radiated power per unit area is proportional to

  1. exp(T)
  2. T⁴ Correct answer.
  3. T
  4. T²

Explanation

The Stefan-Boltzmann law states that emissive power equals σT⁴ for a black body. The fourth-power dependence makes radiation dominate at high temperature. A real surface multiplies this by its emissivity.

Question 8Thermal physics and heat transfer fundamentals

Convective heat transfer is primarily associated with

  1. bulk movement of a fluid carrying energy Correct answer.
  2. molecular vibration with no bulk motion
  3. electromagnetic wave emission from a surface
  4. a phase change at a solid surface

Explanation

Convection combines fluid motion with heat transport, so bulk movement of the fluid is essential. Pure molecular vibration without bulk motion belongs to conduction. Radiation is the mode that involves electromagnetic waves.

Question 9Thermal physics and heat transfer fundamentals

Equal masses of water at 80 °C and 20 °C are mixed in an insulated container. Neglecting losses, the final temperature is

  1. 50 °C Correct answer.
  2. 60 °C
  3. 40 °C
  4. 100 °C

Explanation

With equal masses of the same liquid, the heat lost by the hot water equals the heat gained by the cold water. The final temperature is therefore the arithmetic mean, (80 + 20)/2 = 50 °C. Insulation ensures no energy leaves the system.

Question 10Thermal physics and heat transfer fundamentals

A steel rod 2 m long at 20 °C is heated to 120 °C. Using a linear expansion coefficient of 12 × 10⁻⁶ K⁻¹, its increase in length is about

  1. 2.4 mm Correct answer.
  2. 0.24 mm
  3. 1.2 mm
  4. 4.8 mm

Explanation

Linear expansion follows ΔL = L α ΔT. Here ΔL = 2 × 12 × 10⁻⁶ × 100 = 2.4 × 10⁻³ m, which is 2.4 mm. The temperature change is 100 K, not 120 K.

Question 11Thermal physics and heat transfer fundamentals

Two plane slabs are in series, with thermal resistances 0.5 K/W and 1.5 K/W. The total thermal resistance is

  1. 3.0 K/W
  2. 1.0 K/W
  3. 2.0 K/W Correct answer.
  4. 0.33 K/W

Explanation

Thermal resistances in series add directly, so R_total = 0.5 + 1.5 = 2.0 K/W. The heat flow is the overall temperature difference divided by this total resistance. Resistances in parallel would instead combine reciprocally.

Question 12Thermal physics and heat transfer fundamentals

Newton's law of cooling states that the rate of heat loss from a body is proportional to

  1. the inverse of the temperature difference
  2. the body's absolute temperature
  3. the square of the temperature difference
  4. the body-to-surroundings difference Correct answer.

Explanation

Newton's law of cooling uses a linear driving force equal to the difference between body and ambient temperatures. It is an approximation valid for modest temperature differences with convection and radiation combined. The proportionality constant depends on surface area and the heat transfer coefficient.

Question 13Thermal physics and heat transfer fundamentals

Wien's displacement law relates the peak emission wavelength of a black body to its absolute temperature as

  1. λmax² T = constant
  2. λmax / T = constant
  3. λmax T² = constant
  4. λmax T = constant Correct answer.

Explanation

Wien's law states that the peak wavelength multiplied by absolute temperature is a constant of about 2.9 × 10⁻³ m·K. Peak wavelength therefore falls as temperature rises. This is why hotter objects glow at shorter wavelengths.

Question 14Thermal physics and heat transfer fundamentals

When ice at 0 °C melts, the energy supplied is used mainly to

  1. increase the pressure of the surroundings
  2. increase the kinetic energy of the molecules
  3. break bonds so the phase can change Correct answer.
  4. raise the temperature of the ice

Explanation

During melting the temperature stays at 0 °C while energy goes into overcoming the bonding that holds the solid structure together. This is latent heat, not sensible heat. Molecular kinetic energy, and hence temperature, does not rise until all the ice has melted.

