Answer: AB

A - True: This is the definition of interference - "the meeting of two or more waves along the same medium."

B - True: These are the two possible types of interference.

C - False: When interference occurs, there are two possible results: a resulting wave with a greater displacement than either of the original waves (constructive interference) or a resulting wave with a smaller displacement than one or both of the original waves (destructive interference)

D - False: This is an example of constructive interference leading to a resulting wave with a greater displacement than the individual wave; a "super-trough" would be formed.

E - False: Crest meeting crest and trough meeting trough are examples of constructive interference. These special cases result in the formation of antinodal points - points of maximum displacement. But more generally, constructive interference will occur anytime a wave with a "positive" (up or right or ...) displacement meets another wave with a "positive" displacement OR when a wave with a "negative" (down or left or ...) displacement meets another wave with a "negative" displacement. When the displacements of the two interfering waves are in the same direction at a given point, then constructive interference occurs at that point.

F - False: There is plenty of experimental and observable evidence that light undergoes destructive interference. The best evidence from our studies in class are the dark fringes of a two-point interference pattern. These dark fringes are the result of the destructive interference of light.

Answer: ABDH

A - True: This is exactly what we have observed through computer animations, video segments, transparency overlays, and the actual experiment.

B - True: This is exactly what we observed when we performed Young's experiment.

C - False: The equation relating the variables of Young's experiment can be rearranged to the following form:

Now one notices that y is inversely related to d. So if the slit separation distance (d) is decreased, the distance between nodal and antinodal lines (related to y) would be increased.

D - True: Young's equation is often written as

From the equation, one notices that wavelength (W) is directly related to y. So if the wavelength (W) is decreased, the distance between nodal and antinodal lines (related to y) would be decreased.

E - False: Antinodal points are points of maximum displacement; for a light interference pattern, these are the brightest points.

F - False: Nodal points are points of no displacement or no disturbance; for a light light interference pattern, these are the darkest points.

G - False: In this case the path difference is 1 wavelength; when two waves traveling to the same point have a difference in distance traveled of 1 wavelength, then a crest of one wave would meet up with a crest of the second wave. This condition leads to constructive interference and an antinodal point is formed.

H - True: In this case the path difference is 1.5 wavelengths; when two waves traveling to the same point have a difference in distance traveled of 1.5 wavelengths, then a crest of one wave would meet up with a trough of the second wave. This condition leads to destructive interference and a nodal point is formed.

I - False: In this case the path difference is 1 wavelength; when two waves traveling to the same point have a difference in distance traveled of 1 wavelength, then a crest of one wave would meet up with a crest of the second wave. This condition leads to constructive interference and an antinodal point is formed.

J - False: The first nodal line is designated as m = 0.5; the path difference is 4 cm. Substituting into the equation PD = m•Wavelength and solving for wavelength yields a value of 8 cm.

Answer: ABCFIJ

A - True: An antinode is a point where a crest meets a crest or a trough meets a trough; both are examples of constructive interference.

B - True: A node is a point where a crest meets a trough; this is an example of destructive interference and leads to a location of no displacement.

C - True: Nodal points all lie along lines. Question #21 illustrates this well.

D - False: The central line - that is, the line extending outward from the midpoint between the two sources - is a line upon which antinodes are formed; it is called an antinodal line. Question #21 illustrates this well.

E - False: Nodal lines are formed as a result of destructive interference. If projected onto a screen, the nodal points would appear as the darkest points on the interference pattern.

F - True: The path difference for points on the central antinodal line would be given be the equation: PD = m•W where W=wavelength and m=0 (for the central antinodal line). Substituting into this equation yields PD = 0•W which would be 0.

G - False: The path difference for points on the first antinodal line would be given be the equation: PD = m•W where W=wavelength and m=1 (for the first antinodal line). So the path difference for the first antinodal line would always be 1•W; but it would only be 1 cm for the case in which the wavelength is 1 cm.

