Transcript No Slide Title

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.
• Light is a form of electromagnetic radiation.
• The electromagnetic spectrum includes more than
visible light.
Characteristics of Light
The Electromagnetic Spectrum
Characteristics of Light
Electromagnetic Waves, continued
• Electromagnetic waves vary depending on frequency
and wavelength.
• All electromagnetic waves move at the speed of light.
The speed of light, c, equals
c = 3.00  108 m/s
• Wave Speed Equation
c = fl
speed of light = frequency  wavelength
Characteristics of Light
Electromagnetic Waves, continued
• Waves can be approximated
as rays. This approach to
analyzing waves is called
Huygens’ principle.
• Lines drawn tangent to the
crest (or trough) of a wave
are called wave fronts.
• In the ray approximation,
lines, called rays, are drawn
perpendicular to the wave
front.
Characteristics of Light
Electromagnetic Waves, continued
• Illuminance decreases as the square of the distance
from the source.
• The rate at which light is emitted from a source is
called the luminous flux and is measured in lumens
(lm).
Flat Mirrors
Reflection of Light
• Reflection is the change in direction of an
electromagnetic wave at a surface that causes it to
move away from the surface.
• The texture of a surface affects how it reflects light.
– Diffuse reflection is reflection from a rough, texture
surface such as paper or unpolished wood.
– Specular reflection is reflection from a smooth,
shiny surface such as a mirror or a water surface.
Flat Mirrors
Reflection of Light, continued
• The angle of incidence is the the angle between a
ray that strikes a surface and the line perpendicular
to that surface at the point of contact.
• The angle of reflection is the angle formed by the
line perpendicular to a surface and the direction in
which a reflected ray moves.
• The angle of incidence and the angle of reflection are
always equal.
Flat Mirrors
Flat Mirrors
• Flat mirrors form virtual images that are the same
distance from the mirror’s surface as the object is.
• The image formed by rays that appear to come from
the image point behind the mirror—but never really
do—is called a virtual image.
• A virtual image can never be displayed on a physical
surface.
Flat Mirrors
Image Formation by a Flat Mirror
Curved Mirrors
Concave Spherical Mirrors
• A concave spherical mirror is a mirror whose
reflecting surface is a segment of the inside of a
sphere.
• Concave mirrors can be used to form real images.
• A real image is an image formed when rays of light
actually pass through a point on the image. Real
images can be projected onto a screen.
Curved Mirrors
Image Formation by a Concave Spherical Mirror
Curved Mirrors
Concave Spherical Mirrors, continued
• The Mirror Equation relates object distance (p),
image distance (q), and focal length (f) of a spherical
mirror.
1 1 1
 
p q f
1
1
1


object distance image distance focal length
Curved Mirrors
Concave Spherical Mirrors, continued
• The Equation for Magnification relates image height
or distance to object height or distance, respectively.
h'
q
–
h
p
image height
image distance
magnification 
=–
object height
object distance
M
Curved Mirrors
Concave Spherical Mirrors, continued
• Ray diagrams can be used for checking values
calculated from the mirror and magnification
equations for concave spherical mirrors.
• Concave mirrors can produce both real and virtual
images.
Curved Mirrors
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.
Curved Mirrors
Sample Problem, continued
Imaging with Concave Mirrors
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.
Curved Mirrors
Sample Problem, continued
Imaging with Concave Mirrors
2. Draw a ray diagram using the rules for drawing
reference rays.
Curved Mirrors
Sample Problem, continued
Imaging with Concave Mirrors
3. Use the mirror equation to relate the object and
image distances to the focal length.
1 1 1
 
p q f
4. Use the magnification equation in terms of object
and image distances.
q
M–
p
Curved Mirrors
Sample Problem, continued
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
Curved Mirrors
Sample Problem, continued
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

–

–

q 10.0 cm 30.0 cm
cm
cm
cm
q  15 cm
q
15 cm
M– –
 –0.50
p
30.0 cm
Curved Mirrors
Sample Problem, continued
6. Evaluate your answer in terms of the image
location and size.
The image appears between the focal point (10.0
cm) and the center of curvature (20.0 cm), as
confirmed by the ray diagram. The image is smaller
than the object and inverted (–1 < M < 0), as is also
confirmed by the ray diagram. The image is
therefore real.
Curved Mirrors
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.
Curved Mirrors
Image Formation by a Convex Spherical Mirror
Curved Mirrors
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.
Curved Mirrors
Sample Problem, continued
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=?
Curved Mirrors
Sample Problem, continued
Convex Mirrors
Diagram:
Curved Mirrors
Sample Problem, continued
Convex Mirrors
2. Plan
Choose an equation or situation: Use the mirror
equation and the magnification formula.
1 1 1
h'
q
 
and
M –
p q f
h
p
Rearrange the equation to isolate the unknown:
1 1 1
 –
p f q
and
p
h  – h'
q
Curved Mirrors
Sample Problem, continued
Convex Mirrors
3. Calculate
Substitute the values into the equation and solve:
1
1
1

