1.today: review of material for the exam (chapters 9,10,&13) 1.dec. 14: exam 3 grades posted;...

38
1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions; • Problems (2-3 per chapter); Information/ preparation: http://www.colorado.edu/physics/phys1230/ phys1230_fa11/ Exams.htm Practicing problems; Reading Material; Solutions will be posted on the web page soon after the exam;

Upload: beryl-ross

Post on 23-Dec-2015

219 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

1. Today: Review of material for the exam (chapters 9,10,&13)

1. Dec. 14: Exam 3 grades posted;

2. Dec. 15: Final grades posted;

• Exam:• Multiple choice questions;• Problems (2-3 per chapter);

• Information/preparation:http://www.colorado.edu/physics/phys1230/phys1230_fa11/Exams.htm

• Practicing problems;• Reading Material;• Solutions will be posted on the web

page soon after the exam;

Page 2: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

The same color sensation can often be produced by 2 or more different intensity distribution curves

• Here is an intensity distribution curve which gives us the sensation of yellow

• Here is a different intensity distribution curve which also gives us the same sensation of yellow

• The two colors described by the two different intenstiy curves are called metamers

Page 3: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Hue, Saturation and Brightness (HSB): One way to use 3 numbers to specify a color

instead of using an intensity-distribution curve

• Color tree (e.g. Fig. 9.5 in book)

• Moving up the tree increases the lightness of a color

• Moving around a circle of given radius changes the hue of a color

• Moving along a radius of a circle changes the saturation (vividness) of a color

• These three coordinates can be described in terms of three numbers

• Photoshop: uses H, S and B

lightness

hue

saturation

Page 4: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;
Page 5: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Red, green and blue (RGB): RGB is another way to use 3 numbers to specify a color instead of using an intensity-distribution curve or HSB

• In addition to using Hue, Saturation and Brightness (HSB);

• Many (but not all) colors can be described in terms of the relative intensities of a light mixture of a certain wavelength red, wavelength green and wavelength blue lights

• 650-nm red• 530-nm green• 460-nm blue

• These are called the additive primaries• The mixing of the additive primaries is

called additive mixing• Additive mixing is usually done by mixing

primary color lights with different intensities but there are other ways to be discussed later

• Demonstrate with Physics 2000

cyan magenta

yellow650-nm red530-nm green530-nm green

460-nm blue

http://www.colorado.edu/physics/2000/tv/colortv.html

Page 6: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Complementary additive colors

• Definition of complementary color (for additive mixtures):

• The complement of a color is a second color.

• When the second color is additively mixed to the first, the result is white.

• Blue & yellow are complementary B + Y = W.

• Green & magenta are complementary G + M = W

• Cyan and red are complementary C + R = W

• Magenta is not a wavelength color— it is not in the rainbow

• There is at most one wavelength complementary color for each wavelength color (Fig 9.9)

white

cyan

red

magenta

greengreenyellow

blue

Page 7: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Additive mixing of colored light primaries

Blue added togreen = cyan.

Green added tored = yellow.

Red added toblue = magenta.

Page 8: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Complementary colored lights(additive mixing)

Blue (primary)and yellow.

Green (primary)and magenta.

Red (primary)and cyan.

Page 9: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Chromaticity diagrams: Yet another way to represent colors by (3) numbers

• The chromaticity diagram is in many ways similar to a color tree

• A chromaticity diagram has a fixed brightness or lightness for all colors

• Wavelength colors are on the horseshoe rim but non-wavelength colors like magenta are on the flat part of the rim

• Inside are the less saturated colors, including white at the interior

less saturated colors

saturated wavelength

colors

saturated non-wavelength

colors

Page 10: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Using the chromaticity diagram to identify colors

• The numbers that we use to identify a color are its x-value and y-value inside the diagram and a z-value to indicate its brightness or lightness

• x and y specify the chromaticity of a color

• Example: Apple pickers are told around the country that certain apples are best picked when they are a certaim red (see black dot)

• Since the chromaticity diagram is a world standard the company can tell its employees to pick when the apples have chromaticity

• x = 0.57• y = 0.28

• The "purest" white is at x = 0.33 and y = 0.33

• Chromaticity diagram can be related to colors in Photoshop

Page 11: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Using the chromaticity diagram to understand the result of additive mixing of colors

• An additive mixture of two wavelength colors lies along the line joining them

• Example: The colors seen by mixing 700 nm red and 500 nm green lie along the line shown

• Where along the line is the color of the mixture?

• Answer depends on the relative intensities of the 700 nm red and the 500 nm green.