H - False: The first antinodal line is numbered as the m=1 line. The path difference relates to the wavelength (W) by the equation PD = m•W. Substituting m=1 and PD=5 cm into this equation yields a wavelength value of 5 cm. The second antinodal line is numbered as the m=2 line. Re-using the equation for this line with m=2 and W=5 cm yields a path difference of 10 cm.

I - True: The first antinodal line is numbered as the m=1 line. The path difference relates to the wavelength (W) by the equation PD = m•W. Substituting m=1 and PD=5 cm into this equation yields a wavelength value of 5 cm. Re-using the equation for the third antinodal line with m=3 and W=5 cm yields a path difference of 15 cm.

J - True: The logic on this question is similar to the above question. The first antinodal line is numbered as the m=1 line. The path difference relates to the wavelength (W) by the equation PD = m•W. Substituting m=1 and PD=6 cm into this equation yields a wavelength value of 6 cm. The second nodal line is numbered as m=1.5. Re-using the equation for the second nodal line with m=1.5 and W=6 cm yields a path difference of 9 cm.

K - False: The first nodal line is numbered the m=0.5 line. If the path difference for a point on this line is 4 cm, then the wavelength is 8 cm (using the PD = m•W equation). The third nodal line is numbered as the m=2.5 line. Using the same equation to find the path difference yields a value of 20 cm.

Answer: DG

a. - False: Young's experiment supports the wave-nature of light. Waves interfere and Young's experiment provided clear evidence that light undergoes interference.

b. - False: There are two requirements for the light which is utilized in Young's experiment: the two light sources must be coherent and monochromatic. Monochromatic means that the light sources must provide light of the same wavelength (and a single wavelength); using a white light bulb would produce light of many wavelengths. Second, coherent means that the light from the two sources must be vibrating together, experiencing a crest at the same time and a trough at the same time. Using two light bulbs (as opposed to a single light source shining on a double slit) would likely result in incoherent light.

c. - False: If two light bulbs emitting monochromatic light of the same color were used, one of the two requirements would be met. Yet there would still be the problem of incoherence. See explanation to part b.

d. - True: There are two geometric requirements for Young's experiment: the screen distance (L) must be much greater than the slit separation distance (d) and the slit separation distance must be much greater than the wavelength. That is L >>> d and d >>> W.

e. - False: Vice versa; d >>> W. See explanation to part d.

f. - False: Thomas Young used the equation W = y•d/m•L. Measurement of y, d, m, and L is much more practical since the size of these quantities is much larger. The error introduced in the measurement would not overwhelm the precision of the wavelength measurement. On the other hand, a measurement of the path difference would be very difficult since the only way to achieve this measurement is to measure the two distances. Given the fact that the slits are so close together, these two distances are so nearly identical that the error introduced in the measurement of one distance would overwhelm the actual difference in distance between the two measurements. That's why Young had to derive the equation W = y•d/m•L.

g. - True: Measuring the wavelength of a visible light wave was one of the main outcomes of Young's experiment.

a. electromagnetic

b. transverse

c. unpolarized

d. polarized

Answer: D

Unpolarized light is light whose vibrations are in a multitude of directions. To simplify matters, unpolarized light is light which can be thought of as vibrating in a vertical and a horizontal plane. If one of these planes of vibration is removed, then light would be vibrating in a single plane and said to be "polarized."

a. electromagnetic

b. transverse

c. unpolarized

d. polarized

Answer: C

Unpolarized light is light whose vibrations are in a multitude of directions.

a. translation

b. interference

c. polarization

d. refraction

Answer: C

Polarization is defined as the process of transforming unpolarized light (light whose vibrations are in a multitude of planes) into polarized light (light which can be thought of as vibrating in a single plane).

Answer: A

Polaroid filters have the effect of polarizing light - that is, aligning their vibrations into a specific plane. They can be thought of as performing this feat by removing the vibrations which occur within a plane perpendicular to the transmission axis.

a. polarized

b. unpolarized

c. entirely blocked

d. returned to its original state.

