Unit 7 Waves AS/A Level Physics Cambridge CAIE 9702

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everything on the syllabus like you can

see in the figure

hey describing waves recall that waves

transfer energy without transferring

matter here is the graphs of a

displacement distance graph of a

progressive wave this level represents

the equilibrium position of the wave

these distances represent the amplitude

of the wave these distances represent

the wavelength of the wave so we can

conclude

the amplitude of a wave is the maximum

displacement from the equilibrium

position the amplitude represents energy

carried by the wave a wave with a higher

amplitude carries more energy the

wavelength of a wave is the distance

between consecutive identical points

such as two peaks or two troughs it is

measured in

meters here is the displacement time

graph of a progressive wave these

distances also represent presentent the

amplitude of the wave these distances

represent the time period of the wave so

we can conclude the period of a wave is

the time taken for one complete

vibration or for one wave pass a given

point it is measured in

seconds the frequency of a wave is the

number of vibrations per second or

numbers of waves that pass a given point

per second it is measured in hertz if a

water wave travel n waves in t

seconds its frequency is n / T its

period is T divided by n so frequency is

1 / period the speed of a wave is the

distance traveled by the wave front per

unit time it is measured in me/ second

we can derive the wave equation as

follows when a wave travels one complete

wavelength the distance it moves is

equal to its wavelength and the time

taken is equal to one period from the

equation speed equal the distance

divided by time substituting the

distance equals wavelength Lambda and

the time equals period T since 1 /

period is equal to

frequency so V equals Lambda F therefore

the speed of wave is equal to the

product of the wavelength and the

frequency

exam style question one the two graphs

represent the same wave graph one shows

the variation with time of the

displacement at a particular distance

graph two shows the variation with

distance of the displacement at one

instant what is the speed of the wave

from graph one we can determine the

period of the wave which is 0.5

seconds we can calculate the frequency

using f = 1 / T substituting the period

T = 0.5

seconds we get the frequency f = 2 Herz

from graph 2 we can determine the

wavelength of the wave which is 60

cm we can calculate the speed the wave

using V equals F Lambda substituting the

frequency f = 2 Herz and wavelength

Lambda equal 60 cm

we solve the speed equals 120

cm/s so D is

correct

ex exam style question two the graph

shows the variation with time of the

displacement of an electromagnetic wave

at a point the wave is traveling in a

vacuum what is the amplitude and what is

the wavelength of the wave from graph

can determine the amplitude of the wave

which is three arbitrary units from

graph can also determine the period of

the wave which is 20 microc or 20 * 10^

-6 seconds the speed of an

electromagnetic wave in a vacuum is 3 *

10^ 8

m/s we can calculate the frequency using

FAL 1 / T substituting the period T = 20

* 10 power -6 seconds we get the

frequency F = 50,000 Hertz we can

calculate the speed the wave using V

equal F Lambda substituting the speed V

= 3 * 10^ 8 m/ and frequency F equal

50,000 Herz we solve the wavelength

Lambda equal 6,000

M therefore the amplitude is three

arbitrary units and the wavelength is

6,000

M phase and path difference the phase

difference describes the difference in

position within a cycle between two

waves or two parts of the same wave

expressed as an angle usually in radians

or

degrees the path difference is the

difference in distance traveled by two

waves or two parts of the same wave

expressed in terms of

wavelength phase and path difference of

two positions of the same wave as a wave

oscillates upwards downwards and then

returns to its starting point it

completes one full cycle creating a

waveform that spans one

wavelength consider points A B C D E F G

H and I on a wave as shown their

corresponding phase angles are a equal

0° B =

45° C = 90° d =

135° e =

180° F =

225° G =

270° H = 315° and ials

360° from the graph one complete

wavelength Lambda corresponds to a phase

change of

360° or 2 pi

radians so the relationship between half

difference Delta Lambda and phase

difference Delta Theta is Delta Lambda

over Lambda equal Delta Theta over

360° for example points B and C the

phase difference equals 90 - 45 is equal

to

45° 45° is equivalent to pi/ 4 radians

the path difference equals 45 Lambda /

360 which is Lambda over 8 this

describes that point B and C are 45° out

of phase points C and G the phase