–
p –8.00 cm –4.44 cm
1 –0.125 –0.225 0.100

–

p
cm
cm
cm
p  10.0 cm
Curved Mirrors
Sample Problem, continued
Convex Mirrors
3. Calculate, continued
Substitute the values for p and q to find the magnification of the image.
q
–4.44 cm
M– –
M  0.444
p
10.0 cm
Substitute the values for p, q, and h’ to find the height
of the object.
p
10.0 cm
h  – h'  –
(2.50 cm)
q
–4.44 cm
h  5.63 cm
Curved Mirrors
Parabolic Mirrors
• Images created by spherical mirrors suffer from
spherical aberration.
• Spherical aberration occurs when parallel rays far
from the principal axis converge away from the
mirrors focal point.
• Parabolic mirrors eliminate spherical aberration. All
parallel rays converge at the focal point of a parabolic
mirror.
Curved Mirrors
Spherical Aberration and Parabolic Mirrors
Color and Polarization
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).
Color and Polarization
Color, continued
• Subtractive primary colors filter out all light when
combined.
• Pigments can be produced by combining subtractive
colors (magenta, yellow, and cyan).
Color and Polarization
Polarization of Light Waves
• Linear polarization is the alignment of electromagnetic 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.
Color and Polarization
Linearly Polarized Light
Color and Polarization
Aligned and Crossed Polarizing Filters
Aligned Filters
Crossed Filters
Color and Polarization
Polarization of Light Waves
• Light can be polarized by reflection and scattering.
• At a particular angle, reflected light is polarized
horizontally.
• The sunlight scattered by air molecules is polarized
for an observer on Earth’s surface.
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
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
Multiple Choice, continued
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.
Multiple Choice, continued
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.
Multiple Choice, continued
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.
Multiple Choice, continued
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.
Multiple Choice, continued
4. Which of the following processes does not linearly
polarize light?
F. scattering
G. transmission
H. refraction
J. reflection
Multiple Choice, continued
4. Which of the following processes does not linearly
polarize light?
F. scattering
G. transmission
H. refraction
J. reflection
Multiple Choice, continued
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.
Multiple Choice, continued
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.
Multiple Choice, continued
Use the ray diagram below
to answer questions 5–7.
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
Multiple Choice, continued
Use the ray diagram below
to answer questions 5–7.
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
Multiple Choice, continued
Use the ray diagram below
to answer questions 5–7.
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
Multiple Choice, continued
Use the ray diagram below
to answer questions 5–7.
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
Multiple Choice, continued
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
Multiple Choice, continued
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
Multiple Choice, continued
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
Multiple Choice, continued
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
Multiple Choice, continued
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
Multiple Choice, continued
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.
Short Response, continued
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?
Short Response, continued
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
Short Response, continued
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?
Short Response, continued
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.
Extended Response, continued
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?
Extended Response, continued
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
Extended Response, continued
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.
16. What is the focal
length of the mirror?
Extended Response, continued
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.
16. What is the focal
length of the mirror?
Answer: 10.0 cm
Extended Response, continued
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.
17. What is the
magnification of the
image?
Extended Response, continued
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.
17. What is the
magnification of the
image?
Answer: –0.500
Extended Response, continued
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.
18. If the candle is 12 cm
tall, what is the image
height?
Extended Response, continued
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.
18. If the candle is 12 cm
tall, what is the image
height?
Answer: –6.0 cm
Extended Response, continued
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.
19. Is the image real or
virtual? Is it upright or
inverted?
Extended Response, continued
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.
19. Is the image real or
virtual? Is it upright or
inverted?
Answer: real; inverted
Refraction
Refraction of Light
• The bending of light as it travels from one medium to
another is call refraction.
• As a light ray travels from one medium into another
medium where its speed is different, the light ray will
change its direction unless it travels along the
normal.
Refraction
Refraction of Light, continued
• Refraction can be explained in terms of the wave
model of light.
• The speed of light in a vacuum, c, is an important
constant used by physicists.
• Inside of other mediums, such as air, glass, or
water, the speed of light is different and is usually
less than c.
Refraction
The Law of Refraction
• The index of refraction for a substance is the ratio
of the speed of light in a vacuum to the speed of light
in that substance.
c
n
v
speed of light in a vacuum
index of refraction 
speed of light in medium
Refraction
Indices of Refraction for Various Substances
Refraction
The Law of Refraction, continued
• When light passes from a medium with a smaller
index of refraction to one with a larger index of
refraction (like from air to glass), the ray bends
toward the normal.
• When light passes from a medium with a larger
index of refraction to one with a smaller index of
refraction (like from glass to air), the ray bends
away from the normal.
Refraction
Refraction
Refraction
The Law of Refraction, continued
• Objects appear to be in different positions due to
refraction.
• Snell’s Law determines the angle of refraction.
ni sin i  nr sinr
index of refraction of first medium  sine of the angle of incidence =
index of refraction of second medium  sine of the angle of refraction
Refraction
Image Position for Objects in Different Media
Refraction
Sample Problem
Snell’s Law
A light ray of wavelength 589 nm (produced by a
sodium lamp) traveling through air strikes a smooth,
flat slab of crown glass at an angle of 30.0º to the
normal. Find the angle of refraction, r.
Refraction
Sample Problem, continued
Snell’s Law
Given:
i = 30.0º
ni = 1.00
Unknown: r = ?
Use the equation for Snell’s law.
nr = 1.52
ni sin i  nr sinr
 ni

