An electromagnetic wave oscillating electric and magnetic ...3. Which of the following correctly...

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Section 1 Characteristics of

Light

Electromagnetic Waves

• An electromagnetic wave is a wave that consists of

oscillating electric and magnetic fields, which radiate

outward from the source at the speed of light

c=(3 x108 m/s)• Light is a form of electromagnetic radiation.

• The electromagnetic spectrum includes more than

visible light.



Electromagnetic Spectrum




Light

Electromagnetic Waves, continued

• Electromagnetic waves vary depending on frequency and wavelength.

• Wave Speed Equation

c = flspeed of light = frequency wavelength



Electromagnetic Waves


Light




Light

• Huygens’ principle Waves can be approximated as rays.

• Lines drawn tangent to the crest (or trough) of a wave are called wave fronts.

• Rays are drawn normal (perpendicular to wave fronts)




Light

• luminous flux (P)The rate at which light is emitted unit =lumens (lm).

• Illuminance (E) decreases as the square of the distance from the source. Another inverse square law 1/r2

• E=P/(4π r2)



Section 2 Flat MirrorsReflection of Light

• Reflection =change in

direction of a wave at a

surface.

• The texture of a surface

affects how it reflects

light.

– Diffuse reflection

– Specular reflection



• Opaque= does not transmit light

through the material. (reflected only)

• Translucent= Light passes through the

medium but is distorted.

• Transparent= material with no light

distortion.

• Transparent and translucent transmit

and reflect light at the same time.



Section 2 Flat Mirrors

Reflection of Light, continued

• Normal= line perpendicular

to a surface.

• Angle of incidence θ

=angle between a ray

striking and the normal.

• Angle of reflection θ’

formed by the normal and

the reflected rays.

θ = θ’



Chapter 13

Angle of Incidence and Angle of Reflection





Chapter 13

Flat Mirrors Virtual Images

• Flat mirrors form virtual images =same distance from

the mirror’s surface as the object and is upright.

• image forms behind mirror

• A virtual image cannot be projected.



Drawing ray diagrams1. Reflected light rays can be

extended behind the mirror.

2. Ray intersections represent

locations of image formation.

3. At least 2 rays must be drawn

for reference.



Chapter 13

Image Formation by a Flat Mirror




Chapter 13

Comparing Real and Virtual Images




Section 3 Curved Mirrors

Chapter 13

Concave Spherical Mirrors

• A concave spherical mirror The reflecting surface

is on the inside of a sphere.

• Concave mirrors can be used to form real and virtual

images depending on the position of the object being

reflected.

• A real image is an image formed when rays of light

pass through a point on the image. Real images can

be projected onto a screen.



Chapter 13

Image Formation by a Concave Spherical Mirror





Chapter 13

Concave Spherical Mirrors, continued

• The Mirror Equation relates object distance (p),

image distance (q), and focal length (f) of a spherical

mirror.

1

p

1

q

1

f

1

object distance

1

image distance

1

focal length




Chapter 13

Concave Spherical Mirrors, continued

• The Equation for Magnification relates image height

or distance to object height or distance, respectively.

'–

image height image distancemagnification = –

object height object distance

h qM

h p



Chapter 13

Rules for Drawing Reference Rays for Mirrors




Regular p 465 honors 605



Chapter 13

Ray Tracing for a Concave Spherical Mirror




Sign Convention

• + q= real image in front of mirror

• -q= virtual image behind the mirror

• +M= virtual image (upright)

• -M= real image inverted




Chapter 13

Sample Problem

Imaging with Concave Mirrors

A concave spherical mirror has a focal length of

10.0 cm. Locate the image of a pencil that is

placed upright 30.0 cm from the mirror. Find the

magnification of the image. Draw a ray diagram to

confirm your answer.



• Succeed




Chapter 13

Sample Problem, continued


1. Determine the sign and magnitude of the focal

length and object size.

f = +10.0 cm p = +30.0 cm

The mirror is concave, so f is positive. The object is

in front of the mirror, so p is positive.




Chapter 13



2. Draw a ray diagram using the rules for drawing

reference rays.




Chapter 13



3. Use the mirror equation to relate the object and

image distances to the focal length.