Question 15Thermal physics and heat transfer fundamentals

Kirchhoff's law for a surface in thermal equilibrium implies that, at a given wavelength and temperature,

  1. emissivity is unrelated to absorptivity
  2. emissivity equals absorptivity Correct answer.
  3. emissivity is always zero
  4. absorptivity is always one

Explanation

Kirchhoff's law states that good absorbers are good emitters, so spectral emissivity equals spectral absorptivity at the same condition. A black body has both equal to one. This equality is what makes radiation exchange calculations consistent.

Question 16Thermal physics and heat transfer fundamentals

A wall has series resistances per unit area of 0.2 (inner convection), 0.6 (conduction) and 0.2 K·m²/W (outer convection). Its overall heat transfer coefficient is

  1. 2.0 W/(m²·K)
  2. 1.0 W/(m²·K) Correct answer.
  3. 0.5 W/(m²·K)
  4. 0.2 W/(m²·K)

Explanation

The total resistance per unit area is 0.2 + 0.6 + 0.2 = 1.0 K·m²/W. The overall coefficient is the reciprocal, U = 1/1.0 = 1.0 W/(m²·K). Adding resistances before inverting is essential when convection and conduction act in series.

Question 17Thermal physics and heat transfer fundamentals

The heat capacity of a body is best described as

  1. the energy needed to raise the whole body by 1 K Correct answer.
  2. the energy passing per unit area per kelvin
  3. the energy needed to melt the body completely
  4. the energy needed to raise 1 kg of it by 1 K

Explanation

Heat capacity is an extensive property: the energy required for a 1 K rise of the entire body. It equals mass times specific heat for a uniform material. The per-kilogram quantity is specific heat capacity, not heat capacity.

Question 18Thermal physics and heat transfer fundamentals

Thermal diffusivity is given by the expression

  1. ρc/k
  2. kρc
  3. c/(ρk)
  4. k/(ρc) Correct answer.

Explanation

Thermal diffusivity is conductivity divided by volumetric heat capacity, α = k/(ρc). It measures how quickly a temperature disturbance spreads through a material. High conductivity or low heat capacity per unit volume gives faster diffusion.

Question 19Thermal physics and heat transfer fundamentals

For a pure substance at constant pressure, the temperature during a phase change

  1. falls steadily
  2. rises steadily
  3. stays constant Correct answer.
  4. oscillates around a mean

Explanation

At constant pressure a pure substance changes phase at a fixed temperature while latent heat is absorbed or released. The added energy changes the phase rather than the molecular kinetic energy. The temperature only changes once the phase change is complete.

Question 20Thermal physics and heat transfer fundamentals

A small surface of area A, emissivity ε and absolute temperature T exchanges radiation with a large enclosure at absolute temperature Ts. The net radiative heat loss is

  1. εσA(T² − Ts²)
  2. εσA(T⁴ − Ts⁴) Correct answer.
  3. εσA(T − Ts)
  4. σA(T⁴ + Ts⁴)

Explanation

Net radiation from a small body to a large enclosure is the difference between emitted and absorbed power, εσA(T⁴ − Ts⁴). The fourth-power terms come from the Stefan-Boltzmann law. A sum of the two powers would not represent a net exchange.

Question 21Thermal physics and heat transfer fundamentals

The lumped capacitance method for transient cooling of a solid is valid when the

  1. Fourier number is below 0.1
  2. Biot number is below about 0.1 Correct answer.
  3. Reynolds number is below 2000
  4. Biot number is above about 10

Explanation

A small Biot number means internal conduction resistance is negligible compared with surface convection resistance. The body can then be treated as having a single uniform temperature. Above the usual threshold of about 0.1, internal temperature gradients must be resolved.