1.00

r  sin   sini    sin–1 
sin30.0º


 1.52

 nr

–1
r  19.2º
Thin Lenses
Types of Lenses
• A lens is a transparent object that refracts light rays
such that they converge or diverge to create an
image.
• A lens that is thicker in the middle than it is at the rim
is an example of a converging lens.
• A lens that is thinner in the middle than at the rim is
an example of a diverging lens.
Thin Lenses
Types of Lenses, continued
• The focal point is the location where the image of an
object at an infinite distance from a converging lens if
focused.
• Lenses have a focal point on each side of the lens.
• The distance from the focal point to the center of the
lens is called the focal length, f.
Thin Lenses
Lenses and Focal Length
Thin Lenses
Types of Lenses, continued
• Ray diagrams of thin-lens systems help identify
image height and location.
• Rules for drawing reference rays
Thin Lenses
Characteristics of Lenses
• Converging lenses can produce real or virtual images
of real objects.
• The image produced by a converging lens is real and
inverted when the object is outside the focal point.
• The image produced by a converging lens is virtual
and upright when the object is inside the focal point.
Thin Lenses
Characteristics of Lenses, continued
• Diverging lenses produce virtual images from real
objects.
• The image created by a diverging lens is always a
virtual, smaller image.
Thin Lenses
The Thin-Lens Equation and Magnification
• The equation that relates object and image distances
for a lens is call the thin-lens equation.
• It is derived using the assumption that the lens is very
thin.
1 1 1
 
p q f
1
distance from object to lens

1
distance from image to lens

1
focal length
Thin Lenses
The Thin-Lens Equation and Magnification,
continued
• Magnification of a lens depends on object and image
distances.
h'
q
M
–
h
p
image height
distance from image to lens
magnification =
–
object height
distance from object to lens
Thin Lenses
The Thin-Lens Equation and Magnification,
continued
• If close attention is
given to the sign
conventions defined in
the table, then the
magnification will
describe the image’s
size and orientation.
Thin Lenses
Sample Problem
Lenses
An object is placed 30.0 cm in front of a converging
lens and then 12.5 cm in front of a diverging lens.
Both lenses have a focal length of 10.0 cm. For both
cases, find the image distance and the magnification.
Describe the images.
Thin Lenses
Sample Problem, continued
Lenses
1. Define
Given:
Unknown:
fconverging = 10.0 cm
pconverging = 30.0 cm
fdiverging = –10.0 cm
pdiverging = 12.5 cm
qconverging = ?
Mconverging = ?
qdiverging = ?
Mdiverging = ?
Thin Lenses
Sample Problem, continued
Lenses
1. Define, continued
Diagrams:
Thin Lenses
Sample Problem, continued
Lenses
2. Plan
Choose an equation or situation: The thin-lens
equation can be used to find the image distance, and
the equation for magnification will serve to describe
the size and orientation of the image.
1 1 1
 
p q f
q
M–
p
Thin Lenses
Sample Problem, continued
Lenses
2. Plan, continued
Rearrange the equation to isolate the unknown:
1 1 1
 –
q f p
Thin Lenses
Sample Problem, continued
Lenses
3. Calculate
For the converging lens:
1 1 1
1
1
2
 – 
–

q f p 10.0 cm 30.0 cm 30.0 cm
q  15.0 cm
q
15.0 cm
M– –
p
30.0 cm
M  –0.500
Thin Lenses
Sample Problem, continued
Lenses
3. Calculate, continued
For the diverging lens:
1 1 1
1
1
22.5
 – 
–

