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RRHS Physics Unit 4Slide #1
Module 4.1 – Introduction to Waves
Waves are caused by vibrations, such as objects undergoing simple harmonic motion. Although water waves, sound waves, springs, and light all seem very different, they share many properties that can be explained using a wave model. This module introduces trainees to some general wave properties which will later be applied to specific types of waves.
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RRHS Physics Unit 4Slide #2
Period and Frequency
• Period – time it takes for one complete cycle t
TN
• Frequency – number of cycles per unit timeN
ft
1fT
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RRHS Physics Unit 4Slide #3
ExampleA child swings back and forth on a swing 15 times in 30.0 s. Determine the frequency and period of the swing.
15
30.0
?
?
N
t s
T
f
30.0
15
2.0
tT
N
s
15
30.0
0.50
Nf
t
Hz
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RRHS Physics Unit 4Slide #4
Wave Terminology• Wave – a disturbance that transfers energy
Waves
MechanicalWaves
ElectromagneticWaves
TransverseWaves
LongitudinalWaves
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RRHS Physics Unit 4Slide #5
Mechanical Waves• Transverse Wave
• Longitudinal Wave
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RRHS Physics Unit 4Slide #6
Wave Terminology
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RRHS Physics Unit 4Slide #7
1D Wave Properties
Wave Simulator
• Wave speed depends only on the medium
Heavy medium Light MediumUpright reflected wave
Light medium Heavy MediumInverted reflected wave
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RRHS Physics Unit 4Slide #8
Superposition1. When waves collide they simply pass through one
another unchanged. They continue on as if there were no interaction.
2. While the waves overlap, they temporarily produce a resultant wave due to interference. The displacement of the medium is the sum of the displacements of each component wave Constructive Interference Destructive Interference
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RRHS Physics Unit 4Slide #9
SuperpositionConstructive Interference Destructive Interference
Node
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RRHS Physics Unit 4Slide #10
Resonance and Standing Waves
• Resonance – achieved when energy is added to a system at the same frequency as its natural frequency;
• Results in maximum amplitude.
• Standing Wave – example of resonance
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RRHS Physics Unit 4Slide #11
Check Your Learning1. The ? of a wave depends only on the medium in which it is travelling.
a. Frequency
b. Period
c. Speed
d. Wavelength
(c) speed
2. When a wave passes from one medium to another, the ? must stay the same.
a. Amplitude
b. Frequency
c. Speed
d. Wavelength
(b) frequency•
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RRHS Physics Unit 4Slide #12
Check Your Learning3. A wave in which the medium moves parallel to the medium is called a ?
wave.
a. Electromagnetic
b. Longitudinal
c. Mechanical
d. Transverse
(b) longitudinal
4. The vertical distance from the top of a crest to the bottom of a trough is 34.0 cm. The amplitude of this wave is
a. 8.5 cm
b. 17.0 cm
c. 34.0 cm
d. 68.0 cm
(b) 17.0 cm
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RRHS Physics Unit 4Slide #13
Check Your Learning5. A pulse goes into a medium that is less dense. The reflected pulse is
a. Faster
b. Inverted
c. Larger
d. Upright
(d) upright
6. Resonance occurs when one object causes a second object to vibrate. The second object must have the same natural
a. Amplitude
b. Frequency
c. Speed
d. Wavelength
(b) frequency
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RRHS Physics Unit 4Slide #14
Check Your Learning7. A wave source has a period of 0.20 s. What is the frequency?
a. 0.20 Hz
b. 1.0 Hz
c. 5.0 Hz
d. 20. Hz
(c) 5.0 Hz
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RRHS Physics Unit 4Slide #15
Wave Equation• Wave velocity – the velocity at which the wave crests (or
any other part of the wave) move; • not the same as the velocity of a particle of the medium.
dv
t
1
vT
T
v f
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RRHS Physics Unit 4Slide #16
Example 1
A hiker shouts toward a vertical cliff 685 m away. The echo is heard 4.00 s later. The wavelength of the sound is 0.750 m.a. What is the speed of sound in air?
b. What is the frequency?
c. What is the period of the wave?
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RRHS Physics Unit 4Slide #17
Solutiona. 685
2.00
0.750
?
d m
t s
m
v
685
2.00
343 /
dv
t
m s
b.
343 (0.750)
457
v f
f
f Hz
c.
3
1
1
457
2.19 10
Tf
s
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RRHS Physics Unit 4Slide #18
Example 2Water waves have a wavelength of 3.2 m and a frequency of 0.78 Hz.
a. At what rate does a stationary boat bob up and down?
b. If the boat starts moving into the waves (in the opposite direction to) at a speed of 5.0 m/s, at what rate will the boat bob up and down now?