Answer: C

The first filter serves the role of blocking one-half the light; the horizontal vibrations would emerge from the filter and the vertical vibrations would be blocked. The second filter would allow the vertical vibrations to pass through if there were any. However, since the vertical vibrations have already been filtered out, there is no light remaining after the second filter is used. This combination of two filters serves to block all the light.

Answer: AB

The use of a filter, the reflection of light off nonmetallic surfaces and the use of a birefringent material are all means of polarizing light. Refraction at an air-water surface would change the speed and the direction of light but would not have any effect upon its vibrational orientation. Turning a light on and off at a high frequency would only annoy or impress those present in the room. And light interference could create a pattern of bright and dark spots but would not have any effect upon light's vibrational orientation.

Answer: C

When light reflects off a road surface, a portion of the light vibrations becomes oriented in a plane which is horizontal to the road surface. This polarization often leads to an annoying glare. The glare can be reduced by blocking the polarized light. Since the light is polarized horizontally (assuming a horizontal road way - a good assumption), the sunglasses should be capable of blocking horizontal light and allowing the vertical vibrations to be transmitted. Selecting sunglasses C would make accomplish this feat.

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).

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.

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.

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.

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.

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

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.

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.

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.

Answers: See diagram above for a visual representation of many of the answers.

a. ACDEHLMOPQ are all antinodal points which lie on antinodal lines. They are formed as a result of either a crest meeting a crest (two thick lines) or a trough meeting a trough (two thin lines). At the moment in time depicted in the diagram, point A is not a crest-crest or a trough-trough interference point. However, as time progresses and the circular waves continue their motion outwards from the source, point A (and all points on the antinodal lines) will be locations of crest-crest (or trough-trough) interference. At some instant in time, all points falling upon antinodal lines will be locations of constructive interference.

b. BFGJKNRS are all nodal points. They are formed as a result of a crest (thick line) meeting a trough (thin line).

c. CDEHLMOPQ (and possibly A) are all formed by constructive interference. They are all antinodal points and as such, they are the result of constructive interference. Point A is certainly not an antinodal point; however it is likely the result of constructive interference - two waves meeting with their displacement in the same direction (just not two crests or two troughs).

d. CL are on the central antinodal line. The central antinodal line is marked on the diagram above as AN₀. It is the line which extends from the midpoint of the line connecting the two sources.

e. EHPQ are on the first antinodal line. They lie on the first antinodal line to the left or right of the central antinodal line. See diagram above.

f. ADMO are on the second antinodal line. They lie on the second antinodal line to the left or right of the central antinodal line. See diagram above.

g. There is no third antinodal line. See diagram above.

h. FG are on the first nodal line. They lie on the first nodal line to the left or right of the central antinodal line. See diagram above.

i. BKR are on the second nodal line. They lie on the second nodal line to the left or right of the central antinodal line. See diagram above.

j. JNS are on the third nodal line. They lie on the third nodal line to the left or right of the central antinodal line. See diagram above.

Answer: W = 11.0 cm

Given: S₁A = 49.5 cm and S₂A = 60.5 cm and m=1 (the point is on the first antinodal line to the right of center)

Find: W (wavelength)

Strategy: Find the path difference (PD) from the two distances and then use the PD = m • W equation to calculate the wavelength.

PD = | S₂A - S₁A | = | 60.5 cm - 49.5 cm | = 11.0 cm

Now substitute into the path difference-wavelength equation and solve for wavelength (W):

11.0 cm = 1 • W

W = 11.0 cm

An alternative strategy involves recognizing from the diagram that point A is a distance of 4.5 wavelengths from point S₁. Thus, the distance 49.5 cm equals 4.5 • W. Solving for W yields 11.0 cm. The same strategy can be used for the distance from S₂ to point A, yielding the same answer.

Answer: 6.28 cm

Given: S₁B = 50.4 cm and S₂B = 34.5 cm and m=2.5 (the point is on the third nodal line to the right of center)

Find: W (wavelength)

Strategy: Find the path difference (PD) from the two distances and then use the PD = m • W equation to calculate the wavelength.