q f p –10.0 cm 12.5 cm 125 cm
q  –5.56 cm
q
–5.56 cm
M– –
p
12.5 cm
M  0.445
Thin Lenses
Sample Problem, continued
Lenses
4. Evaluate
These values and signs for the converging lens
indicate a real, inverted, smaller image. This is
expected because the object distance is longer than
twice the focal length of the converging lens. The
values and signs for the diverging lens indicate a
virtual, upright, smaller image formed inside the focal
point. This is the only kind of image diverging lenses
form.
Thin Lenses
Eyeglasses and Contact Lenses
• The transparent front of the eye, called the cornea,
acts like a lens.
• The eye also contains a crystalline lens, that further
refracts light toward the light-sensitive back of the
eye, called the retina.
• Two conditions, myopia and hyperopia, occur when
light is not focused properly retina. Converging and
diverging lenses can be used to correct these
conditions.
Thin Lenses
Farsighted and Nearsighted
Thin Lenses
Combination of Thin Lenses
• An image formed by a lens can be used as the object
for a second lens.
• Compound microscopes use two converging lenses.
Greater magnification can be achieved by combining
two or more lenses.
• Refracting telescopes also use two converging
lenses.
Optical Phenomena
Total Internal Reflection
• Total internal reflection can occur when light moves
along a path from a medium with a higher index of
refraction to one with a lower index of refraction.
• At the critical angle, refracted light makes an angle
of 90º with the normal.
• Above the critical angle, total internal reflection
occurs and light is completely reflected within a
substance.
Optical Phenomena
Total Internal Reflection, continued
• Snell’s law can be used to find the critical angle.
nr
sinC 
ni
for ni  nr
index of refraction of second medium
sine  critical angle  
index of refraction of first medium
• Total internal reflection occurs only if the index of
refraction of the first medium is greater than the index
of refraction of the second medium.
Optical Phenomena
Total Internal Reflection
Optical Phenomena
Atmospheric Refraction
• Refracted light can
create a mirage.
• A mirage is produced by
the bending of light rays
in the atmosphere
where there are large
temperature differences
between the ground
and the air.
Optical Phenomena
Dispersion
• Dispersion is the
process of separating
polychromatic light into
its component
wavelengths.
• White light passed
through a prism
produces a visible
spectrum through
dispersion.
Optical Phenomena
Rainbows
Optical Phenomena
Lens Aberrations
• Chromatic aberration
is the focusing of
different colors of light
at different distances
behind a lens.
• Chromatic aberration
occurs because the
index of refraction
varies for different
wavelengths of light.
Multiple Choice
1. How is light affected by an increase in the index of
refraction?
A. Its frequency increases.
B. Its frequency decreases.
C. Its speed increases.
D. Its speed decreases.
Multiple Choice, continued
1. How is light affected by an increase in the index of
refraction?
A. Its frequency increases.
B. Its frequency decreases.
C. Its speed increases.
D. Its speed decreases.
Multiple Choice, continued
2. Which of the following conditions is not necessary for
refraction to occur?
F. Both the incident and refracting substances must
be transparent.
G. Both substances must have different indices of
refraction.
H. The light must have only one wavelength.
J. The light must enter at an angle greater than 0°
with respect to the normal.
Multiple Choice, continued
2. Which of the following conditions is not necessary for
refraction to occur?
F. Both the incident and refracting substances must
be transparent.
G. Both substances must have different indices of
refraction.
H. The light must have only one wavelength.
J. The light must enter at an angle greater than 0°
with respect to the normal.
Multiple Choice, continued
Use the ray diagram
below to answer
questions 3–4.
3. What is the focal length
of the lens?
A. -12.5 cm
B. -8.33 cm
C. 8.33 cm
D. 12.5 cm
Multiple Choice, continued
Use the ray diagram
below to answer
questions 3–4.
3. What is the focal length
of the lens?
A. -12.5 cm
B. -8.33 cm
C. 8.33 cm
D. 12.5 cm
Multiple Choice, continued
Use the ray diagram
below to answer
questions 3–4.
4. What is true of the
image formed by the
lens?
F. real, inverted, and
enlarged
G. real, inverted, and
diminished
H. virtual, upright, and
enlarged
J. virtual, upright, and
diminished
Multiple Choice, continued
Use the ray diagram
below to answer
questions 3–4.
4. What is true of the
image formed by the
lens?
F. real, inverted, and
enlarged
G. real, inverted, and
diminished
H. virtual, upright, and
enlarged
J. virtual, upright, and
diminished
Multiple Choice, continued
5. A block of flint glass with an index of refraction of 1.66
is immersed in oil with an index of refraction of 1.33.
How does the critical angle for a refracted light ray in
the glass vary from when the glass is surrounded by
air?
A. It remains unchanged.
B. It increases.
C. It decreases.
D. No total internal reflection takes place when the
glass is placed in the oil.
Multiple Choice, continued
5. A block of flint glass with an index of refraction of 1.66
is immersed in oil with an index of refraction of 1.33.
How does the critical angle for a refracted light ray in
the glass vary from when the glass is surrounded by
air?
A. It remains unchanged.
B. It increases.
C. It decreases.
D. No total internal reflection takes place when the
glass is placed in the oil.
Multiple Choice, continued
6. Which color of light is most refracted during
dispersion by a prism?
F. red
G. yellow
H. green
J. violet
Multiple Choice, continued
6. Which color of light is most refracted during
dispersion by a prism?
F. red
G. yellow
H. green
J. violet
Multiple Choice, continued
7. If an object in air is viewed from beneath the surface
of water below, where does the object appear to be?
A. The object appears above its true position.
B. The object appears exactly at its true position.
C. The object appears below its true position.
D. The object cannot be viewed from beneath the
water’s surface.
Multiple Choice, continued
7. If an object in air is viewed from beneath the surface
of water below, where does the object appear to be?
A. The object appears above its true position.
B. The object appears exactly at its true position.
C. The object appears below its true position.
D. The object cannot be viewed from beneath the
water’s surface.
Multiple Choice, continued
8. The phenomenon called “looming” is similar to a
mirage, except that the inverted image appears
above the object instead of below it.What must be
true if looming is to occur?
F. The temperature of the air must increase with
distance above the surface.
G. The temperature of the air must decrease with
distance above the surface.
H. The mass of the air must increase with distance
above the surface.
J. The mass of the air must increase with distance
above the surface.
Multiple Choice, continued
8. The phenomenon called “looming” is similar to a
mirage, except that the inverted image appears
above the object instead of below it.What must be
true if looming is to occur?
F. The temperature of the air must increase with
distance above the surface.
G. The temperature of the air must decrease with
distance above the surface.
H. The mass of the air must increase with distance
above the surface.
J. The mass of the air must increase with distance
above the surface.
Multiple Choice, continued
9. Light with a vacuum wavelength of 500.0 nm passes
into benzene, which has an index of refraction of 1.5.
What is the wavelength of the light within the
benzene?
A. 0.0013 nm
B. 0.0030 nm
C. 330 nm
D. 750 nm
Multiple Choice, continued
9. Light with a vacuum wavelength of 500.0 nm passes
into benzene, which has an index of refraction of 1.5.
What is the wavelength of the light within the
benzene?
A. 0.0013 nm
B. 0.0030 nm
C. 330 nm
D. 750 nm
Multiple Choice, continued
10. Which of the following is not a necessary condition
for seeing a magnified image with a lens?
F. The object and image are on the same side of the
lens.
G. The lens must be converging.
H. The observer must be placed within the focal
length of the lens.
J. The object must be placed within the focal length
of the lens.
Multiple Choice, continued
10. Which of the following is not a necessary condition
for seeing a magnified image with a lens?
F. The object and image are on the same side of the
lens.
G. The lens must be converging.
H. The observer must be placed within the focal
length of the lens.
J. The object must be placed within the focal length
of the lens.
Short Answer
11. In both microscopes and telescopes, at least two
converging lenses are used: one for the objective and
one for the eyepiece. These lenses must be
positioned in such a way that the final image is virtual
and very much enlarged. In terms of the focal points
