Whiteness and Yellowness Indices Questions and Answers

Answer: A

The two chromaticity terms are zero and Y is 100, so W is exactly 100. The perfect diffuser is the anchor of the scale, and values above 100 indicate a sample bluer than that reference rather than one reflecting more total light.

Answer: B

Values above 100 arise from the positive chromaticity terms and are entirely normal for whitened samples. They signify a blue shift, not a violation of energy conservation, since Y itself remains at or below one hundred.

Answer: C

Whiteness alone cannot distinguish a greenish white from a reddish one of the same rank. Tint measures displacement perpendicular to the preferred whiteness direction, with negative values greener and positive values redder.

Answer: D

Correct answer: reddish in its cast relative to the preferred white direction

Work it through the tint equation \( T = 900(x_n - x) - 650(y_n - y) \). A reddish sample has a higher \( x \) and a lower \( y \) than the neutral point, so the first bracket is negative and the second positive - and with the minus sign in front of it, both terms come out negative. A greener sample gives the opposite signs and a positive tint. Negative is therefore the red side and positive the green.

Answer: A

Correct answer: 1 and 2 only

Statements 1 and 2 are correct: the equation is defined for D65 only, and it is valid only over a bounded range of whiteness and tint, typically about 40 to 5Y minus 280 with a tint between plus and minus 3. Statement 3 is the opposite of the truth - the formula was fitted to near-white samples, and applied to a strongly coloured one it returns a number that means nothing.

Answer: B

The coefficients were derived by regression against visual judgements of near white samples. Extrapolating beyond the fitted region gives numbers that no longer track perception, which is why the limits are stated with the formula.

Answer: C

All the formulae start from the same tristimulus values but combine them with different coefficients. A Berger value and a CIE value for one sample are simply different numbers, so only ranks within one formula are comparable.

Answer: D

Berger whiteness is of the form Y + aZ - bX. Adding Z rewards blue reflectance while subtracting X penalises the red content that makes a sample look creamy, so the expression tracks the blue shift preferred by observers.

Answer: A

Stensby whiteness is written as L minus a multiple of b plus a multiple of a, using the Hunter opponent coordinates. The negative weight on b rewards blueness and the positive weight on a rewards a slight red rather than green cast.

Answer: B

Correct answer: a negative coefficient so that a bluer sample scores more highly

In opponent coordinates a negative \( b \) is blue and a positive \( b \) yellow, so subtracting a multiple of \( b \) rewards a sample sitting on the blue side of neutral and penalises a yellow one. That is the whole basis of whitening: yellowness is what makes a substrate look unwhite, and a small blue shift is read by the eye as whiter rather than as blue.