Question 22Thermal physics and heat transfer fundamentals

For steady radial conduction through a cylindrical shell of length L, inner radius r1, outer radius r2 and conductivity k, the thermal resistance is

  1. ln(r2/r1)/(2πkL) Correct answer.
  2. (r2 − r1)/(kA)
  3. ln(r1/r2)/(2πkL)
  4. 2πkL/ln(r2/r1)

Explanation

Integrating Fourier's law in cylindrical coordinates gives a resistance containing the logarithm of the radius ratio divided by 2πkL. The area through which heat flows increases with radius, which produces the logarithmic form. The planar formula using constant area is only an approximation for thin walls.

Question 23Thermal physics and heat transfer fundamentals

The efficiency of a cooling fin is defined as the

  1. fin conductivity divided by the convection coefficient
  2. fin length divided by the fin thickness
  3. actual heat transfer divided by the ideal isothermal-fin value Correct answer.
  4. fin surface area divided by the base area

Explanation

Fin efficiency compares the heat actually removed by the fin with the heat that would be removed if the whole fin were at the base temperature. It is always below one because the fin temperature falls towards its tip. The comparison uses the same convection coefficient and base temperature in both cases.

Question 24Thermal physics and heat transfer fundamentals

The log mean temperature difference is used in heat exchanger analysis because

  1. the stream-to-stream difference varies along the unit Correct answer.
  2. the flow is always fully turbulent
  3. the overall heat transfer coefficient varies along the exchanger
  4. the two fluids always have equal heat capacities

Explanation

The local temperature difference between hot and cold streams changes from one end of the exchanger to the other. Integrating the local heat transfer over the whole area produces a logarithmic mean of the terminal differences. Using a simple arithmetic mean would misstate the duty whenever the terminal differences differ significantly.

Question 25Thermal physics and heat transfer fundamentals

A sunlit surface reaches a steady temperature when the

  1. surface emissivity has fallen to zero
  2. absorbed solar flux plus convective gain equals emitted radiation Correct answer.
  3. convective heat transfer at the surface is zero
  4. absorbed solar flux equals emitted radiation alone

Explanation

At steady state the energy entering the surface must balance the energy leaving it. Solar absorption and any convective gain from warmer air are balanced by the emitted radiation. Omitting convection would overpredict the surface temperature whenever the air is warmer than the surface.

Question 26Thermal physics and heat transfer fundamentals

In Fourier's law of heat conduction, the negative sign indicates that heat flows

  1. only when the body is at a uniform temperature
  2. at a rate that is independent of the temperature difference
  3. from the colder region to the hotter region
  4. in the direction of decreasing temperature Correct answer.

Explanation

Fourier's law is written q'' = −k dT/dx, where the gradient is measured in the direction of increasing temperature. Energy therefore moves in the opposite direction to the gradient, that is, from hot towards cold. The magnitude of the heat flux always depends on the conductivity and on the steepness of the gradient.

Question 27Thermal physics and heat transfer fundamentals

Which mode of heat transfer can occur between two bodies at different temperatures separated by a perfect vacuum?

  1. Forced convection
  2. Thermal radiation Correct answer.
  3. Natural convection
  4. Conduction through the intervening space

Explanation

Thermal radiation is carried by electromagnetic waves, which propagate without any material medium. Conduction needs direct molecular contact or a conducting medium, and both natural and forced convection require a moving fluid.

Question 28Thermal physics and heat transfer fundamentals

A temperature difference of 25 K corresponds to a temperature difference of

  1. 298 °C
  2. 25 °C Correct answer.
  3. 0 °C
  4. 25 °C multiplied by 273

Explanation

The kelvin and the degree Celsius have exactly the same size, because the two scales differ only by an offset of 273.15. Differences in temperature are therefore numerically identical on both scales. Only absolute values, not intervals, require adding or subtracting the offset.

Question 29Thermal physics and heat transfer fundamentals

A solid is said to be at steady state when

  1. its temperature is uniform throughout at all times
  2. its internal energy is decreasing at a constant rate
  3. the temperature at every point is constant in time Correct answer.
  4. no heat crosses its boundaries

Explanation

Steady state means the time derivative of temperature vanishes at every location, so a measurement at a fixed point would not change. Heat may still flow steadily through the body and the temperature may vary from point to point. Uniform temperature is a stronger and separate condition.