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RRHS Physics Unit 4Slide #19
Solutiona. Since the boat is not moving, it will bob up and down at
the same frequency as the waves.
0.78f Hz
b. 3.2
0.78
5.0 /
?
w
w
b
b
m
f Hz
v m s
f
(3.2)(0.78)
2.5 /
w w wv f
m s
5.0 2.5
7.5 /
rel b wv v v
m s
7.5 (3.2)
2.3
rel b
b
b
v f
f
f Hz
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RRHS Physics Unit 4Slide #20
Check Your LearningWater waves in a lake travel 4.4 m in 1.8 s. The period of oscillation is 1.2 s. What is the speed and wavelength of the water waves?
4.4
1.8
1.2
?
?
d m
t s
T s
v
4.4
1.8
2.4 /
dv
t
m s
1
1
1.20.83
fT
Hz
2.4 (0.83)
2.9
v f
m
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RRHS Physics Unit 4Slide #21
Reflection
qi
normal line
incident ray
qi
normal line
incident ray
qr
reflected ray
qi = q
r
Law of Reflection
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RRHS Physics Unit 4Slide #22
Refraction• Refraction – change in speed while going from one
medium to another results in a change of direction of the wave
pavement (fast) mud (slow)
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RRHS Physics Unit 4Slide #23
Refraction
qi
normal line
incident ray
qr
partial reflected ray
refracted ray
qR
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RRHS Physics Unit 4Slide #24
Diffraction• Diffraction – waves bend around the edges of the barrier
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RRHS Physics Unit 4Slide #25
Diffraction
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RRHS Physics Unit 4Slide #26
Interference
nodal line
antinodal line
constructive interference (antinodal lines)
destructive interference (nodal lines)
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RRHS Physics Unit 4Slide #27
Check Your Learning1. The direction a wave moves is
a. Parallel to the wavefronts.
b. Perpendicular to the wavefronts.
c. In the direction of increasing density.
d. In the direction of increasing wavelength.
(b) Perpendicular to the wavefronts.
2. The process by which a wave bounces off an obstacle in its path is called
a. Diffraction
b. Reflection
c. Refraction
d. Superposition
(b) Reflection
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RRHS Physics Unit 4Slide #28
Check Your Learning3. The bending of waves as they go through a small opening is called
a. Diffraction
b. Reflection
c. Refraction
d. Superposition
(a) Diffraction
4. The bending of waves as they go from one medium to a new medium is called
a. Diffraction
b. Reflection
c. Refraction
d. Superposition
(c) Refraction
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RRHS Physics Unit 4Slide #29
Check Your Learning5. When two waves interfere with one another, the word interfere means
a. One wave prevents the other wave from finishing its cycle.
b. One wave stops moving while the other passes.
c. The motion we observe is the sum of the motions of the two individual waves.
d. The wave with the larger amplitude grows and the wave with the smaller amplitude shrinks.
(c) The motion we observe is the sum of the motions of the two individual waves.
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RRHS Physics Unit 4Slide #30
Module SummaryIn this module you have learned that• Mechanical waves need a medium while electromagnetic
ones do not.• Mechanical waves can be transverse or longitudinal.• The correct terminology to use when describing waves,
such as: period, frequency, crest, trough, amplitude, wavelength
• The speed of a wave depends only upon the medium in which it is travelling.
• Waves will be both reflected and transmitted at a boundary.
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RRHS Physics Unit 4Slide #31
Module Summary• The frequency of a wave does not change when going
from one medium to another one.• When waves interfere with one another, they can
interfere constructively or destructively before passing through one another unchanged.
• A standing wave is an example of resonance in a medium.
• All waves are governed by the wave equation
v f
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RRHS Physics Unit 4Slide #32
Module Summary• All two-dimensional waves obey the law of reflection,
which states that the angle of incidence is equal to the angle of reflection
• All two-dimensional waves undergo refraction, diffraction, and interference.
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RRHS Physics Unit 4Slide #33
Module 4.2 – Sound Waves
In this module, the wave properties studied in module 7.2 will be looked at in greater depth as they apply to sound waves. Although these properties can be observed in general with all waves, they are often easily observable and can be demonstrated using these specific types of waves. Sound waves are used in a variety of techniques in exploring for oil and minerals.