difference equals

270 - 90 is equal to

180°

180° is equivalent to PI

radians the path difference equals 180

Lambda / 360 which is Lambda / 2 this

describes that point C and G are exactly

out of phase or

antiphase points e and G the phase

difference equals

270 minus 180 is equal to

90° 90° is equivalent to PI / 2

radians the path difference equals 90

Lambda /

360 which is Lambda over 4 this

describes that point B and C are 90° out

of phase points a and I the phase

difference equals

360° or

0°

360° is equivalent to 2 pi radians and

0° equals 0 radian the path difference

equals Lambda this describes that point

a and I are in Phase Point C and F the

phase difference =

225 - 90 is equal to

135°

135° is equivalent to 3 pi over 4

radians the path difference equals 135

Lambda /

360 which is 3 Lambda over 8 this

describes that point C and F are

135° out of phase

phase and path difference between two

waves when comparing two waves the phase

difference describes the difference in

their positions within their respective

Cycles expressed as an angle usually in

degrees or

radians the path difference describes

the difference in the distances the two

waves have traveled expressed in terms

of

wavelength for first

example both Progressive waves are

moving to the right

at the instant shown we choose

corresponding points on both waves that

are at the same stage of their cycle

such as both at a peak both at a trough

or both crossing the equilibrium

position in the same direction we choose

these points on both waves like this the

phase angle on the first wave is

0° the phase angle on the second wave is

180° so the phase difference between two

waves equal 180 - 0 which is

180° which is equivalent to PI

radians the path difference equals 180

Lambda over

360 which is Lambda / 2 4 second

example both Progressive waves are

moving to the right at the instant shown

we choose these points on both waves

like this the phase angle on the first

wave is

0° the phase angle on the second wave is

270° so the phase difference between two

waves equals 270 - 0 which is

270° which is equivalent to 3 pi/ 2

radians the path difference equals 270

Lambda over

360 which is 3 Lambda / 2

exam style question three two balls

float on the surface of the sea the

balls are separated by a distance of 130

m a wave travels on the surface of the

sea so that the balls move vertically up

and down the distance between a Crest

and an adjacent trough of the wave is

0.9

M what is the phase difference between

the two balls the path difference

between the two balls is is given as

1.30 M the distance between a Crest and

an adjacent trough is

0.90 M this means that half a wavelength

is equal to

0.90

M therefore the full wavelength is 2 *

0.90 m = 1.80 m when the wave travels

one wavelength it oscillates for one

complete cycle one complete wavelength

length 1.80 M corresponds to one

complete cycle which is equivalent to a

phase difference of

360° if the path difference equals 1.30

m what is Phase difference Delta Theta

to find the phase difference Delta Theta

corresponding to a path difference of

130 M we can set up a proportion like

this we solve the phase difference Delta

Theta equal 200

60° so D is

correct exam style question four two

Progressive waves meet at a fixed Point

P the variation with time of the

displacement of each wave at Point p is

shown in the graph what is the phase

difference between the two waves at

Point P from the graph we can determine

the period T of both waves both waves

have the same period t equal 0.8 seconds

to determine the time difference we

identify corresponding points on each

wave's graph we can choose the points

where each wave crosses the zero

displacement line going

downwards so the time difference between

these corresponding waves equals 0.4

minus 0.1 is equal to 0.3 seconds the

time passed one period it oscillates for

one complete cycle one complete period

0.8 seconds corresponds to one complete

cycle which is equivalent to a phase

difference of

360° if the time difference equals 0.3

seconds what is Phase difference Delta

Theta to find the phase difference Delta

Theta corresponding to a time difference

of 0.3 seconds we can set up a

proportion like this

we solve the phase difference Delta

theta equals

135° so C is

correct investigate frequency an

amplitude using an oscilloscope or

C although we cannot see an actual sound

wave we can see an image or

representation of it by connecting a

microphone to a piece of apparatus

called an

oscilloscope the the Y gain or voltage

gain setting on an oscilloscope controls

the voltage scale on the vertical axis

the Y gain is usually given in volts per

division in this case we set the Y gain

to 2 volts per division the time base

setting controls the time scale on the