of the two lenses, how must the lenses be
positioned?
Short Answer, continued
11. In both microscopes and telescopes, at least two
converging lenses are used: one for the objective and
one for the eyepiece. These lenses must be
positioned in such a way that the final image is virtual
and very much enlarged. In terms of the focal points
of the two lenses, how must the lenses be
positioned?
Answer: The focal point of the objective must lie within
the focal point of the eyepiece.
Short Answer, continued
12. A beam of light passes from the fused quartz of a
bottle (n = 1.46) into the ethyl alcohol (n = 1.36) that
is contained inside the bottle. If the beam of the light
inside the quartz makes an angle of 25.0° with
respect to the normal of both substances, at what
angle to the normal will the light enter the alcohol?
Short Answer, continued
12. A beam of light passes from the fused quartz of a
bottle (n = 1.46) into the ethyl alcohol (n = 1.36) that
is contained inside the bottle. If the beam of the light
inside the quartz makes an angle of 25.0° with
respect to the normal of both substances, at what
angle to the normal will the light enter the alcohol?
Answer: 27.0°
Short Answer, continued
13. A layer of glycerine (n = 1.47) covers a zircon slab
(n = 1.92). At what angle to the normal must a beam
of light pass through the zircon toward the glycerine
so that the light undergoes total internal reflection?
Short Answer, continued
13. A layer of glycerine (n = 1.47) covers a zircon slab
(n = 1.92). At what angle to the normal must a beam
of light pass through the zircon toward the glycerine
so that the light undergoes total internal reflection?
Answer: 50.0º
Extended Response
14. Explain how light passing through raindrops is
reflected and dispersed so that a rainbow is
produced. Include in your explanation why the lower
band of the rainbow is violet and the outer band is
red.
Extended Response, continued
14. Explain how light passing through raindrops is
reflected and dispersed so that a rainbow is
produced. Include in your explanation why the lower
band of the rainbow is violet and the outer band is
red.
Answer: There are three effects—a refraction, a
reflection, and then a final refraction. The light of
each wavelength in the visible spectrum is refracted
by a different amount: the red light undergoes the
least amount of refraction, and the violet light
undergoes the most. (Answer continued on next
slide.)
Extended Response, continued
14. Answer (continued): At the far side of the raindrop, the
light is internally reflected and undergoes refraction again
when it leaves the front side of the raindrop. Because of
the internal reflection, the final dispersion of the light is
such that the violet light makes an angle of 40° with the
incident ray, and the red light makes an angle of 42° with
the incident ray. For an observer, the upper edge of the
rainbow has the color of the light that bends farthest from
the incident light, so the outer band of the rainbow is red.
Similarly, the lower edge has the color of the light that
bends least from the incident light, so the inner band is
violet. The net effect is that the ray that is refracted the
most ends up closest to the incident light, that is, the
smallest angular displacement.
Extended Response, continued
Use the ray diagram below
to answer questions 15–18.
A collector wishes to
observe a coin in detail
and so places it 5.00 cm in
front of a converging lens.
An image forms 7.50 cm in
front of the lens, as shown
in the figure below.
15. What is the focal
length of the lens?
Extended Response, continued
Use the ray diagram below
to answer questions 15–18.
A collector wishes to
observe a coin in detail
and so places it 5.00 cm in
front of a converging lens.
An image forms 7.50 cm in
front of the lens, as shown
in the figure below.
15. What is the focal
length of the lens?
Answer: 15 cm
Extended Response, continued
Use the ray diagram below
to answer questions 15–18.
A collector wishes to
observe a coin in detail
and so places it 5.00 cm in
front of a converging lens.
An image forms 7.50 cm in
front of the lens, as shown
in the figure below.
16. What is the magnification of the coin’s image?
Extended Response, continued
Use the ray diagram below
to answer questions 15–18.
A collector wishes to
observe a coin in detail
and so places it 5.00 cm in
front of a converging lens.
An image forms 7.50 cm in
front of the lens, as shown
in the figure below.
16. What is the magnification of the coin’s image?
Answer: 1.5
Extended Response, continued
Use the ray diagram below
to answer questions 15–18.
A collector wishes to
observe a coin in detail
and so places it 5.00 cm in
front of a converging lens.
An image forms 7.50 cm in
front of the lens, as shown
in the figure below.
17. If the coin has a
diameter of 2.8 cm, what is
the diameter of the coin’s
image?
Extended Response, continued
Use the ray diagram below
to answer questions 15–18.
A collector wishes to
observe a coin in detail
and so places it 5.00 cm in
front of a converging lens.
An image forms 7.50 cm in
front of the lens, as shown
in the figure below.
17. If the coin has a
diameter of 2.8 cm, what is
the diameter of the coin’s
image?
Answer: 4.2 cm
Extended Response, continued
Use the ray diagram below
to answer questions 15–18.
A collector wishes to
observe a coin in detail
and so places it 5.00 cm in
front of a converging lens.
An image forms 7.50 cm in
front of the lens, as shown
in the figure below.
18. Is the coin’s image
virtual or real? upright or
inverted?
Extended Response, continued
Use the ray diagram below
to answer questions 15–18.
A collector wishes to
observe a coin in detail
and so places it 5.00 cm in
front of a converging lens.
An image forms 7.50 cm in
front of the lens, as shown
in the figure below.
18. Is the coin’s image
virtual or real? upright or
inverted?
Answer: virtual; upright
Interference
Combining Light Waves
• Interference takes place only between waves with the
same wavelength. A light source that has a single
wavelength is called monochromatic.
• In constructive interference, component waves
combine to form a resultant wave with the same
wavelength but with an amplitude that is greater than
the either of the individual component waves.
• In the case of destructive interference, the resultant
amplitude is less than the amplitude of the larger
component wave.
Interference
Interference Between Transverse Waves
Interference
Combining Light Waves, continued
• Waves must have a constant phase difference for
interference to be observed.
• Coherence is the correlation between the phases of
two or more waves.
– Sources of light for which the phase difference is
constant are said to be coherent.
– Sources of light for which the phase difference is
not constant are said to be incoherent.
Lasers
Incoherent and Coherent Light
Interference
Demonstrating Interference
• Interference can be demonstrated by passing light
through two narrow parallel slits.
• If monochromatic light is used, the light from the two
slits produces a series of bright and dark parallel
bands, or fringes, on a viewing screen.
Interference
Conditions for Interference of Light Waves
Interference
Demonstrating Interference, continued
• The location of interference
fringes can be predicted.
• The path difference is the
difference in the distance
traveled by two beams when
they are scattered in the same
direction from different points.
• The path difference equals
dsin.
Interference
Demonstrating Interference, continued
• The number assigned to interference fringes with
respect to the central bright fringe is called the order
number. The order number is represented by the
symbol m.
• The central bright fringe at q = 0 (m = 0) is called the
zeroth-order maximum, or the central maximum.
• The first maximum on either side of the central
maximum (m = 1) is called the first-order maximum.
Interference
Demonstrating Interference, continued
• Equation for constructive interference
d sin  = ±ml
m = 0, 1, 2, 3, …
The path difference between two waves =
an integer multiple of the wavelength
• Equation for destructive interference
d sin  = ±(m + 1/2)l m = 0, 1, 2, 3, …
The path difference between two waves =
an odd number of half wavelength
Interference
Sample Problem
Interference
The distance between the two slits is 0.030 mm. The
second-order bright fringe (m = 2) is measured on a
viewing screen at an angle of 2.15º from the central
maximum. Determine the wavelength of the light.
Interference
Sample Problem, continued
Interference
1. Define
Given:
d = 3.0  10–5 m
m=2
 = 2.15º
Unknown:
Diagram:
l=?
Interference
Sample Problem, continued
Interference
2. Plan
Choose an equation or situation: Use the equation
for constructive interference.
d sin  = ml
Rearrange the equation to isolate the unknown:
d sin
l
m
Interference
Sample Problem, continued
Interference
3. Calculate
Substitute the values into the equation and solve:
3.0  10