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RRHS Physics Unit 4Slide #34
Sound Waves• Mechanical Wave (longitudinal)• Series of compressions and rarefactions
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RRHS Physics Unit 4Slide #35
Sound Properties
Sound Property Physical Wave Characteristic
Loudness Amplitude
Pitch Frequency
Quality Wave Form (multiple resonant frequencies)
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RRHS Physics Unit 4Slide #36
Range of Hearing
• Human Hearing
20 Hz 20000 Hzinfrasonic ultrasonic
• Range decreases as we age
• Many animals can hear above our range of hearing
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RRHS Physics Unit 4Slide #37
Speed of Sound• Mechanical waves need a medium• Medium determines speed of sound
Material Speed of Sound (m/s)
Air (at 0oC and 101 kPa) 331
Helium (at 0oC and 101 kPa)
965
Fresh water (at 20oC) 1482
Copper 5010
Steel 5960
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RRHS Physics Unit 4Slide #38
Speed of Sound in Air
331 0.59sv T
# o
s
vMach
v
Mach Number
Supersonic – Mach Number is greater than one
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RRHS Physics Unit 4Slide #39
ExampleA plane is flying at a speed of 855 m/s. If the air temperature is 12oC,
a. What is the speed of sound?
b. What is the Mach number for the plane?
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RRHS Physics Unit 4Slide #40
Solutiona.
12
?s
T C
v
331 0.59
331 0.59(12)
338 /
sv T
m s
b. 855 /
338 /
# ?
o
s
v m s
v m s
Mach
#
855
338
2.53
o
s
vMach
v
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RRHS Physics Unit 4Slide #41
Doppler Effect
Observer A hears a higher frequency
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RRHS Physics Unit 4Slide #42
Sonic Boom
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RRHS Physics Unit 4Slide #43
Check Your Learning1. Which of the following is NOT a property of sound?
a. Amplitude
b. Frequency
c. Mass
d. Wavelength
(c) Mass
2. The average human ear cannot hear frequencies above
a. 20 Hz
b. 2000 Hz
c. 20000 Hz
d. 200000Hz
(c) 20000 Hz
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RRHS Physics Unit 4Slide #44
Check Your Learning3. When we describe something as supersonic we mean it is
a. Faster than the speed of sound
b. Higher in frequency than 20000 Hz
c. Lower in frequency than 20 Hz
d. Slower than the speed of sound
(a) Faster than the speed of sound
4. When the amplitude of a sound wave increases,
a. The wavelength of the sound decreases
b. The sound gets louder
c. The pitch increases
d. The speed of sound increases
(b) The sound gets louder
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RRHS Physics Unit 4Slide #45
Check Your Learning5. Sound is a longitudinal wave because
a. The oscillations in pressure are in the same direction as the wave moves.
b. The oscillations in pressure are perpendicular to the direction that the wave moves.
c. The wavelength is long compared to light waves.
d. The wavelength is always longer than the amplitude.
(a) The oscillations in pressure are in the same direction as the wave moves.
6. The wavelength of a sound wave can be calculated by
a. Multiplying the amplitude by the frequency
b. Dividing the amplitude by the frequency
c. Multiplying the speed by the frequency
d. Dividing the speed by the frequency
(d) Dividing the speed by the frequency (the wave equation)
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RRHS Physics Unit 4Slide #46
Check Your Learning7. The speed of sound in air at 7.0oC is
a. 331 m/s
b. 332 m/s
c. 335 m/s
d. 338 m/s
(c) 335 m/s (using v=331+0.59T)
8. A person is behind an ambulance as it moves away from her. The pitch of the sound that she hears is
a. Lower than if the ambulance was stationary.
b. The same as if the ambulance was stationary.
c. Higher than if the ambulance was stationary.
(a) Lower than if the ambulance was stationary, since the wavelength will be larger behind the ambulance.
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RRHS Physics Unit 4Slide #47
Closed Air Columns
• Standing wave in a closed air column requires a node at the closed end and an antinode at the open end of the air column
14
34
Full Standing Wave
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RRHS Physics Unit 4Slide #48
Resonant Lengths
11 4L
32 4L
53 4L
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RRHS Physics Unit 4Slide #49
Example 1Calculate the first 3 resonant lengths for a 512 Hz tuning fork, assuming that the air temperature is 20.0oC.
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RRHS Physics Unit 4Slide #50
Solution
1
2
3
20.0
512
?
?
?
T C
f Hz
L
L
L
331 0.59
331 0.59(20.0)
343 /
sv T
m s
343 (512)
0.670
v f
m
11 4
14 (0.670)
0.168
L
m
32 4
34 (0.670)
0.503
L
m
53 4
54 (0.670)
0.838
L
m
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RRHS Physics Unit 4Slide #51
Fixed Length Closed Air Column
114L 5
34L 324L
FundamentalFrequency
FirstOvertone
SecondOvertone
Multiple frequencies produced
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RRHS Physics Unit 4Slide #52
Example 2A 15.0 cm test tube is blown across so that it resonates. If the air temperature is 20.0oC, calculate the fundamental frequency and the first two overtones.
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RRHS Physics Unit 4Slide #53
Solution
1
2
3
15.0 0.150
20.0
?
?
?