horizontal axis the time base is usually

given in second per division in this

case we set the time base to 5

milliseconds per division when a sound

wave from from a source such as a tuning

fork enters the microphone the

oscilloscope draws the longitudinal

sound wave as a transverse wave which

allows us to see features such as the

waves amplitude and frequency more

easily from the trace drawn on the

screen we can measure the vertical

distance corresponding to the amplitude

a in this example the amplitude is two

divisions so the amplitude a of

Soundwave in volts can be calculated by

multiplying the vertical distance a in

divisions by the Y gain which is two

divisions multiplied by 2 volts per

division so the amplitude a equals 4

volts if the Y gain is increased the

vertical distance a of the wave form on

the screen decreases but the actual

amplitude of the sound wave in volts

Remains the Same from the trace drawn on

the screen we can measure the time for

one complete vibration or one complete

wave this is called the time period of

the Wave t in this example peak-to Peak

distance is four

divisions so the frequency F of the

sound wave can be determined using the

equation F = 1/ period T the period T

equals peak-to Peak distance time time

base so the period T equals 4 divisions

Time 5 milliseconds per division we get

the period t = 20 milliseconds or

0.02

seconds therefore the frequency f = 1 /

0.02 is equal to 50 htz if the time base

is increased decreasing the peak-to peak

distance and creating more waveform but

the frequency and period of the sound

wave remain the

same exam style question five a

microphone connected to the Y plates of

a cathode ray oscilloscope C is placed

in front of a

loudspeaker the trace on the screen of

the cro is shown the time base setting

is 0.5 milliseconds per cimeter and the

Y plate sensitivity is 0.2 Ms per

cimeter what is the frequency of the

sound from the loudspeaker and what is

the amplitude of the trace on the

cro from the graph the peak to PE Peak

distance is measured to be 6

cm the period T equals Peak to Peak

distance time time

base so T = 6 cm * 0.6 milliseconds per

CM we get the period t equal

0.03 seconds we can calculate the

frequency f using the equation F = 1/

period T substituting T = 0.0 03

seconds so the frequency f equals 330 HZ

for two significant figures from the

graph the peak amplitude a is measured

to be 3

cm the amplitude a equals Peak amplitude

a * y gain so amplitude a = 3 cm * 0.2 m

per

cimeter we get the amplitude a equal 0

0.6

mols so a is

correct exam style question six a cathod

ray oscilloscope C is used to display a

wave of frequency 5

KZ the display is shown what is the

timebase setting of The

cro we can calculate the period T using

the equation T = 1 / period f

substituting the frequency F = 5,000

Hertz we get the period T =

0.002 seconds from the graph the peak-to

peak distance is measured to be 2

cm or we can measure this distance

instead peak-to Peak distance because it

is easier to measure scale the period T

of the wave equals peak-to Peak distance

time time

base substituting the period T =

0.2 seconds peak-to Peak distance equals

2

cm we get the time base equals

0.001 seconds per

cimeter

0.1 seconds per cimeter is equivalent to

100 microc per

cimeter so B is

correct Progressive wave and its

intensity a progressive wave transfers

energy from one point to another

intensity of progressive wave intensity

of wave is the energy transferred per

unit time per unit area at right angles

to the wave velocity the energy

transferred per unit time is the power

transferred therefore the equation of

the intensity of wave can be written as

I equals P / s where I is the intensity

in wat per square meter p is the power

in watts and S is the surface area of

the sphere in me squared the surface

area s of the sphere equals 4 Pi s r

where R is the radius of the

sphere the intensity I of a wave is

proportional to the amplitude a^ squ of

the wave for a given power P we can

conclude that the intensity 1 is

directly proportional to the square of

amplitude a and inversely proportional

to surface area s and therefore

inversely proportional to the square of

the

radius this describes a wave source that

emits power P transferring it outward

spherically at a distance are from the

source the surface area is

s the wave intensity is I and the wave

amplitude is a at a distance two are

from the source the surface area s is

directly proportional to the square of

the distance r as our increases two

times our squared increases four

times so the surface area s increases

four times the intensity I is inversely

proportional to surface area S as the

area s increases four times so the

intensity I decreases four times the

intensity I is directly proportional to

the square of the amplitude a as I