l
–5

m  sin2.15º 
2
l  5.6  10 –7 m  5.6  102 nm
l  5.6  102 nm
Interference
Sample Problem, continued
Interference
4. Evaluate
This wavelength of light is in the visible spectrum.
The wavelength corresponds to light of a yellowgreen color.
Diffraction
The Bending of Light Waves
• Diffraction is a change in the direction of a wave
when the wave encounters an obstacle, an opening,
or an edge.
• Light waves form a diffraction pattern by passing
around an obstacle or bending through a slit and
interfering with each other.
• Wavelets (as in Huygens’ principle) in a wave front
interfere with each other.
Diffraction
Destructive Interference in Single-Slit Diffraction
Diffraction
The Bending of Light Waves, continued
• In a diffraction pattern, the central maximum is twice
as wide as the secondary maxima.
• Light diffracted by an obstacle also produces a
pattern.
Diffraction
Diffraction Gratings
• A diffraction grating uses diffraction and interference
to disperse light into its component colors.
• The position of a maximum depends on the
separation of the slits in the grating, d, the order of
the maximum m,, and the wavelength of the light, l.
d sin  = ±ml
m = 0, 1, 2, 3, …
Diffraction
Constructive Interference by a Diffraction Grating
Diffraction
Sample Problem
Diffraction Gratings
Monochromatic light from a helium-neon laser (l =
632.8 nm) shines at a right angle to the surface of a
diffraction grating that contains 150 500 lines/m. Find
the angles at which one would observe the first-order
and second-order maxima.
Diffraction
Sample Problem, continued
Diffraction Gratings
1. Define
Given:
l = 632.8 nm = 6.328  10–7 m
m = 1 and 2
1
1
d

m
lines 150 500
150 500
m
Unknown:
1 = ?
2 = ?
Diffraction
Sample Problem, continued
Diffraction Gratings
1. Define, continued
Diagram:
Diffraction
Sample Problem, continued
Diffraction Gratings
2. Plan
Choose an equation or situation: Use the
equation for a diffraction grating.
d sin  = ±ml
Rearrange the equation to isolate the unknown:
 ml 
  sin 

d


–1
Diffraction
Sample Problem, continued
Diffraction Gratings
3. Calculate
Substitute the values into the equation and solve:
For the first-order maximum, m = 1:


–7

l
6.328

10
m


–1
–1
1  sin    sin 

1
d
 

m 
 150 500

1  5.465º
Diffraction
Sample Problem, continued
Diffraction Gratings
3. Calculate, continued
For m = 2:
–1  2l 
 2  sin  
 d 