L cm m
T C
f
f
f
331 0.59
331 0.59(20.0)
343 /
sv T
m s
114
114
1
0.150
0.600
L
m
324
324
2
0.150
0.200
L
m
534
534
3
0.150
0.120
L
m
1 1
1
1
343 (0.600)
572
v f
f
f Hz
2 2
2
2
343 (0.200)
1720
v f
f
f Hz
3 3
3
3
343 (0.120)
2860
v f
f
f Hz
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RRHS Physics Unit 4Slide #54
Overtones and Harmonics• Harmonics – Whole number multiples of the fundamental
frequency • For closed air columns, only the odd number
harmonics are present.
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RRHS Physics Unit 4Slide #55
Check Your LearningAn organ pipe is 80.0 cm long. If the temperature is 23 ºC , what are the fundamental frequency and first three audible overtones if the pipe is closed at one end?
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RRHS Physics Unit 4Slide #56
Check Your Learning
1
2
3
4
23
80.0
?
T C
L cm
f
f
f
f
331 0.59
331 0.59(23)
345 /
sv T
m s
114
114
1
0.800
3.20
L
m
1 1
1
1
345 (3.20)
108
v f
f
f Hz
2 1
3 1
4 1
3 324
5 540
7 756
f f Hz
f f Hz
f f Hz
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RRHS Physics Unit 4Slide #57
Open Air Columns• Open at both ends• Antinode required at both ends
12
11 2L
2L 3
3 2L
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RRHS Physics Unit 4Slide #58
Fixed Length Open Air Columns
112L 2L 3
32L
FundamentalFrequency
FirstOvertone
SecondOvertone
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RRHS Physics Unit 4Slide #59
Example 3Assuming an air temperature of 20.0oC, calculate the fundamental frequency and the first two overtones for a 30.0 cm long open air column.
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RRHS Physics Unit 4Slide #60
Solution
1
2
3
20.0
30.0 0.300
?
?
?
T C
L cm m
f
f
f
331 0.59
331 0.59(20.0)
343 /
sv T
m s
112
112
1
0.300
0.600
L
m
2
2
2
0.300
0.300
L
m
332
332
3
0.300
0.200
L
m
1 1
1
1
343 (0.600)
572
v f
f
f Hz
2 2
2
2
343 (0.300)
1140
v f
f
f Hz
3 3
3
3
343 (0.200)
1720
v f
f
f Hz
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RRHS Physics Unit 4Slide #61
Overtones and Harmonics
• For open air columns, all of the harmonics are present.
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RRHS Physics Unit 4Slide #62
Check Your LearningAn organ pipe is 80.0 cm long. If the temperature is 23°C, what are the fundamental frequency and first three audible overtones if the pipe is open at both ends?
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RRHS Physics Unit 4Slide #63
Check Your Learning
1
2
3
4
23
80.0
?
T C
L cm
f
f
f
f
331 0.59
331 0.59(23)
345 /
sv T
m s
112
112
1
0.800
1.60
L
m
1 1
1
1
345 (1.60)
216
v f
f
f Hz
2 1
3 1
4 1
2 432
3 648
4 864
f f Hz
f f Hz
f f Hz
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RRHS Physics Unit 4Slide #64
Beat Frequency
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RRHS Physics Unit 4Slide #65
Beat Frequency
• Beat frequency is the absolute value of the difference between the two sources:
2 1| |beatf f f
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RRHS Physics Unit 4Slide #66
ExampleA 512.0 Hz tuning fork is sounded at the same time as a key on a piano. You count 23 beats over 8.0 s. What are the possible frequencies of the piano key?
1 512.0
23
8.0beats
f Hz
N
t s
23
8.02.9
beats
Nf
t
Hz
2 1
2
| |
2.9 512.0beatf f f
f
2
2
2.9 512.0
514.9
f
f Hz
2
2
2.9 512.0
509.1
f
f Hz
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RRHS Physics Unit 4Slide #67
Check Your Learning A guitar string produces a beat frequency of 4 Hz when sounded with a 350 Hz tuning fork and a beat frequency of 9 Hz when sounded with a 355 Hz tuning fork. What is the frequency of the string?
Using the first beat frequency, the possible frequencies of the string are 346 Hz or 354 Hz. Using the second beat frequency, the possible frequencies of the string are 346 Hz or 364 Hz. Since the only frequency in common is 346 Hz, this must be the frequency of the string.