–q

Mp

1

p

1

q

1

f

4. Use the magnification equation in terms of object

and image distances.




Chapter 13


5. Rearrange the equation to isolate the image

distance, and calculate. Subtract the reciprocal of

the object distance from the reciprocal of the focal

length to obtain an expression for the unknown

image distance.

1 1 1

–q f p




Chapter 13


Substitute the values for f and p into the mirror

equation and the magnification equation to find the

image distance and magnification.

1 1 1 0.100 0.033 0.067– –

10.0 cm 30.0 cm cm cm cm

15 cm

15 cm– – –0.50

30.0 cm

q

q

qM

p




Chapter 13

Convex Spherical Mirrors

• A convex spherical mirror is a mirror whose

reflecting surface is outward-curved segment of a

sphere.

• Light rays diverge upon reflection from a convex

mirror, forming a virtual image that is always smaller

than the object.



Chapter 13

Image Formation by a Convex Spherical Mirror





Chapter 13

Sample Problem

Convex Mirrors

An upright pencil is placed in front of a convex

spherical mirror with a focal length of 8.00 cm. An

erect image 2.50 cm tall is formed 4.44 cm behind

the mirror. Find the position of the object, the

magnification of the image, and the height of the

pencil.




Chapter 13


Convex Mirrors

Given:

Because the mirror is convex, the focal length is

negative. The image is behind the mirror, so q is

also negative.

f = –8.00 cm q = –4.44 cm h’ = 2.50 cm

Unknown:

p = ? h = ?




Chapter 13


Convex Mirrors

2. Plan

Choose an equation or situation: Use the mirror

equation and the magnification formula.

1 1 1

– and – 'p

h hp f q q

1 1 1 '

and –h q

Mp q f h p

Rearrange the equation to isolate the unknown:




Chapter 13


Convex Mirrors

3. Calculate

Substitute the values into the equation and solve:

1 1 1–

–8.00 cm –4.44 cm

1 –0.125 –0.225 0.100–

cm cm cm

10.0 cm

p

p

p




Chapter 13


Convex Mirrors

3. Calculate, continued

Substitute the values for p and q to find the magnifi-cation of the image.

10.0 cm– ' – (2.50 cm)

–4.44 cm

5.63 cm

ph h

q

h

Substitute the values for p, q, and h’ to find the height of the object.

–4.44 cm

– – 0.44410.0 cm

qM M

p



Chapter 13

Ray Tracing for a Convex Spherical Mirror





Chapter 13

Parabolic Mirrors

• Images created by spherical mirrors suffer from

spherical aberration.

• parallel rays far from the principal axis converge

away from the mirrors focal point.

• Parabolic mirrors eliminate spherical aberration. All

rays converge at the focal point.



Chapter 13

Spherical Aberration and Parabolic Mirrors




Chapter 13

Reflecting Telescope




Section 4 Color and Polarization

Chapter 13

Objectives

• Recognize how additive colors affect the color of

light.

• Recognize how pigments affect the color of reflected

light.

• Explain how linearly polarized light is formed and

detected.




Chapter 13

Color

• Additive primary colors produce white light when

combined.

• Light of different colors can be produced by adding

light consisting of the primary additive colors (red,

green, and blue).



Chapter 13

Additive Color Mixing





Chapter 13

Color, continued

• Subtractive primary colors filter out all light when

combined.

• Pigments can be produced by combining subtractive

colors (magenta, yellow, and cyan).



Chapter 13

Subtractive Color Mixing




Wavelength’s of visible spectrum

• Use nm remember 1X10-9

• Red = 700 nm

• Blue=450 nm

• Violet = 420 nm

• Why is the sky blue?






Chapter 13

Polarization of Light Waves

• Linear polarization is the alignment of electro-

magnetic waves in such a way that the vibrations of

the electric fields in each of the waves are parallel to

each other.

• Light can be linearly polarized through transmission.

• The line along which light is polarized is called the

transmission axis of that substance.



