light waves and color review with answers

7/25/2019 Light Waves and Color Review With Answers

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The Physics Classroom » The Review Session » Light and Color » Light Waves

and Color Review with Answers

Interference, Polarization and Color

Review

Navigate to Answers for:

Questions #1-#11

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Part A: Multiple Choice

1. Which of the following statements are true statements

about interference?

a. Interference occurs when two (or more) waves meet

while traveling along the same medium.

b. Interference can be constructive or destructive.

c. Interference of two waves at a given location results in

the formation of a new wave pattern which has a

greater amplitude than either of the two interfering

waves.d. The meeting of a trough of one wave with a trough of

another wave results in destructive interference.

e. The only way for two waves to interfere constructively

is for a crest to meet a crest or a trough to meet a

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f. It is only a theory that light can interfere destructively;

the theory is based on the assumption that light is a

wave and most waves exhibit this behavior.

Experimental evidence supporting the theory has not

yet been observed.

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.

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about two-point light source interference patterns?

a. Two-point light source interference patterns consist of

alternating nodal and antinodal lines.

b. If projected onto a screen, two-point light source

interference patterns would be viewed as alternating

bright and dark spots with varying gradients of light

intensity in between.

c. As the distance between the sources is decreased, the

distance between the nodal and antinodal lines is

decreased.

d. As the wavelength of the laser light is decreased, the distance between the nodal and antinodal lines is

decreased.

e. A nodal point would be formed if a trough of one wave

meets a trough of another wave.

f. Antinodal points are points where the medium is

undergoing no vibrational motion.

g. Suppose point P is a point where a wave from one

source travels a distance of 2.5 wavelengths before

meeting up with a wave from another source which

travels a distance of 3.5 wavelengths. Point P would be

a nodal point.

h. Suppose point Q is a point where a wave from one

source travels a distance of 2 wavelengths before

meeting up with a wave from another source which

travels a distance of 3.5 wavelengths. Point Q would

be a nodal point.

i. Suppose point R is a point where a wave from one

source travels a distance of 2 wavelengths before

meeting up with a wave from another source which travels a distance of 3 wavelengths. Point R would be a

nodal point.

j. If the path difference for points on the first nodal line is

4 cm, then the wavelength would be 6 cm. (NOTE: the

first nodal line is considered to be the first nodal line to

the left or right from the central antinodal line.)

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:

y = m • L • W / d ... (where W=wavelength).

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

W = y • d / (m • L) ... (where W=wavelength).

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.

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Two Point Source Interference







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about nodal and antinodal points in light interference

patterns?

a. Antinodes result from constructive interference.

b. Nodes result from destructive interference.

c. The nodal points on an interference pattern are

positioned along lines; these lines are called nodal lines.

d. The central line on the interference pattern is a nodal

line.

e. Points on nodal lines would be represented by bright

spots if projected onto a screen.

f. The path difference for points on the central antinodal

line would be 0.

g. The path difference for points on the first antinodal line

would be 1 cm.

h. (This question presumes that the interference pattern is a water interference pattern.) If the path difference for

points on the first antinodal line is 5 cm, then the path

difference for points on the second antinodal line

would be 7 cm.

i. (This question presumes that the interference pattern is

a water interference pattern.) If the path difference for


difference for points on the third antinodal line would

be 15 cm.

j. (This question presumes that the interference pattern is



difference for points on the second nodal line would be

9 cm. (NOTE: the second nodal line is considered to be

the second nodal line to the left or right from the

central antinodal line.)

k. (This question presumes that the interference pattern is


points on the first nodal line is 4 cm, then the path difference for points on the third nodal line would be

12 cm. (NOTE: the third nodal line is considered to be

the third nodal line to the left or right from the central





antinodal line.)

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



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







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.

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about Thomas Young's experiment?

a. Young's experiment provided evidence that light

exhibits particle-like behavior.

b. Young's experiment depends upon the use of white

light from two sources.

c. The two sources of light in Young's experiment could be

two different light bulbs.

d. For Young's equation to be geometrically valid, the

distance from the sources to the screen must be much

greater than the slit separation distance.e. For Young's equation to be geometrically valid, the

wavelength of the light must be much greater than the

slit separation distance.

f. Thomas Young measured the distance from an

antinodal point (of known number) to each of the two

sources, computed a path difference and calculated

the wavelength of light.

g. Thomas Young was able to determine the wavelength

of a light wave.

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.

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5. Light which is vibrating in a single plane is referred to as

_____ light

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."

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Polarization

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6. Light which is vibrating in a variety of planes is referred

to as _____ light

a.

electromagnetic

b.

transverse

c.

unpolarized

d.

polarized

Answer: C

Unpolarized light is light whose vibrations are in a

multitude of directions.