346f Hz
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RRHS Physics Unit 4Slide #68
Module SummaryIn this module you learned that • Sound waves are longitudinal mechanical waves.• Sounds can be distinguished by loudness, pitch, and
quality.• Sound travels through air with a speed given by
331 0.59sv T
• The mach number of an object can be calculated using
# o
s
vMach
v
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RRHS Physics Unit 4Slide #69
Module Summary• The Doppler Effect and sonic booms can be explained
using wave theory.• Air columns resonate at their natural frequencies.• Closed air columns resonate at their fundamental
frequency when
14L
• Open air columns resonate at their fundamental frequency when
12L
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RRHS Physics Unit 4Slide #70
Module Summary• All of the harmonics are present in open air columns.• Only the odd harmonics are present in closed air
columns.• When two frequencies are close but not the exact same,
beats will be heard with a frequency of
2 1| |beatf f f
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RRHS Physics Unit 4Slide #71
Module 4.3 – Electromagnetic Waves
The wave model of light will be applied to electromagnetic waves to further study wave properties such as reflection, refraction, and diffraction. A brief introduction to geometric optics is also included. An understanding of light is important as it applies to one of the principle means through which we obtain information, using both instruments and our sense of sight.
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RRHS Physics Unit 4Slide #72
Light as a Wave• Two basic methods of transferring energy:
– Particles – for example, a baseball travelling through the air has kinetic energy which can be transferred to another object in a collision.
– Waves – water waves transfer energy to the shore and cause erosion.
• Newton proposed a particle model• Christian Huygens proposed a wave model• Newton’s model initially accepted
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RRHS Physics Unit 4Slide #73
Light as a Wave• Huygens model began to gain more acceptance for the
following reasons. – double slit experiment to show that light passing through two
slits demonstrated the same interference pattern as two sources of water waves;
– speed of light was shown to be lower in water than in air; this supported Huygen's theory of refraction and contradicted Newton's theory of refraction.
• Huygens wave model replaced by Electromagnetic Wave Model
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RRHS Physics Unit 4Slide #74
Electromagnetic Spectrum
• Current model of light incorporates both waves and particles
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RRHS Physics Unit 4Slide #75
Reflection of Light
Regular Reflection
Smooth Surface Rough Surface
Diffuse reflection
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RRHS Physics Unit 4Slide #76
Plane Mirror
Virtual Image
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RRHS Physics Unit 4Slide #77
Speed of Light• Speed accurately determined around 1900 by Michelson
83.00 10 /c m s
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RRHS Physics Unit 4Slide #78
ExampleUsing the accepted value for the speed of light, calculate the minimum frequency that would have been needed for light to be reflected into the eye of the observer in Michelson’s apparatus.
8
5
3.00 10 /
70. 7.0 10
?
v m s
d km m
f
48
4
7.0 103.00 10
2.3 10
dv
t
t
t s
18
4 182.33 10
0.00186
t T
T
T s
1
1
0.00186
540
fT
Hz
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RRHS Physics Unit 4Slide #79
Check Your Learning1. Why is it better when the pages of a book are rough rather than
smooth and glossy?
Rough pages allow light to undergo diffuse reflection, meaning the light is not all reflected in the same direction. This reduces glare from the page.
2. A particular nearsighted person can only see clearly 0.50 m from their face. How far from a plane mirror should they be to see their image clearly?
They should be 0.25 m (or less) from the mirror. Because their image is the same distance behind the mirror as they are in front of it, the total distance from the person to their image will be 0.50 m.
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RRHS Physics Unit 4Slide #80
Check Your Learning3. What is the angle of incidence if the angle between a reflected ray
and the mirror is 34o?
If the angle between the reflected ray and the mirror is 34o, the angle of reflection (the angle with the normal) is 56o (90-34). The angle of incidence must therefore be 56o.
4. The moon is 3.85×108 m away from the earth. How long does it take light reflected from the moon to reach the earth?
8
8
3.85 10
3.00 10 /
?
d m
v m s
t
88 3.85 10
3.00 10
1.28
dv
t
t
t s
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RRHS Physics Unit 4Slide #81
Coin in a Cup Demo
Can See Coin Cannot See Coin Can See Coin because of refraction
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RRHS Physics Unit 4Slide #82
Index of Refraction• Index of refraction (n) defined as
cn
v
Substance Index of RefractionVacuum 1.00Air 1.00Water 1.33Ethyl alcohol 1.36Quartz 1.46Plexiglass 1.51Crown Glass 1.52Flint Glass 1.65Diamond 2.42
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RRHS Physics Unit 4Slide #83
Example 1Calculate the speed of light in water.
8
1.33
3.00 10 /
?
water
water
n
c m s
v
8
8
3.00 101.33
2.26 10 /
waterwater
water
water
cn
v
v
v m s
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RRHS Physics Unit 4Slide #84
Snell’s Law
qi
normal line
incident ray
refracted ray
qR
incident medium
refracting medium
sin sini i R Rn nq q
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RRHS Physics Unit 4Slide #85
Example 2A ray of light (travelling in air) has an angle of incidence of 30.0o on a block of quartz and an angle of refraction of 20.0o. What is the index of refraction for this block of quartz?
