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CIE Colour Systems Questions and Answers

71.

Assertion (A): Black and white have the same chromaticity coordinates.
Reason (R): Chromaticity is normalised for luminance and therefore carries no lightness information.

Answer: C

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the mechanism. Chromaticity coordinates are ratios of the tristimulus values to their sum, so multiplying all three by any factor leaves them unchanged. Black and white differ only in that scaling, so they share a chromaticity - which is precisely why a chromaticity diagram cannot represent a colour on its own.

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72.

Consider the following statements about chromaticity diagrams.
1. They carry no information about luminance or lightness.
2. All neutral samples map to the same single point.
3. They fully specify the appearance of a surface colour.

Answer: D

The first two are correct and identify the limitation. The third is wrong, since lightness is essential to appearance and is entirely absent. Two coordinates cannot describe three dimensions. Lightness is entirely absent from the diagram.

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Answer: A

It divides the chromaticity space into named regions for communication. It is a naming convention rather than a measurement system. It aids communication rather than calculation. Names are assigned to regions of the space.

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Answer: B

The shifts in the test samples are expressed as distances in that space. The choice of space influences the resulting index value. Distances must be measured in a defined space. The space chosen affects the index obtained.

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Answer: C

Correct answer: the Munsell atlas of physical colour samples

The Munsell system is a set of physical samples arranged by perceived hue, value and chroma, developed independently and long before the CIE existed. It is widely used and can be related to CIE coordinates, but it is not a CIE recommendation. The other three are.

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Answer: D

The mapping between XYZ and CIELAB is one to one for a given white. Recovering the spectral reflectance is a quite different and impossible problem. The XYZ to CIELAB map is one to one. Spectral recovery is a quite different matter.

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77.

Consider the following statements about invertibility.
1. CIELAB can be converted back to XYZ given the white point.
2. XYZ cannot be converted back into spectral reflectance.
3. Spectral reflectance can be recovered from CIELAB values.

Answer: A

Correct answer: 1 and 2 only

Statements 1 and 2 are correct: the CIELAB transformation is invertible given the white point, but the tristimulus integration is not - a great many reflectance curves give the same three numbers, which is exactly what metamerism means. Statement 3 follows the same lost information one step further and is equally impossible.

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78.

Assertion (A): Information is lost in going from spectrum to tristimulus values but not from XYZ to CIELAB.
Reason (R): The first reduces many numbers to three while the second is a one to one mapping.

Answer: B

Both are true and the reason gives where the loss occurs, and metamerism follows from the first step. The second step is fully reversible. Only the first discards information. Metamerism follows from that reduction alone.

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Answer: C

Storing the spectrum preserves every future calculation option. Tristimulus values commit irrevocably to one illuminant and one observer. The spectrum preserves every future option. Tristimulus values commit to one condition.

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Answer: D

The construction depends on drawing a line from the white point outward. A different white gives a different intersection with the locus. The construction begins at the white point. A different white gives a different result.

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