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• 1

COMP 546

Lecture 12

Illumination and Reflectance

Tues. Feb. 20, 2018

• Illumination and Reflectance

• Brightness versus Lightness

• Color constancy

• N(x)

L 

𝑁

𝐿

Lambert’s (cosine) Law:

𝐼 𝑋 = 𝑁(𝑋) ∙ 𝐿

• Unit Surface Normal

1

𝜕𝑍 𝜕𝑋

2

+ 𝜕𝑍 𝜕𝑌

2

+ 1

𝜕𝑍

𝜕𝑋 , 𝜕𝑍

𝜕𝑌 ,−1𝑁 ≡

𝑁

4

• Shading on a sunny day

5

cast : 𝑁(𝑋) ∙ 𝐿 > 0 but light is occluded

attached : 𝑁(𝑋) ∙ 𝐿 < 0

𝐿

• sunny day (last lecture)

• sunny day + low relief

• cloudy day

• Examples of low relief surfaces

(un)crumpled paper

• Low relief surface

1

𝜕𝑍 𝜕𝑋

2

+ 𝜕𝑍 𝜕𝑌

2

+ 1

𝜕𝑍

𝜕𝑋 , 𝜕𝑍

𝜕𝑌 ,−1𝑁 ≡

9

𝜕𝑍

𝜕𝑋 ≈ 0

≈0 ≈0

𝜕𝑍

𝜕𝑌 ≈ 0

Thus,

Suppose and are both small.

• Linear shading model for low relief

𝐼 𝑋, 𝑌 ≈ 𝜕𝑍

𝜕𝑋 , 𝜕𝑍

𝜕𝑌 ,−1 ∙ 𝐿𝑋 , 𝐿𝑌 , 𝐿𝑍

10

• one equation per point but two unknowns

• Example: curtains

𝑍 𝑋, 𝑌 = 𝑍0 + 𝑎 sin( 𝑘𝑋 𝑋 )

𝐿 𝑁

𝑋

𝑍

• Example: curtains

𝑍 𝑋, 𝑌 = 𝑍0 + 𝑎 sin( 𝑘𝑋 𝑋 )

𝜕𝑍

𝜕𝑋 = 𝑎 𝑘𝑋 cos 𝑘𝑋 𝑋

𝜕𝑍

𝜕𝑌 = 0,

𝑋

𝑍

• Example: curtains

𝐼 𝑋, 𝑌 ≈ 𝜕𝑍

𝜕𝑋 , 𝜕𝑍

𝜕𝑌 , −1 ∙ 𝐿𝑋 , 𝐿𝑌 , 𝐿𝑍

𝑍 𝑋, 𝑌 = 𝑍0 + 𝑎 sin( 𝑘𝑋 𝑋 )

𝜕𝑍

𝜕𝑋 = 𝑎 𝑘𝑋 cos 𝑘𝑋 𝑋

𝐼 𝑋, 𝑌 = 𝑎 𝑘𝑋 cos 𝑘𝑋 𝑋 𝐿𝑋 − 𝐿𝑍

𝜕𝑍

𝜕𝑌 = 0,

• Example: curtains

𝑍 𝑋, 𝑌 = 𝑍0 + 𝑎 sin( 𝑘𝑋 𝑋 )

𝐼 𝑋, 𝑌 = − 𝐿𝑍 + 𝑎 𝑘𝑋 𝐿𝑋 cos 𝑘𝑋 𝑋

𝐿 = (𝐿𝑋, 𝐿𝑌, 𝐿𝑍)

𝑋

𝑍

𝑁

Q: where do the intensity maxima and minima occur?

• Shading on a cloudy day (my Ph.D. thesis)

15

•  (x)

• Shading on a Cloudy Day

17

• Shading on a Sunny Day

Shading determined by surface normal only.

19

Q: What is the task ? What problem is being solved?

A: Estimate surface slant, tilt, curvature.

How to account for (or estimate) the lighting ?

• Illumination and Reflectance

[Shading and shadowing models assume that intensity variations on a surface are entirely due to illumination. But surfaces have reflectance variations too. ]

• Brightness versus Lightness

• Color constancy

• Which paper is lighter ?

21

Paper on the left is in shadow. It has lower physical intensity and it appears darker. But do both papers seem to be of same (white) material?

• Which paper is lighter ?