1.00
30.0
20.0
?
i
i
R
R
n
n
sin sin
(1.00)sin 30.0 sin 20.0
1.46
i i R R
R
R
n n
n
n
q q
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RRHS Physics Unit 4Slide #86
Check Your LearningThe speed of light in a clear plastic is 1.90×108 m/s. A ray of light travelling through air enters the plastic with an angle of incidence of 22°. At what angle is the ray refracted?
8
8
1.90 10 /
3.00 10 /
1.00
22
?
plastic
i
i
R
v m s
c m s
n
8
8
3.00 10
1.90 101.58
plasticplastic
cn
v
sin sin
1.00sin 22 1.58sin
14
i i R R
R
R
n nq qq
q
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RRHS Physics Unit 4Slide #87
Total Internal Reflection• Total internal reflection can only occur going from a high
index of refraction to a lower index of refraction
higher index of refraction
lower index of refraction
higher index of refraction
lower index of refraction
Critical Angle
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RRHS Physics Unit 4Slide #88
Total Internal Reflection
higher index of refraction
lower index of refraction
Total Internal Reflection
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RRHS Physics Unit 4Slide #89
Total Internal Reflection
Two conditions required for total internal reflection to occur:
1. The light must be travelling from a higher index of refraction to a lower index of refraction.
2. The angle of incidence must be greater than the critical angle, θc, associated with the two materials.
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RRHS Physics Unit 4Slide #90
Example 3What is the critical angle for the interface between air and water?
1.33
1.00
90
?
i
R
R
c
n
n
sin sin
1.33sin (1.00)sin 90
48.8
i i R R
c
c
n nq qq
q
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RRHS Physics Unit 4Slide #91
Fibre Optics
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RRHS Physics Unit 4Slide #92
Check Your LearningThe critical angle for a certain liquid-air surface is 42.9o. What is the index of refraction for the liquid?
?
1.00
42.9
90
i
R
c
R
n
n
sin sin
sin 42.9 1.00sin 90
1.47
i i R R
i
i
n n
n
n
q q
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RRHS Physics Unit 4Slide #93
Double-Slit Diffraction
Central Maximum
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RRHS Physics Unit 4Slide #94
Double Slit Diffraction
dark
Dark Spot – Destructive Interference
Bright Spot – Constructive Interference
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RRHS Physics Unit 4Slide #95
Small Scale
sinn
d
q
Path Difference = nλBright spot
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RRHS Physics Unit 4Slide #96
Large Scale
q
x
central maximum
bright spot
L
tanx
Lq
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RRHS Physics Unit 4Slide #97
Example 1Red light with a wavelength of 685 nm is shone through two small slits. An interference pattern is observed on a screen that is 4.2 m away. The distance between the central maximum and the second order bright spot is 3.2 cm. What was the distance between the two slits?
7
3.2 0.032
4.2
685 6.85 10
2
?
x cm m
L m
nm m
n
d
tan
0.032
4.20.437
x
Lq
7
4
sin
(2)(6.85 10 )sin 0.437
1.8 10
0.18
n
d
d
d m
mm
q
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RRHS Physics Unit 4Slide #98
Diffraction Gratings
sinn
d
q
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RRHS Physics Unit 4Slide #99
Diffraction Gratings
Double Slit
Diffraction Grating
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RRHS Physics Unit 4Slide #100
Example 2Calculate the angle between the central maximum and the first order bright spot for a diffraction grating that has 3800 lines per centimetre on it if monochromatic light with a wavelength of 420 nm is shone on it.
4
1
3800 lines/cm
2.63 10
d
cm
7
4 6
420 4.2 10
2.63 10 2.63 10
1
?
nm m
d cm m
n
q
7
6
sin
(1)(4.2 10 )
2.63 10
9.2
n
d
q
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RRHS Physics Unit 4Slide #101
Check Your LearningLight with a wavelength of 542 nm is shone through a diffraction grating. The third order bright spot is observed to be 74.0 cm away from the central maximum on a screen 8.20 m away. How many lines per cm does the grating have?
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RRHS Physics Unit 4Slide #102
Check Your Learning7542 5.42 10
3
74.0 0.740
8.20
?
nm m
n
x cm m
L m
d
tan
0.740
8.205.16
x
Lq
q
7
5
3
sin
(3)(5.42 10 )sin 5.16
1.81 10
1.81 10
n
d
d
d m
cm
q
3
1552 lines/cm
1.81 10 cm
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RRHS Physics Unit 4Slide #103
Concave Mirrors
FCPrincipal Axis
2
rf
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RRHS Physics Unit 4Slide #104
Ray Diagrams• Consider an object in front of a concave mirror.
FC Principal Axis
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RRHS Physics Unit 4Slide #105
Rule 1• Any ray drawn parallel to the principal axis will reflect
through the focal point.