• Here is what you get when the green is much more intense than the red (a green)

• Here is what you get when the red is much more intense than the green (a red)

• Here is what you get when the red is slightly more intense than the green (a yellow)

Note — this works for addingtwo colors in middle also!

Page 12: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Using the chromaticity diagram to understand complementary colors

• The complement to any wavelength color on the edge of the chromaticity diagram is obtained by drawing a straight line from that color through white to the other edge of the diagram• Example: The complement to

700 nm red is 490 nm cyan• Example: The complement to

green is magenta - a non-wavelength color

Page 13: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Using the chromaticity diagram to find the dominant hue of a color in the interior of the diagram

• To find the dominant hue of the color indicated by the black dot • Draw st. line from white

through the point to get dominant wavelength, and hence, hue (547 nm green)

• Works because additive mixture of white with a fully-saturated (wavelength) color gives the desaturated color of the original point

Page 14: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

• Partitive mixing is another kind of additive color mixing but not achieved by superimposing colored lights!

• Instead, it works by putting small patches of colors next to each other. • From a distance these

colors mix just as though they were colored lights superimposed on each other

• Examples:• Seurat pointillism• Color TV and computer

screens (Physics 2000)• Photoshop example

What is partitive mixing?

Page 15: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;
Page 16: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

A colored filter subtracts colors by absorption.

=

Incident white light Only greengets

through

Cyanfilter subtracts

red

Yellowfilter subtracts

blue

Page 17: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

A colored filter subtracts certain colors by absorption and transmits

the rest

=

Incident white light Magentafilter subtracts

green

Cyanfilter subtracts

red

Only bluegets

through

Page 18: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

A colored filter subtracts colors by absorption.

=

Incident white light Magentafilter subtracts

green

Only redgets

through

Yellowfilter subtracts

blue

Page 19: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;
Page 20: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

What is the effect of combining (sandwiching) different colored filters together?

• Rules for combining the subtractive primaries, cyan, yellow and magenta:• White light passed through

a cyan filter plus a magenta filter appears blue

• White light passed through a yellow filter plus a magenta filter appears red

• White light passed through a yellow filter plus a cyan filter appears green

• Why?

cyan

magenta

yellow

Page 21: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Colored surfaces subtract certain colors by absorbing them, while

reflecting others

Magenta surfaceabsorbs (subtracts)

green.

Green surfaceabsorbs (subtracts)

red and blue (magenta).

White inMagenta out

White inGreen out

Page 22: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Green light on a magenta surface appears colorless because green is

absorbed

Magenta surfaceabsorbs (subtracts)

green.

Green surfaceabsorbs (subtracts)

red and blue (magenta).

Magenta light on a green surface

appears colorless because magenta is

absorbedGreen in

No color

Magenta in

No colo

r

Page 23: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

When looking at a colored object in a colored light source what is the resulting color?

• Rule: Multiply the intensity-distribution of the light source by the reflectance of the colored object to get the intensity distribution of the the illuminated object• Example: Look at a magenta shirt in reflected light from a Cool White fluorescent tube.• It appears grey (colorless)Confirm by multiplying the intensity distribution curve by the reflectance curve to get the new intensity distribution curve for the reflected light

Cool white fluorescent bulbMagenta shirt

You multiply the two y-valuesat each x to get the new curve

this number this number

equals this number equals this number

This number timesThis number times

How the shirtappears in this light

Page 24: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Halftone

• Left: Halftone dots. • Right: How the human eye

would see this sort of arrangement from a sufficient distance or when they are small.

• Resolution: measured in lines per inch (lpi) or dots per inch (dpi); for example, Laser Printer (600dpi)

Page 25: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Color halftoning

Three examples of color halftoning with CMYK separations. From left to right: The cyan separation, the magenta separation, the yellow separation, the black separation, the combined halftone pattern and finally how the human eye would observe the combined halftone pattern from a sufficient distance.

Paper beneath

Printer's ink

Page 26: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Demonstration

Page 27: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Color Liquid Crystal Displays (LCDs)

Page 28: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Chapter 10: We have three different kinds of cones whose responses are mainly at short, intermediate and long wavelengths

• s-cones absorb short wavelength light best, with peak response at 450 nm (blue)

• L-cones absorb long wavelength light best, with peak response at 580 nm (red)

• i-cones absorb intermediate wavelengths best, with peak response at 540 nm (green)

• Light at any wavelength in the visual spectrum from 400 to 700 nm will excite these 3 types of cones to a degree depending on the intensity at each wavelength.