–7
2
6.328

10
m

–1
 2  sin 
1

m
150 500


 2  10.98º






Diffraction
Sample Problem, continued
Diffraction Gratings
4. Evaluate
The second-order maximum is spread slightly more
than twice as far from the center as the first-order
maximum. This diffraction grating does not have high
dispersion, and it can produce spectral lines up to the
tenth-order maxima (where sin  = 0.9524).
Diffraction
Diffraction and Instrument Resolution
• The ability of an optical system to distinguish
between closely spaced objects is limited by the
wave nature of light.
• Resolving power is the ability of an optical
instrument to form separate images of two objects
that are close together.
• Resolution depends on wavelength and aperture
width. For a circular aperture of diameter D:
  1.22
l
D
Diffraction
Resolution of Two Light Sources
Lasers
Lasers and Coherence
• A laser is a device that produces coherent light at a
single wavelength.
• The word laser is an acronym of “light amplification
by stimulated emission of radiation.”
• Lasers transform other forms of energy into coherent
light.
Lasers
Applications of Lasers
• Lasers are used to measure distances with great
precision.
• Compact disc and DVD players use lasers to read
digital data on these discs.
• Lasers have many applications in medicine.
– Eye surgery
– Tumor removal
– Scar removal
Lasers
Components of a
Compact Disc Player
Multiple Choice
1. In the equations for interference, what does the term
d represent?
A. the distance from the midpoint between the two
slits to the viewing screen
B. the distance between the two slits through which a
light wave passes
C. the distance between two bright interference
fringes
D. the distance between two dark interference fringes
Multiple Choice, continued
1. In the equations for interference, what does the term
d represent?
A. the distance from the midpoint between the two
slits to the viewing screen
B. the distance between the two slits through which a
light wave passes
C. the distance between two bright interference
fringes
D. the distance between two dark interference fringes
Multiple Choice, continued
2. Which of the following must be true for two waves
with identical amplitudes and wavelengths to undergo
complete destructive interference?
F. The waves must be in phase at all times.
G. The waves must be 90º out of phase at all times.
H. The waves must be 180º out of phase at all times.
J. The waves must be 270º out of phase at all times.
Multiple Choice, continued
2. Which of the following must be true for two waves
with identical amplitudes and wavelengths to undergo
complete destructive interference?
F. The waves must be in phase at all times.
G. The waves must be 90º out of phase at all times.
H. The waves must be 180º out of phase at all times.
J. The waves must be 270º out of phase at all times.
Multiple Choice, continued
3. Which equation correctly describes the condition for
observing the third dark fringe in an interference
pattern?
A. dsin  = l/2
B. dsin  = 3l/2
C. dsin  = 5l/2
D. dsin  = 3l
Multiple Choice, continued
3. Which equation correctly describes the condition for
observing the third dark fringe in an interference
pattern?
A. dsin  = l/2
B. dsin  = 3l/2
C. dsin  = 5l/2
D. dsin  = 3l
Multiple Choice, continued
4. Why is the diffraction of sound easier to observe than
the diffraction of visible light?
F. Sound waves are easier to detect than visible light
waves.
G. Sound waves have longer wavelengths than
visible light waves and so bend more around barriers.
H. Sound waves are longitudinal waves, which
diffract more than transverse waves.
J. Sound waves have greater amplitude than visible
light waves.
Multiple Choice, continued
4. Why is the diffraction of sound easier to observe than
the diffraction of visible light?
F. Sound waves are easier to detect than visible light
waves.
G. Sound waves have longer wavelengths than
visible light waves and so bend more around barriers.
H. Sound waves are longitudinal waves, which
diffract more than transverse waves.
J. Sound waves have greater amplitude than visible
light waves.
Multiple Choice, continued
5. Monochromatic infrared waves with a wavelength of
750 nm pass through two narrow slits. If the slits are
25 µm apart, at what angle will the fourth order bright
fringe appear on a viewing screen?
A. 4.3º
B. 6.0º
C. 6.9º
D. 7.8º
Multiple Choice, continued
5. Monochromatic infrared waves with a wavelength of
750 nm pass through two narrow slits. If the slits are
25 µm apart, at what angle will the fourth order bright
fringe appear on a viewing screen?
A. 4.3º
B. 6.0º
C. 6.9º
D. 7.8º
Multiple Choice, continued
6. Monochromatic light with a wavelength of 640 nm
passes through a diffraction grating that has 5.0  104
lines/m. A bright line on a screen appears at an angle
of 11.1º from the central bright fringe.What is the
order of this bright line?
F. m = 2
G. m = 4
H. m = 6
J. m = 8
Multiple Choice, continued
6. Monochromatic light with a wavelength of 640 nm
passes through a diffraction grating that has 5.0  104
lines/m. A bright line on a screen appears at an angle
of 11.1º from the central bright fringe.What is the
order of this bright line?
F. m = 2
G. m = 4
H. m = 6
J. m = 8
Multiple Choice, continued
7. For observing the same object, how many times
better is the resolution of the telescope shown on the
left in the figure below than that of the telescope
shown on the right?
A. 4
B. 2
C. 1/2
D. 1/4
Multiple Choice, continued
7. For observing the same object, how many times
better is the resolution of the telescope shown on the
left in the figure below than that of the telescope
shown on the right?
A. 4
B. 2
C. 1/2
D. 1/4
Multiple Choice, continued
8. What steps should you employ to design a telescope
with a high degree of resolution?
F. Widen the aperture, or design the telescope to
detect light of short wavelength.
G. Narrow the aperture, or design the telescope to
detect light of short wavelength.
H. Widen the aperture, or design the telescope to
detect light of long wavelength.
J. Narrow the aperture, or design the telescope to
detect light of long wavelength.
