Interference, Polarization and Color Review

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12. TRUE or FALSE:

White and black are actual colors of light.

a. TRUE

 

b. FALSE

 

Answer: B

Black is the absence of all light. Things appear black when they do not reflect or emit light. White is the presence of all colors of visible light. Objects appear white when they reflect or emit all wavelengths of visible light (or at least three wavelengths - Red, Blue and Green - in equal intensity).



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The Electromagnetic and Visible Spectra


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13. The three primary colors of light are ____.

a. white, black, gray

b. blue, green, yellow

c. red, blue, green

d. red, blue, yellow

e. ... nonsense! There are more than three primary colors of light.


Answer: C

Yes, you must know this one! It forms the basis of most of our logic and reasoning about color, light and the appearance of objects.



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Color Addition


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14. The three secondary colors of light are ____.

a. cyan, magenta, green

b. cyan, magenta, and yellow

c. orange, yellow, violet

d. red, blue, yellow

e. ... nonsense! There are more than three secondary colors of light.


Answer: B

The secondary colors of light are those colors which are formed when two primary colors are mixed in equal amounts. Mixing blue and green light results in cyan light. Mixing red and blue light results in magenta light. And mixing red and green light results in yellow light.



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Color Addition


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15. Combining red and green light (with equal intensity) makes ____ light; combining red and blue light (with equal intensity) makes ____ light; and combining blue and green light (with equal intensity) makes ____ light. Choose the three colors in respective order.

a. brown, purple, aqua

b. brown, magenta, yellow

c. yellow, magenta, brown

d. yellow, magenta, cyan


Answer: D

You must know this for it forms the foundation of much of our reasoning. To assist in recalling the three primary colors of light, three secondary colors of light, and the means by which adding primaries form secondaries, develop some form of graphical reminder such as a color wheel or a diagram like those at the right.



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Color Addition


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16. Demonstrate your understanding of color addition by completing the following color equations. Select colors from the Color Table at the right.

a. Red + Blue = _____

b. Red + Green = _____

c. Green + Blue = _____

d. Red + Blue + Green = _____

e. Blue + Yellow = _____

Answer: See table above.

A. Magenta is a secondary color of light formed by combining red light with blue light in equal amounts. Refer to graphic in previous question.

B. Yellow is a secondary color of light formed by combining red light with green light in equal amounts. Refer to graphic in previous question.

C. Cyan is a secondary color of light formed by combining green light with blue light in equal amounts. Refer to graphic in previous question.

D. White light is formed when all three primary colors of light are combined in equal amounts.

E. Yellow light is a combination of red and green light. So combining blue with yellow light is like combining blue light with red and green light. The result of combining these three primary colors of light is to produce white light.



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Color Addition


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17. Demonstrate your understanding of color subtraction by completing the following color equations. Select colors from the Color Table at the right.

a. White - Blue = _____

b. White - Red = _____

c. White - Green = _____

d. White - Blue - Green = _____

e. White - Yellow = _____

f. Red + Green - Green = _____

g. Yellow - Green = _____

h. Yellow - Red = _____

i. White - Magenta = _____

j. White - Cyan = _____

k. Yellow + Blue - Cyan = _____

l. Yellow + Cyan + Magenta = _____

m. Yellow + Cyan - Magenta = _____

n. Yellow + Cyan - Blue - Red = _____

Answer: See table above.

Each of these questions is best answered by first converting any secondary color of light into a mix of two primary colors of light. Then "do the arithmetic." If the result of the "arithmetic" is a combination of two primary colors, translate the combo into a secondary color of light. Here it goes:

a. White - Blue = R+G+B - B = R+G = Yellow

b. White - Red = R+B+G - R = G+B = cyan

c. White - Green = R+G+B - G = R+B = magenta

d. White - Blue - Green = R+G+B - B - G = R = red

e. White - Yellow = R+G+B - R+G = B = blue

f. Red + Green - Green = R + G - G = R = red

g. Yellow - Green = R+G - G = R = red (Note the similarity to part f.)

h. Yellow - Red = R+G - R = G = green

i. White - Magenta = R+G+B - R+B = G = green

j. White - Cyan = R+G+B - G+B = R = red

k. Yellow + Blue - Cyan = R+G + B - G+B = R = red

(Note the similarity to part j: R+G + B is the same as white; so this question is White - Cyan.)

l. Yellow + Cyan + Magenta = R+G + B+G + R+B = R+R+G+G+B+B = white + white (that is very bright white since there is double the red, green and blue added together)

m. Yellow + Cyan - Magenta = R+G + B+G - R+B = G+G = green

n. Yellow + Cyan - Blue - Red = R+G + G+B - B - R = G+G = green



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18. Sunsets often have a reddish-orange color associated with them. This is attributable to the phenomenon of _____.

a. polarization

b. diffraction

c. dispersion

d. refraction


Answer: B

Sunsets are the result of the longer wavelengths of light diffracting around atmospheric particles and reaching our eyes, giving the reddish-orange appearance. More detail about the phenomenon can be accessed using the Useful Web Link below.