FC Principal Axis
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RRHS Physics Unit 4Slide #106
Rule 2• Because of the law of reflection, the opposite must also
be true. Any ray drawn through the focal point must reflect parallel to the principal axis.
FC Principal Axis
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RRHS Physics Unit 4Slide #107
Rule 3• Any ray that goes through the center of curvature hits the
mirror at a 90o angle, and so reflects back on itself.
FC Principal Axis
Image is real, inverted, larger
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RRHS Physics Unit 4Slide #108
Check Your Learning
Locate and describe the images of the object in each of the following diagrams:a.
FC
FCThe image is inverted, real, and smaller.
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RRHS Physics Unit 4Slide #109
Check Your Learningb.
FC
FC
Notice that in this case, the reflected rays are spreading apart and will not cross. It is necessary to extend the rays behind the mirror until they cross. This image is larger, upright, and virtual.
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RRHS Physics Unit 4Slide #110
Check Your Learningc.
FC
FC
Image is inverted, real, and the same size.
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RRHS Physics Unit 4Slide #111
Mirror Equation
FC
O
O¢
I
I¢
A
B
do
di
ho
hi
o o
i i
h d
h d o o
i
h d f
h f
1 1 1
o o
i
o o i i
o o i i
i o i o i o
i o
d d f
d f
fd d d fd
fd d d fd
fd d fd d fd d
d f d
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RRHS Physics Unit 4Slide #112
Mirror Equation
1 1 1
o if d d i i
o o
h dm
h d
• Image height hi is positive if upright, negative if inverted (relative to the object)
• The image distance di and the object distance do positive if on the reflecting side of the mirror (real) and negative if behind the mirror (virtual)
• The focal length f is positive if on the reflecting side of the mirror, which will always be true for concave mirrors.
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RRHS Physics Unit 4Slide #113
Example 1A concave mirror has a radius of curvature of 12.0 cm. A 1.2 cm tall object is placed a distance of 8.2 cm away from the mirror.
a. Locate the image.
b. Calculate the height of the image.
c. Describe the image.
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RRHS Physics Unit 4Slide #114
Solutiona.
FC Principal Axis
12.0
1.2
8.2
?
o
o
i
r cm
h cm
d cm
d
212.0
26.0
rf
cm
1 1 1
1 1 1
6.0 8.2
22
o i
i
i
f d d
d
d cm
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RRHS Physics Unit 4Slide #115
Solutionb.
22
1.2 8.2
3.2
i i
o o
i
i
h d
h d
h
h cm
c. The image is inverted (because hi is negative), larger (because hi is bigger) and real (because di is positive).
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RRHS Physics Unit 4Slide #116
Convex Mirrors
Principal Axis
F C
• Rays of light diverge as if coming from the focal point
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RRHS Physics Unit 4Slide #117
Convex Mirrors• Rules for drawing rays diagrams that we learned before
are very similar for convex mirrors;• instead of our incoming rays going through the focal
point or the centre of curvature, they simply go toward them (since they are on the other side of the mirror).
Principal Axis
F C
• Image is always upright, smaller, virtual
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RRHS Physics Unit 4Slide #118
Example 2A convex mirror has a radius of curvature of 12.0 cm. A 1.2 cm tall object is placed a distance of 8.2 cm away from the mirror.
a. Locate the image.
b. Calculate the height of the image.
c. Describe the image.
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RRHS Physics Unit 4Slide #119
Solutiona. Remember, since the mirror is convex, the radius of
curvature and the focal length must be negative.
12.0
8.2
1.2
?
o
o
i
r cm
d cm
h cm
d
212.0
26.0
rf
cm
1 1 1
1 1 1
6.0 8.2
3.5
o i
i
i
f d d
d
d cm
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RRHS Physics Unit 4Slide #120
Solutionb.
3.5
1.2 8.2
0.51
i i
o o
i
i
h d
h d
h
h cm
c. The image is upright (because hi is positive), smaller (because hi is smaller) and virtual (because di is negative).
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RRHS Physics Unit 4Slide #121
Check Your LearningA 5.3 cm tall object is placed 6.4 cm away from a spherical mirror. A virtual image is formed 4.2 cm from the mirror.
a. What is the focal length of the mirror?
6.4
4.2
?
o
i
d cm
d cm
f
1 1 1
1 1
6.4 4.2
12
o if d d
f cm
Since the image is virtual, di must be negative
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RRHS Physics Unit 4Slide #122
Check Your Learningb. What kind of mirror is it?
Because the focal length was calculated to be negative, the mirror is convex.
c. What is the height of the image?