22

Image is processed so that the right paper is given same image intensities as left paper. Now, right paper appears to be made of different material. Why?

• 𝐼 𝑥, 𝑦 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑥, 𝑦 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒 (𝑥, 𝑦)

Abstract version

“Real” example

• 𝐼 𝑥, 𝑦 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑥, 𝑦 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒 (𝑥, 𝑦)

luminance

Physical quantities

• low reflectance, high illumination

low reflectance, low illumination

high reflectance, low illumination

𝐼 𝑥, 𝑦 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑥, 𝑦 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒 (𝑥, 𝑦)

• 𝐼 𝑥, 𝑦 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑥, 𝑦 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒 (𝑥, 𝑦)

“lightness”

“brightness”

Perceptual quantities

(no standard term for perceived illumination)

• All four indicated “squares” have same intensity.

What is the key difference between the two configurations ?

• “Lightness” perception

Q: What is the task ?

What problem is being solved?

A:

• “Lightness” perception

Q: What is the task ?

What problem is being solved?

A: Estimate the surface reflectance, by

discounting the illumination.

𝐼 𝑥, 𝑦 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑥, 𝑦 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒 (𝑥, 𝑦)

?

• “Lightness” perception: solution sketch

Q: What is the task ?

What problem is being solved?

A: Estimate the surface reflectance, by

discounting illumination effects.

Compare points that have same illumination.

𝐼 𝑥1, 𝑦1 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒 (𝑥1, 𝑦1)

𝐼 𝑥2, 𝑦2 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒 (𝑥2, 𝑦2) =

• Illumination and Reflectance

• Brightness versus Lightness

• Color constancy

• Recall lecture 3 - color

32

Illumination

Surface Reflectance (fraction)

Absorption by photoreceptors (fraction)

There are three different spectra here.

• LMS cone responses

𝐼 𝑥, 𝑦, 𝜆 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑥, 𝑦, 𝜆 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒 𝑥, 𝑦, 𝜆

Cone response

𝐼 𝑥, 𝑦, 𝜆 𝐶𝐿𝑀𝑆(𝜆) 𝑑𝜆

• Surface Color Perception

34

Q: What is the task? What is the problem to be solved?

A:

illumination

Surface Reflectance

• Surface Color Perception

35

Q: What is the task? What is the problem to be solved?

A: Estimate the surface reflectance, by discounting the illumination.

𝐼 𝑥, 𝑦, 𝜆 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑥, 𝑦, 𝜆 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒 𝑥, 𝑦, 𝜆

illumination

Surface Reflectance

• 𝐼𝑅𝐺𝐵 𝑥, 𝑦 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑅𝐺𝐵 𝑥, 𝑦 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒𝑅𝐺𝐵 (𝑥, 𝑦)

For simplicity, let’s ignore the continuous wavelength 𝜆 and just consider 𝑅𝐺𝐵 (LMS).

𝜆

Cone response

𝐼 𝑥, 𝑦, 𝜆 𝐶𝐿𝑀𝑆(𝜆) 𝑑𝜆

• “Color Constancy”

37

Task: estimate the surface reflectance, by discounting the illumination

𝐼𝑅𝐺𝐵 𝑥, 𝑦 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑅𝐺𝐵 𝑥, 𝑦 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒𝑅𝐺𝐵 (𝑥, 𝑦)

illumination

Surface Reflectance

• Why we need color constancy

• object recognition

• skin evaluation (health, emotion, …)

• food quality

• …

• Example 1: spatially uniform illumination

= *

𝐼𝑅𝐺𝐵 𝑥, 𝑦 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑅𝐺𝐵 𝑥, 𝑦 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒𝑅𝐺𝐵 (𝑥, 𝑦)

Given this, how to estimate this?

= *

𝐼𝑅𝐺𝐵 𝑥, 𝑦 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑅𝐺𝐵 𝑥, 𝑦 ∗ 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑎𝑛𝑐𝑒𝑅𝐺𝐵 (𝑥, 𝑦)

Divide each 𝐼𝑅𝐺𝐵 channel by the max value of 𝐼𝑅𝐺𝐵 in each channel.

When does this give the correct answer?

• Example 2: non-uniform illumination

= *

𝐼𝑅𝐺𝐵 𝑥, 𝑦 = 𝑖𝑙𝑙𝑢𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑅𝐺𝐵 𝑥, 𝑦 ∗ 𝑟𝑒

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