RRHS Physics Unit 4 Slide #1 Module 4.1 – Introduction to Waves Waves are caused by vibrations,...

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.


Period and Frequency

• Period – time it takes for one complete cycle t

TN

• Frequency – number of cycles per unit timeN

ft

1fT


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


Wave Terminology• Wave – a disturbance that transfers energy

Waves

MechanicalWaves

ElectromagneticWaves

TransverseWaves

LongitudinalWaves


Mechanical Waves• Transverse Wave

• Longitudinal Wave


Wave Terminology


1D Wave Properties

Wave Simulator

• Wave speed depends only on the medium

Heavy medium Light MediumUpright reflected wave

Light medium Heavy MediumInverted reflected wave

http://phet.colorado.edu/en/simulation/wave-on-a-string


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


SuperpositionConstructive Interference Destructive Interference

Node


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


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•


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


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


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


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


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?


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


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?


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


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


Reflection

qi

normal line

incident ray

qi

normal line

incident ray

qr

reflected ray

qi = q

r

Law of Reflection


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)


Refraction

qi

normal line

incident ray

qr

partial reflected ray

refracted ray

qR


Diffraction• Diffraction – waves bend around the edges of the barrier


Diffraction


Interference

nodal line

antinodal line

constructive interference (antinodal lines)

destructive interference (nodal lines)


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


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


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.


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.


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


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.


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.


Sound Waves• Mechanical Wave (longitudinal)• Series of compressions and rarefactions


Sound Properties

Sound Property Physical Wave Characteristic

Loudness Amplitude

Pitch Frequency

Quality Wave Form (multiple resonant frequencies)


Range of Hearing

• Human Hearing

20 Hz 20000 Hzinfrasonic ultrasonic

• Range decreases as we age

• Many animals can hear above our range of hearing


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


Speed of Sound in Air

331 0.59sv T

# o

s

vMach

v

Mach Number

Supersonic – Mach Number is greater than one


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?


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


Doppler Effect

Observer A hears a higher frequency


Sonic Boom


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


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


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)


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.


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


Resonant Lengths

11 4L

32 4L

53 4L


Example 1Calculate the first 3 resonant lengths for a 512 Hz tuning fork, assuming that the air temperature is 20.0oC.


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


Fixed Length Closed Air Column

114L 5

34L 324L

FundamentalFrequency

FirstOvertone

SecondOvertone

Multiple frequencies produced


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.


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


Overtones and Harmonics• Harmonics – Whole number multiples of the fundamental

frequency • For closed air columns, only the odd number

harmonics are present.


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?


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


Open Air Columns• Open at both ends• Antinode required at both ends

12

11 2L

2L 3

3 2L


Fixed Length Open Air Columns

112L 2L 3

32L

FundamentalFrequency

FirstOvertone

SecondOvertone


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.


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


Overtones and Harmonics

• For open air columns, all of the harmonics are present.


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?


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


Beat Frequency


Beat Frequency

• Beat frequency is the absolute value of the difference between the two sources:

2 1| |beatf f f


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


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


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


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


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


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.


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


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


Electromagnetic Spectrum

• Current model of light incorporates both waves and particles


Reflection of Light

Regular Reflection

Smooth Surface Rough Surface

Diffuse reflection


Plane Mirror

Virtual Image


Speed of Light• Speed accurately determined around 1900 by Michelson

83.00 10 /c m s


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


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.


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


Coin in a Cup Demo

Can See Coin Cannot See Coin Can See Coin because of refraction


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


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


Snell’s Law

qi

normal line

incident ray

refracted ray

qR

incident medium

refracting medium

sin sini i R Rn nq q


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

qq

sin sin

(1.00)sin 30.0 sin 20.0

1.46

i i R R

R

R

n n

n

n

q q


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

qq

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


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



Critical Angle


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.


Example 3What is the critical angle for the interface between air and water?

1.33

1.00

90

?

i

R

R

c

n

n

qq

sin sin

1.33sin (1.00)sin 90

48.8

i i R R

c

c

n nq qq

q


Fibre Optics


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

qq

sin sin

sin 42.9 1.00sin 90

1.47

i i R R

i

i

n n

n

n

q q


Double-Slit Diffraction

Central Maximum


Double Slit Diffraction

dark

Dark Spot – Destructive Interference

Bright Spot – Constructive Interference


Small Scale

sinn

d

q

Path Difference = nλBright spot


Large Scale

q

x

central maximum

bright spot

L

tanx

Lq


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


Diffraction Gratings

sinn

d

q


Diffraction Gratings

Double Slit

Diffraction Grating


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


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?


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


Concave Mirrors

FCPrincipal Axis

2

rf


Ray Diagrams• Consider an object in front of a concave mirror.

FC Principal Axis


Rule 1• Any ray drawn parallel to the principal axis will reflect

through the focal point.

FC Principal Axis


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


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


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.


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.


Check Your Learningc.

FC

FC

Image is inverted, real, and the same size.


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


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.


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.


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


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).


Convex Mirrors

Principal Axis

F C

• Rays of light diverge as if coming from the focal point


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


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.



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


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).


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


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


Convex (Converging) Lens

F

• Lens is thicker in the middle


Concave (Diverging) Lens

• Lens is thinner in the middle

F


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'


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'


Rule 3• A ray directed to the center of the lens continues in a

straight line.

F

F'

• image is real, inverted, and smaller


Check Your LearningLocate and describe the image in each of the following diagrams:

a. F

F'

F

F'

• Image is upright, larger, and virtual


Check Your Learningb.

F

F'

F

F'

• Image is inverted, larger, and real.


Check Your Learningc.

F

F'

F'

F

• Image is upright, smaller and virtual


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)


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.


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?


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


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


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.


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


The image is upright (because hi is positive), smaller (because hi is smaller than ho), and virtual (because di is negative).


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


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.


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


Module Summary• The power of a lens can be calculated from the focal

length1

Pf

RRHS Physics Unit 4 Slide #1 Module 4.1 – Introduction to Waves Waves are caused by vibrations,...

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Transcript of RRHS Physics Unit 4 Slide #1 Module 4.1 – Introduction to Waves Waves are caused by vibrations,...