Chapter 13

Linearly Polarized Light




Chapter 13

Aligned and Crossed Polarizing Filters


Crossed FiltersAligned Filters



Quantification of Polarization with two films

• Malus’s Law

• I2=I1 cos2θ

• Or

• I2=I1(cos θ)2

• I2=Light intensity passing through the second film

• I1 =Light intensity entering the first film.

• Θ=angle of separation between the films




Chapter 13

Polarization of Light Waves

• Light can also be polarized

by reflection and scattering.



Chapter 13

Polarization by Reflection and Scattering




Multiple Choice

1. Which equation is correct for calculating the focal

point of a spherical mirror?

A. 1/f = 1/p – 1/q

B. 1/f = 1/p + 1/q

C. 1/p = 1/f + 1/q

D. 1/q = 1/f + 1/p

Standardized Test PrepChapter 13



Multiple Choice, continued

1. Which equation is correct for calculating the focal

point of a spherical mirror?

A. 1/f = 1/p – 1/q

B. 1/f = 1/p + 1/q

C. 1/p = 1/f + 1/q

D. 1/q = 1/f + 1/p





2. Which of the following statements is true about the

speeds of gamma rays and radio waves in a

vacuum?

F. Gamma rays travel faster than radio waves.

G. Radio rays travel faster than gamma rays.

H. Gamma rays and radio waves travel at the same

speed in a vacuum.

J. The speed of gamma rays and radio waves in a

vacuum depends on their frequencies.





2. Which of the following statements is true about the

speeds of gamma rays and radio waves in a

vacuum?

F. Gamma rays travel faster than radio waves.

G. Radio rays travel faster than gamma rays.

H. Gamma rays and radio waves travel at the same

speed in a vacuum.

J. The speed of gamma rays and radio waves in a

vacuum depends on their frequencies.





3. Which of the following correctly states the law of reflection?A. The angle between an incident ray of light and the normal to the mirror’s surface equals the angle between the mirror’s surface and the reflected light ray.

B. The angle between an incident ray of light and the mirror’s surface equals the angle between the normal to the mirror’s surface and the reflected light ray.

C. The angle between an incident ray of light and the normal to the mirror’s surface equals the angle between the normal and the reflected light ray.

D. The angle between an incident ray of light and the normal to the mirror’s surface is complementary to the angle between the normal and the reflected light ray.





3. Which of the following correctly states the law of reflection?A. The angle between an incident ray of light and the normal to the mirror’s surface equals the angle between the mirror’s surface and the reflected light ray.

B. The angle between an incident ray of light and the mirror’s surface equals the angle between the normal to the mirror’s surface and the reflected light ray.

C. The angle between an incident ray of light and the normal to the mirror’s surface equals the angle between the normal and the reflected light ray.

D. The angle between an incident ray of light and the normal to the mirror’s surface is complementary to the angle between the normal and the reflected light ray.





4. Which of the following processes does not linearly

polarize light?

F. scattering

G. transmission

H. refraction

J. reflection





4. Which of the following processes does not linearly

polarize light?

F. scattering

G. transmission

H. refraction

J. reflection





5. Which kind of mirror is

shown in the ray

diagram?

A. flat

B. convex

C. concave

D. Not enough

information is available

to draw a conclusion.


Use the ray diagram below

to answer questions 5–7.




5. Which kind of mirror is

shown in the ray

diagram?

A. flat

B. convex

C. concave

D. Not enough

information is available

to draw a conclusion.







6. What is true of the

image formed by the

mirror?

F. virtual, upright, and diminished

G. real, inverted, and diminished

H. virtual, upright, and enlarged

J. real, inverted, and enlarged







6. What is true of the

image formed by the

mirror?

F. virtual, upright, and diminished

G. real, inverted, and diminished

H. virtual, upright, and enlarged

J. real, inverted, and enlarged







7. What is the focal length

of the mirror?

A. –10.0 cm

B. –4.30 cm

C. 4.30 cm

D. 10.0 cm







7. What is the focal length

of the mirror?

A. –10.0 cm

B. –4.30 cm

C. 4.30 cm

D. 10.0 cm







8. Which combination of primary additive colors will

produce magenta-colored light?

F. green and blue

G. red and blue

H. green and red

J. cyan and yellow





8. Which combination of primary additive colors will

produce magenta-colored light?