• Our perception of which color we are seeing (color sensation) is determined by how much S, i and L resonse occurs to light of a particular intensity distribution.

Rule: To get the overall response of each type of cone, multiply the intensity of the light at each wavelength by the response of the cone at that wavelength and then add together all of the products for all of the wavenumbers in the intensity distribution

i-conesL-cones

s-cones

Spectral Spectral response of conesresponse of cones in typical human eye in typical human eye

rela

tive

res

pons

ere

lati

ve r

espo

nse

Page 29: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Examples of two different ways we see whiteLight color Brightness S-cone r esponse I-cone r esponse L-cone response 460 nm blue 1 60 5 2 575 nm yellow 1.66 0 1.66 x 33 1.66 x 35 Mixtur e ( perceived as white) 60 + 0 = 60 5+1.66 x 33 = 60 2+1.66 x 35 = 60

Spectral Spectral response of conesresponse of cones in typical human eye in typical human eye

rela

tive

resp

onse

rela

tive

resp

onse

• Our sensation of color depends on how much total s, i & L cone response occurs due to a light intensity-distribution

• Multiply the intensity distribution curve by each response curve to determine how much total S, i, and L response occurs

• We experience the sensation white when we have equal total s, i & L responses

• There are many ways this can occur!!• E.g., when broadband light enters our eye• Another way to experience white is by viewing a

mixture of blue and yellow• E.g., 460 nm blue of intensity 1 and 575 nm

yellow of intensity 1.66• The blue excites mainly s-cones but also a

bit of i-cones and a bit of L-cones• The yellow excites i-cones and (slightly

more) L-cones but no s-cones• The result is an equal response of s-cones,

i-cones and L-cones (details)

575 nm yellowof intensity 1.66460 nm blue of

intensity 11

1.66

0

Page 30: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

How does a normal person see yellow when only red and green lights are superimposed?

Spectral Spectral response of conesresponse of cones in typical human eye in typical human eye

rela

tive

resp

onse

rela

tive

resp

onse

• Our sensation of yellow depends on a special s, i & L cone response

• We experience the sensation yellow when 575 nm light reaches our eyes

• What really gives us the sensation of yellow is the almost equal response of i and L cones together with no s-cones!!

• Another way to experience yellow is by seeing overlapping red & green lights

• E.g., 530 nm green of intensity 1 and 650 nm red of intensity 2.15

• The green excites mainly i-cones but also L-cones, while the red excites mainly L-cones but also i-cones

• The total respone of s & i-cones due to the spectral green and red is the same as the total response due to spectral yellow

• In general need 3 wavelength lights to mix to any color

575 nm yellowof intensity 1.35

530 nm green of intensity 11

2.15

0

650 nm red of intensity

2.15

Light color

Brightness S-cone response I-cone response L-cone response

530 nm green 1 negligible 41 28

650 nm red 2.15 negligible 2.15 x 2 2.15 x 9

Mixture (perceived as yellowyellow ) negligible 41 +2.15 x 2 =45 28 +2.15 x 9 =47

575 nm yellow 1.35 negligible 1.35 x 33 = 45 1.35 x 35 = 47

Page 31: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

We can verify color naming of hues in terms of the psychological primaries on the chromaticity diagram

• All of the hues can be named qualitatively by how much green, red, blue or yellow is "in" them

• We don't need orange, purple or pink: • orange can be thought of as yellow-red• purple can be thought of as red-blue• pink has the same hue as red but differs

only in lightness• We can break up the diagram into 4

different regions by drawing two lines whose endpoints are the psychological primary hues

• The endpoints of the yellow line are 580 nm "unique" yellow and 475 nm "unique" blue

• One endpoint of the red line is 500 nm "unique" green and the other is "red" (not unique or spectral - really more like magentamagenta)

Greenness &Greenness &yellownessyellowness

Redness &Redness &yellownessyellowness

Redness &Redness &

bluenessblueness

Greenness &

Greenness &blueness

blueness

Page 32: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

What is meant by the opponent nature of red vs green (r-g) perception and of yellow vs blue (y-b) perception.

• Viewing a progression of colors in the direction of the yellow line from 475 nm blue towards 580 nm yellow, we see more yellowness of each color and less blueness.

• We call this perception our y-b channel

• Yellow & blue are opponents • Moving parallel to the red line from

500 nm green towards nonspectral red we see more redness in each color and less greenness.

