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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 = 7.60 cm and m=2.50 (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.