6.4
4.2
5.3
?
o
i
o
i
d cm
d cm
h cm
h
4.2
5.3 6.4
3.5
i i
o o
i
i
h d
h d
h
h cm
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RRHS Physics Unit 4Slide #123
Convex (Converging) Lens
F
• Lens is thicker in the middle
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RRHS Physics Unit 4Slide #124
Concave (Diverging) Lens
• Lens is thinner in the middle
F
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RRHS Physics Unit 4Slide #125
Ray Diagrams – Rule 1• A ray drawn parallel to the axis is refracted by the lens
so that it passes along a line through the focal point.
F
F'
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RRHS Physics Unit 4Slide #126
Rule 2• A ray drawn on a line passing through the other focal
point F’ emerges from the lens parallel to the axis.
F
F'
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RRHS Physics Unit 4Slide #127
Rule 3• A ray directed to the center of the lens continues in a
straight line.
F
F'
• image is real, inverted, and smaller
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RRHS Physics Unit 4Slide #128
Check Your LearningLocate and describe the image in each of the following diagrams:
a. F
F'
F
F'
• Image is upright, larger, and virtual
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RRHS Physics Unit 4Slide #129
Check Your Learningb.
F
F'
F
F'
• Image is inverted, larger, and real.
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RRHS Physics Unit 4Slide #130
Check Your Learningc.
F
F'
F'
F
• Image is upright, smaller and virtual
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RRHS Physics Unit 4Slide #131
Lens Equation
1 1 1
o if d d i i
o o
h dm
h d
• Power of a lens defined as
1P
f
• If f is in metres, P is in diopters (D)
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RRHS Physics Unit 4Slide #132
Lens Equation• The object distance do is positive if it is on the same side
of the lens from which the light is coming (in other words, if it is real)
• The image distance di is positive if it is on the opposite side of the lens from which the light is coming (in other words, if it is real); it is negative if it is on the same side of the lens from which the light is coming (in other words, if it is virtual)
• The height of the image hi is positive if upright and negative if inverted relative to the object.
• The focal length is positive for a convex (converging) lens and negative for a concave (diverging) lens.
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RRHS Physics Unit 4Slide #133
Example 3A certain lens focuses an object 22.5 cm away as an image 33.0 cm on the other side of the lens.
a. Is the image real or virtual?
b. What type of lens is it and what is its focal length?
c. What is the power of the lens?
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RRHS Physics Unit 4Slide #134
Solutiona. Because the image is on the other side of the lens, it
must be real.
b. Because the image is real, the lens must be convex, or diverging (since concave lenses will always give virtual images). A positive focal length will confirm this.
22.5
33.0
?
o
i
d cm
d cm
f
1 1 1
1 1
22.5 33.0
13.4
o if d d
f cm
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RRHS Physics Unit 4Slide #135
Solutionc. To calculate the power, the focal length must be in
metres.
13.4 0.134
?
f cm m
P
1
1
0.134
7.46
Pf
D
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RRHS Physics Unit 4Slide #136
Check Your LearningA concave lens has a focal point 20.0 cm away from the lens. A 2.1 m tall object is placed 3.0 m away from it.
a. Where is the image?
20.0 0.200
3.0
?o
i
f cm m
d m
d
1 1 1
1 1 1
0.200 3.0
0.19
o i
i
i
f d d
d
d m
• The image is located 0.19 m from the lens on the same side as the object.
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RRHS Physics Unit 4Slide #137
Check Your Learningb. How big is the image?
3.0
0.19
2.1
?
o
i
o
i
d m
d m
h m
h
0.19
2.1 3.0
0.13
i i
o o
i
i
h d
h d
h
h m
c. Describe the image.
The image is upright (because hi is positive), smaller (because hi is smaller than ho), and virtual (because di is negative).
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RRHS Physics Unit 4Slide #138
Module SummaryIn this module you learned that
• Light exhibits many wave properties and can be modeled in many situations as a wave.
• Light can undergo both regular reflection (mirrors) or diffuse reflection (rough surfaces)
• Light is just a small part of the electromagnetic spectrum.• The speed of light in a vacuum is
83.00 10 /c m s
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RRHS Physics Unit 4Slide #139
Module Summary• The index of refraction for a medium can be calculated
using cn
v
• The angle of incidence and angle of refraction for a refracting ray of light are related by Snell’s Law
sin sini i R Rn nq q
• When going from a high index of refraction to a low index of refraction, there is a critical angle beyond which light cannot refract. For all angles of incidence greater than this critical angle, total internal reflection occurs.
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RRHS Physics Unit 4Slide #140
Module Summary• Diffraction and interference of light for double slits and
diffraction gratings can be modeled using the equations
sinn
d
q tanx
Lq
• Ray diagrams can be used to locate images in spherical mirrors and lenses.
• The following equations can be applied to problems involving spherical mirrors and lenses:
1 1 1
o if d d i i
o o
h dm
h d
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RRHS Physics Unit 4Slide #141
Module Summary• The power of a lens can be calculated from the focal
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