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Blue Skies and Red Sunsets


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19. A filter serves the function of ____.

a. subtracting color(s) from the light which is incident upon it

b. adding color(s) to the light which is incident upon it

c. removing nicotine from light so that we can live longer lives

d. confusing physics students who are studying color, causing them to live shorter lives

Answer: A

Filters can be thought of as absorbing one or more of the primary colors of light which are incident upon it, allowing remaining colors to be transmitted. For instance, a green filter will absorb all wavelengths except for green light. In this sense, filters subtract colors from the mix of incident light, allowing only selected colors to pass through.



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Color Subtraction


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20. Express your understanding of filters by answering the following questions. Choose the best answer(s) from the Color Table shown at the right.

a. A red filter is capable of transmitting ____ light (if it is incident upon the filter).

b. A blue filter is capable of transmitting ____ light (if it is incident upon the filter).

c. A green filter is capable of transmitting ____ light (if it is incident upon the filter).

d. A red filter will absorb ____ light (if it is incident upon the filter).

e. A blue filter will absorb ____ light (if it is incident upon the filter).

f. A yellow filter will absorb ____ light (if it is incident upon the filter).

g. A magenta filter will absorb ____ light (if it is incident upon the filter).

h. A white object is illuminated with white light and viewed through a green filter. The object will appear _____.

i. A white object is illuminated with white light and viewed through a blue filter. The object will appear _____.

j. A white object is illuminated with white light and viewed through a cyan filter. The object will appear _____.

k. A blue object is illuminated with white light and viewed through a green filter. The object will appear _____.

l. A cyan object is illuminated with white light and viewed through a cyan filter. The object will appear _____.

m. A cyan object is illuminated with white light and viewed through a green filter. The object will appear _____.

n. A yellow object is illuminated with white light and viewed through a green filter. The object will appear _____.

o. A yellow object is illuminated with white light and viewed through a magenta filter. The object will appear _____.

p. A yellow object is illuminated with yellow light and viewed through a yellow filter. The object will appear _____.

q. A yellow object is illuminated with yellow light and viewed through a blue filter. The object will appear _____.

r. A yellow object is illuminated with blue light and viewed through a yellow filter. The object will appear _____.

s. A blue object is illuminated with blue light and viewed through a yellow filter. The object will appear _____.

t. A yellow object is illuminated with yellow light and viewed through a red filter. The object will appear _____.

u. A yellow object is illuminated with yellow light and viewed through a green filter. The object will appear _____.

v. A yellow object is illuminated with green light and viewed through a yellow filter. The object will appear _____.

w. A yellow object is illuminated with green light and viewed through a green filter. The object will appear _____.

x. A yellow object is illuminated with green light and viewed through a red filter. The object will appear _____.

y. A yellow object is illuminated with green light and viewed through a cyan filter. The object will appear _____.

z. A red object is illuminated with yellow light and viewed through a cyan filter. The object will appear _____.

Answer: See sentences above.

Parts a-g target your understanding of the ability of filters to subtract colors of light from the mix of incident light that strikes it. A filter will absorb its complementary color of light. So a yellow filter absorbs blue light since blue is across from it on the color wheel. Whatever light is not absorbed will be transmitted; so yellow filters transmit red and green light (if incident upon it), also known as yellow light.

a. Red filters absorb cyan light (the complementary color of red). If white light (red + blue + green) shines on a red filter and cyan (blue + green) light is absorbed, all that is left to be transmitted is red light.

b. Blue filters absorb yellow light (the complementary color of blue). If white light (red + blue + green) shines on a blue filter and yellow (red + green) light is absorbed, all that is left to be transmitted is blue light.

c. Green filters absorb magenta light (the complementary color of green). If white light (red + blue + green) shines on a green filter and magenta (red + blue) light is absorbed, all that is left to be transmitted is green light.

d. Red filters absorb its complementary color - cyan. So this question could be answered as cyan. And since cyan light consists of blue + green light, this question could also be answered as blue + green.

e. Blue filters absorb its complementary color - yellow. So this question could be answered as yellow. And since yellow light consists of red + green light, this question could also be answered as red + green.

f. Yellow filters absorb its complementary color - blue. So this question must be answered as blue.

g. Magenta filters absorb its complementary color - green. So this question must be answered as green.