Question 30Thermal physics and heat transfer fundamentals

The Nusselt number for convective heat transfer represents

  1. the ratio of convective to conductive heat transfer Correct answer.
  2. the ratio of buoyancy to viscous forces
  3. the ratio of inertial to viscous forces in the flow
  4. the ratio of momentum diffusivity to thermal diffusivity

Explanation

The Nusselt number is defined as Nu = hL/k and measures how much the convective heat transfer coefficient exceeds pure conduction across the same fluid layer. A value near unity indicates conduction-dominated transport, while large values indicate strong convection.

Question 31Thermal physics and heat transfer fundamentals

The Prandtl number of a fluid is the ratio of

  1. inertial forces to viscous forces
  2. buoyancy forces to viscous forces
  3. convective to conductive heat transfer
  4. momentum diffusivity to thermal diffusivity Correct answer.

Explanation

The Prandtl number is Pr = ν/α = μcp/k, comparing how quickly momentum and heat diffuse through the fluid. It therefore controls the relative thickness of the velocity and thermal boundary layers. Liquid metals have very small Prandtl numbers, while oils have large ones.

Question 32Thermal physics and heat transfer fundamentals

In free convection, the Grashof number is important because it measures

  1. the ratio of momentum diffusivity to thermal diffusivity
  2. the relative roughness of the heat transfer surface
  3. the ratio of heat convected to heat conducted
  4. the ratio of buoyancy forces to viscous forces Correct answer.

Explanation

The Grashof number, Gr = gβΔT L³/ν², compares the buoyant force produced by density differences with the viscous forces resisting motion. It plays the role in natural convection that the Reynolds number plays in forced convection. Large Grashof numbers indicate vigorous buoyant motion and hence stronger heat transfer.

Question 33Thermal physics and heat transfer fundamentals

Two large parallel black surfaces are held at 500 K and 300 K. Taking σ = 5.67 × 10⁻⁸ W/(m²·K⁴), the net radiant heat flux exchanged between them is closest to

  1. 4.6 kW/m²
  2. 1.3 kW/m²
  3. 0.9 kW/m²
  4. 3.1 kW/m² Correct answer.

Explanation

For infinite parallel black plates the view factor is unity, so the net flux is q'' = σ(T₁⁴ − T₂⁴). Substituting gives 5.67 × 10⁻⁸ × (6.25 × 10¹⁰ − 8.1 × 10⁹) W/m², which is about 3.1 kW/m².

Question 34Thermal physics and heat transfer fundamentals

A small object is placed inside a large closed isothermal enclosure. The view factor from the object to the enclosure is

  1. 0
  2. 1 Correct answer.
  3. the ratio of the object's area to the enclosure's area
  4. equal to the enclosure's emissivity

Explanation

By the summation rule, all radiation leaving a surface must reach the surfaces of the enclosing space, so the view factor from a convex body to a complete enclosure is unity. Reciprocity then gives the view factor from the large enclosure to the small body as the area ratio A_object/A_enclosure.

Question 35Thermal physics and heat transfer fundamentals

A pipe of outside radius 10 mm is covered with insulation of conductivity 0.15 W/(m·K) and the outer surface has a convection coefficient of 10 W/(m²·K). The critical radius of insulation is

  1. 67 mm
  2. 15 mm Correct answer.
  3. 1.5 mm
  4. 150 mm

Explanation

For a cylindrical layer the critical radius is r_cr = k/h, giving 0.15/10 = 0.015 m, or 15 mm. Because the pipe radius of 10 mm is below this value, adding a thin layer of insulation initially increases the heat loss before the resistance of the layer dominates.

Question 36Thermal physics and heat transfer fundamentals

On the pool boiling curve for water at atmospheric pressure, the maximum heat flux (the critical or burnout heat flux) marks the end of

  1. transition boiling
  2. nucleate boiling Correct answer.
  3. free convection boiling
  4. film boiling

Explanation

The heat flux rises steeply through the nucleate boiling regime as bubbles agitate the liquid, and reaches its maximum when the surface can no longer be wetted reliably by liquid. Beyond that point the surface becomes partly covered by vapour and the flux falls through transition boiling. Film boiling occurs at still higher wall temperatures with a much lower heat flux.

