CIE Colour Systems Questions and Answers

Answer: A

Correct answer: integrating reflectance, illuminant and matching functions together

\[ X = k\int R(\lambda)S(\lambda)\bar{x}(\lambda)\,d\lambda, \] and similarly for Y and Z. All three factors must be present: what the sample reflects, what the illuminant provides, and how the observer responds. Change any one and the tristimulus values change, which is why the illuminant and observer must always be quoted with a result.

Answer: B

Only Y has a direct interpretation, by deliberate design of the system. The chromaticity coordinates separate the colour quality from the brightness. Only Y was given a direct interpretation. X and Z are mathematical conveniences alone.

Answer: C

Both are true and the reason gives the design choice, and X and Z carry no such meaning. It allows photometric and colorimetric quantities to be read from the same calculation. X and Z carry no perceptual meaning at all.

Answer: D

The normalisation removes the luminance and leaves only the colour quality. The third coordinate z follows since the three must sum to one. Two coordinates therefore suffice to fix a chromaticity. The third adds nothing new at all.

Answer: A

Correct answer: 0.3

The three chromaticity coordinates sum to unity by construction, so \[ z = 1 - x - y = 1 - 0.3 - 0.4 = 0.3. \] This is why only two need ever be quoted, and why the diagram is drawn in two dimensions.

Answer: B

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. Since \( x+y+z=1 \), the third coordinate carries no information the first two do not already contain. That is exactly what makes a two-dimensional chromaticity diagram possible - and it is also why chromaticity alone cannot describe a colour, the luminance having been normalised away.

Answer: C

Every physically realisable colour lies on or inside it. The straight line closing the horseshoe is the purple line of non-spectral colours. Every realisable colour lies on or inside it. Outside the boundary lies nothing physical.

Answer: D

The first two are correct and describe the geometry. The third is wrong, since points outside the boundary do not correspond to any realisable colour whatever. The boundary is an absolute limit, not a saturation scale.

Answer: A

Correct answer: a mixture of all three primaries in fixed proportion

The centroid is the point reached with equal weights on the three vertices, so it represents that particular mixture. It is not necessarily the white point, which depends on the relative luminances of the primaries rather than on the geometry of the triangle, and it is certainly not the most saturated colour.

Answer: B

Both are true and the reason gives the geometry, and the primaries form the vertices. No colour outside the triangle can be matched by those primaries. The vertices are the primaries themselves. Nothing outside can be matched additively.