 

Parts h - z target your understanding of color subtraction for both pigments and filters. In each question, there is light incident upon an object. This light can be broken down into primary colors. Some light might be subtracted from this incident mix by either the object or the filter. The only possible color of light that could ultimately pass through the filter and effect the appearance of the object would be one of the primary colors in the incident light. For instance, suppose that an object is illuminated with yellow light (which is a combination of red and green primary colors of light. The object could appear yellow (if neither red nor green are subtracted away), or red (if green light subtracted is taken away) or green (if red light is subtracted away) or black (if both red and green light is subtracted away).

In the explanations below, each question will be approached by identifying the primary colors of light in the incident mix (the light used to illuminate the object) and then primaries will be successively subtracted away by the pigments in the object and by the filter. Here it goes:

h. RGB light (white light) hits a white object; white objects do not subtract (i.e., absorb) any colors; so RGB reflects off the object and heads towards a green filter. Green filters would subtract R and B (when present) and allow G to pass through. So RGB - nothing - GB = R = red.

i. RGB light (white light) hits a white object; white objects do not subtract (i.e., absorb) any colors; so RGB reflects off the object and heads towards a blue filter. Blue filters would subtract R and B (when present) and allow B to pass through. So RGB - nothing - RG = B = blue.

j. RGB light (white light) hits a white object; white objects do not subtract (i.e., absorb) any colors; so RGB reflects off the object and heads towards a cyan filter. Cyan filters would subtract R (when present) and allow G to pass through. So RGB - nothing - R = GB = cyan.

k. RGB light (white light) hits a blue object; blue objects subtract (i.e., absorb) R and G light (when present); so B light reflects off the object and heads towards a green filter. Green filters would subtract R and B (when present) and allow G to pass through; blue light is present so it will be subtracted. So RGB - GB - B = nothing = black.

l. RGB light (white light) hits a cyan object; cyan objects subtract (i.e., absorb) R light (when present); so GB light reflects off the object and heads towards a cyan filter. Cyan filters would subtract R (when present) and allow GB to pass through. So RGB - R = GB = cyan.

m. RGB light (white light) hits a cyan object; cyan objects subtract (i.e., absorb) R light (when present); so GB light reflects off the object and heads towards a green filter. Green filters would subtract RB (when present) and allow G to pass through; B is present so it will be subtracted. So RGB - R - B = G = green.

n. RGB light (white light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off the object and heads towards a green filter. Green filters would subtract RB (when present) and allow G to pass through; R is present so it will be subtracted. So RGB - B - R = G = green.

o. RGB light (white light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off the object and heads towards a magenta filter. Magenta filters would subtract G (when present) and allow RB to pass through; G is present so it will be subtracted. So RGB - B - G = R = red.

p. RG light (yellow light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off the object and heads towards a yellow filter. Yellow filters would subtract B (when present) and allow RG to pass through. So RG - nothing - nothing = RG = yellow.

q. RG light (yellow light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off the object and heads towards a blue filter. Blue filters would subtract RG (when present) and allow B to pass through; R and G are both present so they will be subtracted. So RG - nothing - RG = nothing = black.

r. B light (blue light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so no light light reflects off the object and it wouldn't matter what type of filter is used. This object will appear black. So B - B - nothing = nothing = black.

s. B light (blue light) hits a blue object; blue objects subtract (i.e., absorb) RG light (when present); so B light reflects off the object and heads towards a yellow filter. Yellow filters would subtract B (when present) and allow RG to pass through (if present); neither R nor G are present and the B gets subtracted. So B - nothing - B = nothing = black.

t. RG light (yellow light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off the object and heads towards a red filter. Red filters would subtract GB (when present) and allow R to pass through (if present); G is present so it gets subtracted. So RG - nothing - G = R = red.

u. RG light (yellow light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so RG light reflects off the object and heads towards a green filter. Green filters would subtract RB (when present) and allow G to pass through (if present); R is present so it gets subtracted. So RG - nothing - R = G = green.

v. G light (green light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so G light reflects off the object and heads towards a yellow filter. Yellow filters would subtract B (when present) and allow RG to pass through (if present). So G - nothing - nothing = G = green.

w. G light (green light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so G light reflects off the object and heads towards a green filter. Green filters would subtract RB (when present) and allow G to pass through (if present). So G - nothing - nothing = G = green.

x. G light (green light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so G light reflects off the object and heads towards a red filter. Red filters would subtract GB (when present) and allow R to pass through (if present); G is present so it gets subtracted. So G - nothing - G = nothing = black.

y. G light (green light) hits a yellow object; yellow objects subtract (i.e., absorb) B light (when present); so G light reflects off the object and heads towards a cyan filter. Cyan filters would subtract R (when present) and allow GB to pass through (if present). So G - nothing - nothing = G= green.

z. G light (green light) hits a red object; red objects subtract (i.e., absorb) GB light (when present). G is present so it gets subtracted and it wouldn't matter what filter is used to view this object; there is no light reflecting off the object so it will appear black. So G - G - nothing = nothing = black.



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Color Subtraction


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