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Polarization

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7. Light usually vibrates in multiple vibrational planes. It

can be transformed into light vibrating in a single plane of

vibration. The process of doing this is known as ____.

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).

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Polarization

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8. Light is passed through a Polaroid filter whose transmission axis is aligned horizontally. This will have the

effect of ____.

a. making the light one-half as intense and aligning the

vibrations into a single plane.

b. aligning the vibrations into a single plane without any

effect on its intensity.

c. merely making the light one-half as intense; the

vibrations would be in every direction. d. ... nonsense! This will have no effect on the light itself;

only the filter would be effected.

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 .

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Polarization

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9. Light is passed through a Polaroid filter whose

transmission axis is aligned horizontally. It then passes

through a second filter whose transmission axis is aligned

vertically. After passing through both filters, the light will

be ______.

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.

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Polarization

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10. Which of the following are effective methods of

polarization? Include all that apply.

a. Passing light through a Polaroid filter.

b. Reflection of light off a nonmetallic surface.

c. Passing light from water to air.

d. Passing light through a birefringent material such as

Calcite.

e. Turning the light on and off at a high frequency.

f. Interfering light from one source with a second source.

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.

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Polarization

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11. Consider the three pairs of

sunglasses to the right. Which

pair of glasses is capable of

eliminating the glare from a road

surface? (The transmission axes are shown by the straight lines.)







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.

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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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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.

Useful Web Links

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.

Useful Web Links

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

Useful Web LinksColor Addition | Color Subtraction

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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.

Useful Web Links

Blue Skies and Red Sunsets

[ #1 | #2 | #3 | #4 | #5 | #6 | #7 | #8 | #9 | #10 | #11 | #12 | #13

| #14 | #15 | #16 | #17 | #18 | #19 | #20 | #21 | #22 | #23 | #24 |



http://www.physicsclassroom.com/Class/light/u12l2f.cfm

http://www.physicsclassroom.com/Class/light/u12l2f.cfm







#25 | #26 | #27 | #28 ]

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.

Useful Web LinksColor 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


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


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


_____.

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


_____.

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


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


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


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


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


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


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


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


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


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 .

Useful Web Links





Color Subtraction

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Part B: Diagramming, Analysis,

Calculations

21. Two

point

sources are

vibrating

together

(in phase)

at the

same

frequency to produce a two-point source interference

pattern. The diagram at the right depicts the two-point

source interference pattern. The crests are represented by

thick lines and the troughs by thin lines. Several points on the pattern are marked by a dot and labeled with a letter.

Use the diagram to answer the following questions.

a. Which of the labeled points are antinodal points?

b. Which of the labeled points are nodal points?

c. Which of the labeled points are formed as a result of

constructive interference?

d. Which of the labeled points are located on the central

antinodal line?

e. Which of the labeled points are located on the first

antinodal line?

f. Which of the labeled points are located on the second

antinodal line?

g. Which of the labeled points are located on the third

antinodal line?

h. Which of the labeled points are located on the first

nodal line (using the notation that the first nodal line is the nodal line directly to the left or the right of the central

antinodal line)?

i. Which of the labeled points are located on the second







nodal line (using the notation that the second nodal line is

the second nodal line directly to the left or the right of the

central antinodal line)?

j. Which of the labeled points are located on the third

nodal line (using the notation that the third nodal line is

the third nodal line directly to the left or the right of the

central antinodal line)?

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 AN . 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.

0





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.

Useful Web Links

Anatomy of a Two-Point Source Interference Pattern

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#13 | #14 | #15 | #16 | #17 | #18 | #19 | #20 | #21 | #22 | #23 |

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22. Consider the interference

pattern at the right. (The

crests are represented by thick

lines and the troughs by thin

lines.) If the distance from S

to point A is 49.5 cm and the distance from S to point A

is 60.5 cm, then what is the wavelength?

Answer: W = 11.0 cm

Given: S A = 49.5 cm and S A = 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 = | S A - S A | = | 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

1

2

1 2

2 1







An alternative strategy involves recognizing from the

diagram that point A is a distance of 4.5 wavelengths from

point S . 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 S to point A, yielding the same

answer.

Useful Web Links

Anatomy of a Two-Point Source Interference Pattern | The

Path Difference

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| #14 | #15 | #16 | #17 | #18 | #19 | #20 | #21 | #22 | #23 | #24 |

#25 | #26 | #27 | #28 ]

23. Consider the

interference pattern at

the right. (The crests

are represented by

thick lines and the

troughs by thin lines.)

If the distance from S to point B is 50.4 cm and the distance from S to point A is 34.5 cm, then what is the

wavelength?