F. green and blue

G. red and blue

H. green and red

J. cyan and yellow





9. What is the frequency of an infrared wave that has a

vacuum wavelength of 5.5 µm?

A. 165 Hz

B. 5.5 1010 Hz

C. 5.5 1013 Hz

D. 5.5 1016 Hz





9. What is the frequency of an infrared wave that has a

vacuum wavelength of 5.5 µm?

A. 165 Hz

B. 5.5 1010 Hz

C. 5.5 1013 Hz

D. 5.5 1016 Hz





10. If the distance from a light source is increased by a

factor of 5, by how many times brighter does the light

appear?

F. 25

G. 5

H. 1/5

J. 1/25





10. If the distance from a light source is increased by a

factor of 5, by how many times brighter does the light

appear?

F. 25

G. 5

H. 1/5

J. 1/25




Short Response

11. White light is passed through a filter that allows only

yellow, green, and blue light to pass through it. This

light is then shone on a piece of blue fabric and on a

piece of red fabric. Which colors do the two pieces of

fabric appear to have under this light?




Short Response, continued

11. White light is passed through a filter that allows only

yellow, green, and blue light to pass through it. This

light is then shone on a piece of blue fabric and on a

piece of red fabric. Which colors do the two pieces of

fabric appear to have under this light?

Answer:

The blue fabric appears blue. The red fabric

appears black.





12. The clothing department of a store has a mirror that consists of three flat mirrors, each arranged so that a person standing before the mirrors can see how an article of clothing looks from the side and back. Suppose a ray from a flashlight is shined on the mirror on the left. If the incident ray makes an angle of 65º with respect to the normal to the mirror’s surface, what will be the angle q of the ray reflected from the mirror on the right?





12. The clothing department of a store has a mirror that consists of three flat mirrors, each arranged so that a person standing before the mirrors can see how an article of clothing looks from the side and back. Suppose a ray from a flashlight is shined on the mirror on the left. If the incident ray makes an angle of 65º with respect to the normal to the mirror’s surface, what will be the angle q of the ray reflected from the mirror on the right?


Answer: 65º




13. X rays emitted from material around compact

massive stars, such as neutron stars or black holes,

serve to help locate and identify such objects. What

would be the wavelength of the X rays emitted from

material around such an object if the X rays have a

frequency of 5.0 1019 Hz?





13. X rays emitted from material around compact

massive stars, such as neutron stars or black holes,

serve to help locate and identify such objects. What

would be the wavelength of the X rays emitted from

material around such an object if the X rays have a

frequency of 5.0 1019 Hz?

Answer: 6.0 10–12 m = 6.0 pm




Extended Response

14. Explain how you can use a piece of polarizing

plastic to determine if light is linearly polarized.




Extended Response, continued

14. Explain how you can use a piece of polarizing

plastic to determine if light is linearly polarized.

Answer: Polarized light will pass through the plastic

when the transmission axis of the plastic is parallel

with the light’s plane of polarization. Rotating the

plastic 90º will prevent the polarized light from

passing through the plastic, so the plastic appears

dark. If light is not linearly polarized, rotating the

plastic 90º will have no effect on the light’s intensity.





15. What is the distance

between the surface of

the mirror and the

image?


Use the ray diagram below to

answer questions 15–19.

A candle is placed 30.0 cm from

the reflecting surface of a

concave mirror. The radius of

curvature of the mirror is 20.0 cm.




15. What is the distance

between the surface of

the mirror and the

image?

Answer: 15.0 cm











16. What is the focal

length of the mirror?











16. What is the focal

length of the mirror?

Answer: 10.0 cm











17. What is the

magnification of the

image?











17. What is the

magnification of the

image?

Answer: –0.500











18. If the candle is 12 cm

tall, what is the image

height?











18. If the candle is 12 cm

tall, what is the image

height?

Answer: –6.0 cm











19. Is the image real or

virtual? Is it upright or

inverted?











19. Is the image real or

virtual? Is it upright or

inverted?

Answer: real; inverted








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An electromagnetic wave oscillating electric and magnetic ...3. Which of the following correctly...

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