PD = | S₂B - S₁B | = | 34.5 cm - 50.4 cm | = 15.7 cm

Now substitute into the path difference-wavelength equation and solve for wavelength (W):

15.7 cm = 2.50 • W

W = 6.28 cm

An alternative strategy involves recognizing from the diagram that point B is a distance of 8 wavelengths from point S₁. Thus, the distance 50.4 cm equals 8 • W. Solving for W yields 6.28 cm. The same strategy can be used for the distance from S₂ to point B, yielding the same answer.

Answer: 84.6 cm

Given: S₁C = 65.6 cm and W = 7.6 cm and m=2.5 (third nodal line)

Find: S₂C

Strategy: Find the path difference (PD) using the equation PD = m • W. The path difference signifies the difference in distance from the sources to the nearest point. The S₂C distance is larger than the S₁C distance by an amount equal to the path difference.

First find the path difference:

PD = m • W = 2.50 • 7.60 cm = 19.0 cm

Now add the path difference to the S₁C distance to determine S₂C.

S₂C = S₁C + PD = 65.6 cm + 19.0 cm = 84.6 cm

Answer: wavelength = 7.87 cm; S₂D = 59.0 cm

Given: S₁D = 47.2 cm and m = 1.50 (the second nodal line to the left of the central antinodal line)

Find: W and S₂D

Strategy: Since neither a wavelength or a path difference is given or implicitly stated, the diagram will have to be used to determine the wavelength. The wavelength will then be used to determine the path difference and the path difference will be used to find the S₂D distance.

From the diagram, it is observed that the point D is exactly 6 full wavelengths from S₁. So S₁D = 6.00 • W. Substituting and solving for W yields the following:

47.2 cm = 6.00 • W

W = (47.2 cm) / 6.00 = 7.87 cm

Now the path difference can be found using the relationship PD = m • W where m = 1.50 and W = 7.87 cm. Substituting and solving for PD yields

PD = m • W = 1.50 • (7.87 cm) = 11. 8 cm

This means that the point S₂ is 11.8 cm further from the point D than S₁'s distance from point D. So adding 11.8 cm to 47.2 cm yields 59.0 cm.

Answer: 402 nm

Given: d = 2.50 x 10^-2 mm; L = 8.20 m; m = 3; y = 39.6 cm

Find: wavelength (W)

Strategy: Substitute into Young's equation and solve for W; be very careful with units - in fact, first perform conversions to get all quantities in unit of meters. Once the W is calculated, convert it to nanometers.

First the conversions of all given quantities to meters yields:

d = 2.50 x 10^-5 m; L = 8.20 m; y = 0.396 m

Now substitute into Young's equation:

W = y • d / (m • L) = (0.396 m) • (2.50 x 10^-5 m) / [(3) • (8.20 m)] = 4.02 x 10^-7 m

Now convert to meters using the conversion factor: (1•10⁹ nm/1 m). This yields 402 nm as the answer.

Answer: 1.5 x 10^-6 m

Given: W = 4.02 x 10^-7 m; L = 4.5 m; m = 1; y = 1.2 m

Find: d

Strategy: Use Young's equation to solve for the unknown quantity.

Rearrange Young's equation to produce an equation with d expressed in terms of the known quantities.

d = m • L • W / y

Substitute and solve

d = (1) • (4.5 m) • (4.02 x 10^-7 m) / (1.2 m) = 1.5 x 10^-6 m

Answer: See table above

In the first column, if a secondary color of light is shown, it is translated into the equivalent primaries. These primaries will strike the paper and may or may not be absorbed. The color which a paper pigment will absorb is the complementary color; this color is typically expressed in terms of the equivalent primary colors of light. The subtraction process is then done to determine what primary color of light is reflected. This/these reflected primaries determine the color appearance of the paper. They are added (if there are more than 2) to determine the resulting appearance. As an example of the entire process, consider row i:

Row i: Yellow light is equivalent to red and green (RG). Red paper contains pigments capable of absorbing both green and blue light if present. Only green light is present, so it is absorbed. So the subtraction process is

Red light is reflected; this gives the paper the appearance of red.

The same process can be performed for all other parts of this question.