Answer: 6.28 cm

Given: S B = 50.4 cm and S B = 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 = | S B - S B | = | 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

1

2

1

2

1 2

2 1











W = 6.28 cm

An alternative strategy involves recognizing from the

diagram that point B is a distance of 8 wavelengths from

point S . 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 S to point B, yielding the same answer.

Useful Web Links


Path Difference

[ #1 | #2 | #3 | #4 | #5 | #6 | #7 | #8 | #9 | #10 | #11 | #12 | #13

| #14 | #15 | #16 | #17 | #18 | #19 | #20 | #21 | #22 | #23 | #24 |

#25 | #26 | #27 | #28 ]

24. Two point sources are vibrating in phase to produce an

interference pattern. The wavelength of the waves is 7.60

cm. Point C is a point on the third nodal line. The distance

from S (the nearest source) to point C is 65.6 cm.

Determine the distance from S to point C.

Answer: 84.6 cm

Given: S C = 65.6 cm and W = 7.6 cm and m=2.5 (third

nodal line)

Find: S C

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 S C

distance is larger than the S C 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 S C distance to

determine S C.

S C = S C + PD = 65.6 cm + 19.0 cm = 84.6 cm

Useful Web Links


1

2

1

2

1

2

2

1

1

2

2 1















Path Difference

[ #1 | #2 | #3 | #4 | #5 | #6 | #7 | #8 | #9 | #10 | #11 | #12 | #13

| #14 | #15 | #16 | #17 | #18 | #19 | #20 | #21 | #22 | #23 | #24 |

#25 | #26 | #27 | #28 ]

25. Consider the

interference pattern at

the right. (The crests

are represented by thick

lines and the troughs by

thin lines.) The distance from S to point D is 47.2 cm. What is the wavelength?

What is the distance from S to point D? (HINT: Use the

diagram.)

Answer: wavelength = 7.87 cm; S D = 59.0 cm

Given: S D = 47.2 cm and m = 1.50 (the second nodal line

to the left of the central antinodal line)

Find: W and S D

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 S D distance.

From the diagram, it is observed that the point D is exactly 6 full wavelengths from S . So S D = 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 S is 11.8 cm further from the

point D than S 's distance from point D. So adding 11.8

1

2

2

1

2

2

1 1

2

1







cm to 47.2 cm yields 59.0 cm.

Useful Web Links


Path Difference

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#25 | #26 | #27 | #28 ]

26. Laser light is directed towards a pair of slits which are

2.50 x 10 mm apart. The light shines on a screen 8.20 meters away and an interference pattern is observed. A

point on the 3rd antinode is observed to be 39.6 cm away

from the central antinode. What is the wavelength of the

laser light in units of nanometers? (1 m = 10 nm)

Answer: 402 nm

Given: d = 2.50 x 10 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 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 m)

/ [(3) • (8.20 m)] = 4.02 x 10 m

Now convert to meters using the conversion factor: (1•10

nm/1 m). This yields 402 nm as the answer.

Useful Web Links

Young's Equation | Young's Experiment

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http://www.physicsclassroom.com/Class/light/u12l3c.cfm











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27. This same laser light (from #26) is reflected off of the

grooves in a compact disc. The disc is 4.5 meters from the

screen where its interference pattern is projected.

Antinode 1 is found to be 1.2 meters from the central

antinode. What is the spacing between the "grooves" of

the C.D.?

Answer: 1.5 x 10 m

Given: W = 4.02 x 10 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 m) / (1.2 m) =

1.5 x 10 m

Useful Web Links

Young's Equation | Young's Experiment

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28. Different colors of paper are illuminated with different

primary colors of light. Determine the colors of light

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absorbed by the paper (if any), the colors of light reflected

by the paper (if any), and the appearance of the paper.

Color of

Light

Color of

Paper

Colors

Absorbed

Colors

Reflected

Appearance

a. White White None RGB White

b. Cyan

(= GB) White None GB Cyan

c. Yellow

(= RG) White None RG Yellow

d. Red Yellow

None

(B if

present)R Red

e. Red BlueR

(G if

present)None Black

f. Red Cyan R None Black

g. Red Red

None

(GB if

present)R Red

h. Magenta

(= RB) Red

B

(G if

present)

R Red

i. Yellow

(= RG) Red

G

(B if

present)R Red

j. Cyan

(= GB) Red GB None Black

k. Cyan

(= GB) Blue

G

(R if

present)

B Blue

l. Yellow

(= RG ) Blue RG None Black

m. Yellow

(= RG ) Green

R

(B if

present)G Green

n. Yellow

(= RG ) Cyan R G green

o. Yellow(= RG) Magenta G R Red

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

RG - G = R

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.

Useful Web Links

Color Subtraction

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light waves and color review with answers

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