Multiple Choice, continued
8. What steps should you employ to design a telescope
with a high degree of resolution?
F. Widen the aperture, or design the telescope to
detect light of short wavelength.
G. Narrow the aperture, or design the telescope to
detect light of short wavelength.
H. Widen the aperture, or design the telescope to
detect light of long wavelength.
J. Narrow the aperture, or design the telescope to
detect light of long wavelength.
Multiple Choice, continued
9. What is the property of a laser called that causes
coherent light to be emitted?
A. population inversion
B. light amplification
C. monochromaticity
D. stimulated emission
Multiple Choice, continued
9. What is the property of a laser called that causes
coherent light to be emitted?
A. population inversion
B. light amplification
C. monochromaticity
D. stimulated emission
Multiple Choice, continued
10. Which of the following is not an essential
component of a laser?
F. a partially transparent mirror
G. a fully reflecting mirror
H. a converging lens
J. an active medium
Multiple Choice, continued
10. Which of the following is not an essential
component of a laser?
F. a partially transparent mirror
G. a fully reflecting mirror
H. a converging lens
J. an active medium
Short Response
11. Why is laser light useful for the purposes of making
astronomical measurements and surveying?
Short Response, continued
11. Why is laser light useful for the purposes of making
astronomical measurements and surveying?
Answer: The beam does not spread out much or lose
intensity over long distances.
Short Response, continued
12. A diffraction grating used in a spectrometer causes
the third-order maximum of blue light with a
wavelength of 490 nm to form at an angle of 6.33º
from the central maximum (m = 0). What is the
separation between the lines of the grating?
Short Response, continued
12. A diffraction grating used in a spectrometer causes
the third-order maximum of blue light with a
wavelength of 490 nm to form at an angle of 6.33º
from the central maximum (m = 0). What is the
separation between the lines of the grating?
Answer: 7.5 104 lines/m = 750 lines/cm
Short Response, continued
13. Telescopes that orbit Earth provide better images of
distant objects because orbiting telescopes are more
able to operate near their theoretical resolution than
telescopes on Earth. The orbiting telescopes needed
to provide high resolution in the visible part of the
spectrum are much larger than the orbiting
telescopes that provide similar images in the
ultraviolet and X-ray portion of the spectrum. Explain
why the sizes must vary.
Short Response, continued
13. (See previous slide for question.)
Answer: The resolving power of a telescope depends
on the ratio of the wavelength to the diameter of the
aperture. Telescopes using longer wavelength
radiation (visible light) must be larger than those
using shorter wavelengths (ultraviolet, X ray) to
achieve the same resolving power.
Extended Response
14. Radio signals often reflect
from objects and
recombine at a distance.
Suppose you are moving
in a direction
perpendicular to a radio
signal source and its
reflected signal. How
would interference
between these two signals
sound on a radio
receiver?
Extended Response, continued
14. (See previous slide for
question.)
Answer: The interference
pattern for radio signals
would “appear” on a radio
receiver as an alternating
increase in signal intensity
followed by a loss of
intensity (heard as static
or “white noise”).
Extended Response, continued
Base your answers to questions
15–17 on the information below.
In each problem, show all of
your work.
A double-slit apparatus for
demonstrating interference is
constructed so that the slits are
separated by 15.0 µm. A firstorder fringe for constructive
interference appears at an
angle of 2.25° from the zerothorder (central) fringe.
15. What is the wavelength of the light?
Extended Response, continued
Base your answers to questions
15–17 on the information below.
In each problem, show all of
your work.
A double-slit apparatus for
demonstrating interference is
constructed so that the slits are
separated by 15.0 µm. A firstorder fringe for constructive
interference appears at an
angle of 2.25° from the zerothorder (central) fringe.
15. What is the wavelength of the light?
Answer: 589 nm
Extended Response, continued
Base your answers to questions
15–17 on the information below.
In each problem, show all of
your work.
A double-slit apparatus for
demonstrating interference is
constructed so that the slits are
separated by 15.0 µm. A firstorder fringe for constructive
interference appears at an
angle of 2.25° from the zerothorder (central) fringe.
16. At what angle would
the third-order (m = 3)
bright fringe appear?
Extended Response, continued
Base your answers to questions
15–17 on the information below.
In each problem, show all of
your work.
A double-slit apparatus for
demonstrating interference is
constructed so that the slits are
separated by 15.0 µm. A firstorder fringe for constructive
interference appears at an
angle of 2.25° from the zerothorder (central) fringe.
16. At what angle would
the third-order (m = 3)
bright fringe appear?
Answer: 6.77º
Extended Response, continued
Base your answers to questions
15–17 on the information below.
In each problem, show all of
your work.
A double-slit apparatus for
demonstrating interference is
constructed so that the slits are
separated by 15.0 µm. A firstorder fringe for constructive
interference appears at an
angle of 2.25° from the zerothorder (central) fringe.
17. At what angle would
the third-order (m = 3)
dark fringe appear?
Extended Response, continued
Base your answers to questions
15–17 on the information below.
In each problem, show all of
your work.
A double-slit apparatus for
demonstrating interference is
constructed so that the slits are
separated by 15.0 µm. A firstorder fringe for constructive
interference appears at an
angle of 2.25° from the zerothorder (central) fringe.
17. At what angle would
the third-order (m = 3)
dark fringe appear?
Answer: 7.90º