• We call this perception our r-g channel

• Red and green are opponents• The lines cross at white, where both

y-b & r-g are neutralized

Greenness &Greenness &yellownessyellowness

Redness &Redness &yellownessyellowness

Redness &Redness &

bluenessblueness

Greenness &

Greenness &blueness

blueness

y-b

r-g

Page 33: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

How might the three types of cones be "wired" to neural cells to account for our perception of hues in terms of two opponent

pairs of psychological primaries r-g and y-b?

• The 3 kinds of cones are related to r-g and y-b by the way they are connected to neural cells (such as ganglion cells)

• Cones of each kind are attached to 3 different neural cells which control the two chromatic channels, y-b and r-g, and the white vs black channel called the achromatic channel (lightness)

• "wiring" is the following:• When light falls on the L-cones they tell all 3

neural cells to increase the electrical signal they send to the brain

• When light falls on the i-cones they tell the r-g channel cell to decrease (inhibit) its signal but tell the other cells to increase their signal

• When light falls on the s-cones they tell the y-b channel cell to decrease (inhibit) its signal but tell the other cells to increse their signal

s-cone i-cone L-cone

neural cellfor y-b

chromaticchannel

neural cellfor r-g

chromaticchannel

neural cellfor w-blk

achromaticchannel

Electrical signal to brain

+++ ++ ++

Page 34: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

s-cone i-cone L-cone

neural cellfor y-b

chromaticchannel

neural cellfor r-g

chromaticchannel

neural cellfor w-blk

achromaticchannel

Electrical signal to brain

++++ +++

How can this "wiring" work to produce the chromatic channels?

• The neural cell for the y-b chromatic channel has its signal

• inhibited when (bluE) light excites the s-cone INTERPRETED AS BLUE

• enhanced when light excites the i & L cones INTERPRETED AS YELLOW

• The neural cell for the r-g chromatic channel has its signal

• inhibited when (green) light falls on the i-cone INTERPRETED AS GREEN

• enhanced when light excites the s and L coneINTERPRETED AS MAGENTA (Psychological red)

• The neural cell for the achromatic channel has its signal enhanced when light excites any of the cones

Page 35: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Systematic description of color-blindness (no need to

memorize terminology)

• Monochromacy (can match any colored light with any 1 spectral light by adjusting intensity)

• Either has no cones (rod monochromat) or has only 1 of the 3 types of cones working (cone monochromat).

• Sees ony whites, greys, blacks, no hues• Dichromacy (can match any colored light

with 2 spectral lights of different intensities of (rather than the normal 3)

• L-cone function lacking = protanopia• i-cone function lacking = deuteranopia• s-cone function lacking = tritanopia• no y-b channel but all 3 cones OK =

tetartanopia

• Anomalous trichromacy (can match any colored light with 3 spectral lights of different intensities as in normal vision, but still have color perception problems)• Protanomaly

• Shifted L-cone response curve• Deuteranomaly (most common)

• Shifted i-cone response curve• Confusion between red and green.

• Tritanomaly• Yellow-blue problems: probably

defective s-cones• Neuteranomaly

• ineffective r-g channel

Page 36: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

Receptive field of a double-opponent cell of the r-g type

• 2 different ways to2 different ways to INCREASEINCREASE the the signal the ganglion cell sends to brainsignal the ganglion cell sends to brain• RedRed light falling on cones in centercenter

of receptive field attached to ganglion cell

• GreenGreen light on surroundsurround• 2 different ways to2 different ways to decrease the the

signal the ganglion cell sends to the signal the ganglion cell sends to the brainbrain• RedRed light on surroundsurround• Green Green light on light on centercenter

• Electrical signal to brain from ganglion cell is at ambient level when no light is on center or surround

• When signal to brain is INCREASEDwe interpret that as redred

• When signal to brain is decreased we interpret that as greengreen

signal to brain

Page 37: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

We can summarize this by just showing the center & surround of the receptive field and indicating the effect of

red (R) and green (G) on each

• A double-opponent cell differs from a single opponent cell

• In both of them R in the center increases the signal

• In a single-opponent cell G in surround would inhibit signal, whereas in double-opponent cell G enhances

• In a double-opponent cell• R in center enhances signal

(ganglion cell signals red)• G in surround enhances signal

(ganglion cell signals red)• R in surround inhibits signal

(ganglion cell signals green)• G in center inhibits signal

(ganglion cell signals green)

Fictional cell real cell

Page 38: 1.Today: Review of material for the exam (chapters 9,10,&13) 1.Dec. 14: Exam 3 grades posted; 2.Dec. 15: Final grades posted; Exam: Multiple choice questions;

How do 3D movies use

polaroid filters?