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CIE Colour Systems Questions and Answers

Answer: A

Correct answer: the Euclidean distance between the two points

\[ \Delta E^*_{ab} = \sqrt{(\Delta L^*)^2 + (\Delta a^*)^2 + (\Delta b^*)^2}. \] The straight-line distance is meaningful only because the space was designed to be approximately uniform - and the word approximately is what has since given rise to the weighted formulae.

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42.

Assertion (A): Delta E ab is no longer the preferred colour difference measure.
Reason (R): CIELAB is not uniform enough for equal distances to mean equal differences.

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. CIELAB is only roughly uniform, so a distance of one unit is plainly visible near the neutral axis and barely visible at high chroma. CMC, CIE94 and CIEDE2000 all correct for that by weighting the components, and it is those formulae that specifications now cite.

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Answer: C

Correct answer: equal distances correspond only roughly to equal perceived differences

The qualifier is the substance of the claim. CIELAB was a great improvement on the xy diagram, whose ellipses vary by a factor of twenty, but the residual non-uniformity is still large enough to matter commercially - which is exactly why the weighted difference formulae exist.

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44.

Consider the following statements about CIELAB uniformity.
1. It is far more uniform than the xy chromaticity diagram.
2. It remains non-uniform enough to require weighted difference formulae.
3. It is perfectly uniform throughout the colour space.

Answer: D

The first two are correct and describe the position honestly. The third overstates it, and the existence of CIEDE2000 is itself the evidence. Perfect uniformity would need no correction. The weighting formulae are themselves the evidence.

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Answer: A

The cube root style function is applied after dividing by the white. This non-linearity is what makes the space approximately uniform. A linear matrix alone would not suffice. Non-linearity is what produces the uniformity.

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46.

Arrange the steps of computing CIELAB from spectral data in order.
1. divide by the tristimulus values of the white point
2. multiply reflectance by illuminant and matching functions
3. apply the non-linear function to obtain L star, a star and b star
4. sum the products to obtain X, Y and Z

Answer: B

Each step requires the outputs of the one before it. The illuminant and observer enter at the first step and the white point at the third. Each step consumes the output of the last. Illuminant, observer and white all enter in turn.

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Answer: C

The transformation is identical but its inputs are not. Reporting CIELAB values without naming the observer is therefore incomplete. The transform is identical but its inputs differ. Observer must accompany any quoted value.

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Answer: D

Correct answer: the name of the colourant in the sample

A CIELAB value is computed from the reflectance by weighting with an illuminant and an observer and normalising to a white point, so all three of those must be stated for the figure to be reproducible. What the sample is made of does not enter the calculation at all.

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Answer: A

Correct answer: zero for a perfect black to one hundred for the white point

The scale is anchored at both ends: zero for a perfect black and one hundred for the reference white. Values slightly above a hundred are possible for a sample brighter than the reference, such as one carrying a fluorescent whitening agent, but the nominal range is nought to a hundred.

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50.

State whether the following is true or false.
A CIELAB lightness above one hundred is possible for a fluorescent sample.

Answer: B

Such a sample returns more light than the reference white does. The scale is anchored at the white rather than at a physical maximum. Fluorescent samples routinely exceed one hundred. The white is the anchor, not a ceiling.

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