Question 37Thermal physics and heat transfer fundamentals

How much energy is needed to raise the temperature of 2 kg of water from 20 °C to 80 °C? Take the specific heat capacity of water as 4.18 kJ/(kg·K).

  1. 84 kJ
  2. 1004 kJ
  3. 502 kJ Correct answer.
  4. 250 kJ

Explanation

For a single phase with constant specific heat, the energy required is Q = mcΔT. Here ΔT = 60 K, so Q = 2 × 4.18 × 60 ≈ 502 kJ. The kelvin and Celsius intervals are the same size, so the difference may be used directly.

Question 38Thermal physics and heat transfer fundamentals

Two large parallel black plates at fixed temperatures are separated by two thin, opaque, black radiation shields. Compared with the unshielded arrangement, the net radiant heat flux is reduced to about

  1. one quarter
  2. the same value
  3. one half
  4. one third Correct answer.

Explanation

Each shield introduces an additional radiation resistance, so the heat flux falls to 1/(N + 1) of the unshielded value for black surfaces and N shields. With two shields the flux is therefore about one third. This is exploited in multilayer superinsulation for cryogenic vessels.

Question 39Thermal physics and heat transfer fundamentals

A steel sphere 10 mm in diameter (ρ = 8000 kg/m³, c = 500 J/(kg·K)) at 100 °C is quenched in a fluid at 20 °C with h = 100 W/(m²·K). Treating the sphere as spatially uniform, the time for its temperature to fall to 60 °C is closest to

  1. 23 s
  2. 46 s Correct answer.
  3. 133 s
  4. 67 s

Explanation

For a lumped body, (T − T∞)/(Ti − T∞) = exp(−t/τ) with time constant τ = ρVc/(hA) = ρcD/(6h) ≈ 67 s. Setting the ratio to (60 − 20)/(100 − 20) = 0.5 gives t = τ ln 2 ≈ 46 s. The small diameter keeps the Biot number low, so the uniform-temperature assumption is reasonable.

Question 40Thermal physics and heat transfer fundamentals

A condenser in which one stream condenses at essentially constant temperature operates with NTU = 2.0. Its effectiveness is closest to

  1. 0.50
  2. 0.63
  3. 0.86 Correct answer.
  4. 1.00

Explanation

When one stream changes phase, its capacity rate is effectively infinite, so Cmin/Cmax = 0 and the effectiveness relation reduces to ε = 1 − exp(−NTU). With NTU = 2.0 this gives 1 − exp(−2) ≈ 0.86. The condenser can never quite reach the entering temperature of the hot stream in a finite area.

Question 41Infrared equipment, camera setup and measurement techniques

When a thermographer sets the camera emissivity higher than the true emissivity of a target that is warmer than the reflected background, the displayed temperature will generally be:

  1. the same as the true temperature
  2. independent of the emissivity setting
  3. higher than the true temperature
  4. lower than the true temperature Correct answer.

Explanation

The camera calculates target temperature from received radiance using the entered emissivity. If the entered emissivity is too high, the camera treats more of the received signal as target emission and less as reflected background, so it computes a lower target temperature. Emissivity error therefore biases the reading in a predictable direction.

Question 42Infrared equipment, camera setup and measurement techniques

The instantaneous field of view (IFOV) of a thermal camera is best described as:

  1. The temperature difference represented by one palette colour
  2. The total angular width of the detector array
  3. The minimum distance at which the lens can focus
  4. The angular size of one detector pixel Correct answer.

Explanation

IFOV is the angular projection of a single detector element through the lens. It is used with distance to determine the spot size on the target. It is not the full field of view or a focus limit.

Question 43Infrared equipment, camera setup and measurement techniques

Which camera input compensates for infrared radiation emitted by the surroundings and reflected from a low-emissivity target?

  1. Object distance
  2. Reflected temperature Correct answer.
  3. Atmospheric transmittance
  4. Relative humidity

Explanation

Reflected temperature represents the apparent temperature of the radiation reflected from the target and its surroundings. It is especially important for low-emissivity surfaces, where reflected radiation forms a large part of the received signal. Humidity, distance, and transmittance correct other effects.

Question 44Infrared equipment, camera setup and measurement techniques

Before recording a temperature measurement, the most important camera adjustment is to:

  1. Increase the frame rate
  2. Focus the target sharply Correct answer.
  3. Set the palette to rainbow
  4. Activate digital zoom

Explanation

An out-of-focus image spreads target radiation across pixels and corrupts quantitative readings. Accurate focus must be established before spot, box, or isotherm measurements are trusted. Palette and zoom are display choices and do not improve measurement accuracy.

Question 45Infrared equipment, camera setup and measurement techniques

To reduce the effect of solar reflection when inspecting a building facade outdoors on a clear day, the thermographer should:

  1. Move closer until the spot fills the display
  2. Set emissivity to 1.0
  3. Change the viewing angle or timing Correct answer.
  4. Switch to a narrower temperature range

Explanation

Solar reflection can dominate the apparent temperature of a facade, and its strength changes with viewing geometry and time of day. Surveying from a different angle or when the surface is not directly sunlit reduces this reflected component. Emissivity and range settings do not remove reflected solar radiation.

Question 46Infrared equipment, camera setup and measurement techniques

In thermal camera terminology, NETD is:

  1. The number of pixels in the detector array
  2. The time required for the detector to stabilize
  3. The smallest resolvable temperature difference Correct answer.
  4. The maximum temperature the detector can measure

Explanation

NETD describes the noise-equivalent temperature difference, the smallest temperature difference the camera can distinguish from noise. A lower NETD indicates better thermal sensitivity. It is not a maximum range, pixel count, or warm-up time.

Question 47Infrared equipment, camera setup and measurement techniques

If the distance from the camera to a target is doubled while the lens and detector are unchanged, the diameter of the spot projected by one detector element on the target will:

  1. Double Correct answer.
  2. Quadruple
  3. Halve
  4. Remain the same

Explanation

Spot size is approximately IFOV multiplied by distance. Because IFOV is fixed for a given lens and detector, doubling the distance doubles the projected spot diameter. The effect is linear with distance.

Question 48Infrared equipment, camera setup and measurement techniques

The temperature span control on a thermal camera determines:

  1. The frame rate of the detector
  2. The distance entered for atmospheric correction
  3. The temperature range mapped across the palette Correct answer.
  4. The emissivity assigned to the target

Explanation

Span sets the width of the temperature interval that the palette represents, while level sets its centre. Together they control how temperature differences appear on the display. Span is a display and analysis setting, not an emissivity or distance input.

Question 49Infrared equipment, camera setup and measurement techniques

A measurement is made with a fixed-focus lens at a distance closer than its minimum focus distance. The most likely result is:

  1. A higher apparent emissivity
  2. A wider field of view and a larger spot
  3. Blurring and unreliable readings Correct answer.
  4. An increase in NETD

Explanation

Each lens has a minimum focus distance, and using it closer than that limit prevents a sharp image. Blur spreads radiation between pixels, so spot and area measurements become unreliable. The field of view and emissivity are not corrected by being too close.

Question 50Infrared equipment, camera setup and measurement techniques

A target at 40 m has a diameter of 0.2 m. Using a camera with an IFOV of 1.0 mrad, approximately how many detector pixels span the target diameter?

  1. 20 pixels
  2. 5 pixels Correct answer.
  3. 2 pixels
  4. 0.5 pixel

Explanation

At 40 m, an IFOV of 1.0 mrad projects a spot of about 0.04 m. Dividing the 0.2 m target diameter by 0.04 m gives 5 pixels across the target. This calculation shows whether the target is large enough for a reliable spot measurement.