decreases four times the amplitude

squared decreases four times so

amplitude also decreases two times at a

distance 3 R from The Source the surface

area s is directly proportional to the

square of the distance r as our

increases three times R squ increases 9

times so the surface area s increases N9

times the intensity I is inversely

proportional to surface area S as the

area s increases nine times so the

intensity I decreases nine times the

intensity I is directly proportional to

the square of the amplitude a as I

decreases N9 times the amplitude squar

decreases N9 times so amplitude also

decreases three

times exam style question seven a sound

wave consists of a series of moving

pressure variations from the normal

constant air pressure the graph shows

these pressure variations for two waves

at one instant in time wave 1 has an

intensity of 1.6 * 10^ -6 per square me

what is the intensity of Wave 2 the

amplitude a of wave 1 is equal to 2 *

10^2 pascals the amplitude a of Wave 2

is equal to 3 * 10^ are -2

pascals so the amplitude of Wave 2 is

1.5 times of the amplitude of wave 1 the

intensity 1 is directly proportional to

the amplitude squared as the amplitude

increases 1.5 times the amplitude

squared increases 2.25 times so the

intensity increases 2.25

times therefore the intensity of Wave 2

is 2 2.25 times of the intensity of wave

1 so the intensity of Wave 2 is equal to

2.25 * 1.6 * 10^

-6 we get the intensity of Wave 2 is

equal to 3.6 * 10^ -6 watt per square

meter so C is

correct exam style question 8 light wave

of amplitude a is incident normally on a

surface of area s the power per unit

area reaching the surface is p this mean

that the intensity I equals P du to the

power per unit area is the intensity the

amplitude of the light wave is increased

to 2 a the light is then focused onto a

smaller area 1/3 of s what is the power

per unit area on this smaller area the

power per per unit area is the intensity

I the intensity I is directly

proportional to the amplitude squared as

the amplitude increases two times the

amplitude squared increases four times

so the intensity I increases four times

the intensity I is inversely

proportional to the surface area s 1/3

of area s meaning that the surface area

s decreases three times so the intensity

I increases three times therefore the

intensity I increases 12

times so the intensity I equal

12p so C is

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correct hey everyone welcome to PL

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videos in this one I covered everything

on the syllabus like you can see in the

figure waves are classified into two

types transverse waves and longitudinal

waves a transverse ver wave is a wave in

which the direction of vibration is

perpendicular to the direction of energy

transfer or wave

propagation as the particle oscillates

up and down for one complete oscillation

it creates one wave form as shown and

repeats

continuously this level is the

equilibrium position around which the

particle

oscillates this direction is the

direction of energy transfer wave travel

or wave

propagation this is a crest of a

transverse wave this is a trough of a

transverse wave the maximum displacement

or amplitude a is the distance from the

equilibrium position to a Crest or

trough the wavelength Lambda is the

distance of one complete wave for

example water waves seismic secondary

waves Slinky waves electromagnetic waves

in the figure shown the water particles

move upward and downward not along the

direction the wave travels as

demonstrated by the red ball in the

diagram which only moves up and down

repeatedly as the wave

passes when a transverse wave travels to

the right as shown the wave forms show

how the wave travels over time we see

that the point a moves upward to its

maximum displacement and then downward

and point B moves downward to its

maximum displacement and then upward

a longitudinal wave is a wave in which

the direction of vibration is parallel

to the direction of energy transfer or

wave

propagation shaking a slinky spring

forward and backward causes compressions

and rare factions to propagate along the

spring at a compression is where wave

particles are close together and high

pressure or high density at a rif action

is where wave particles are far apart

and low pressure or low density the

wavelength Lambda is the distance of one

complete wave which is the distance

between consecutive compressions or

consecutive rare

factions the particle oscillates around

the equilibrium position like this for

example sound waves Slinky waves and

seismic primary waves as a sound wave

travels to the right air particles

oscillate back and forth about their

equilibrium positions

this represents the equilibrium

positions of air particles in the

absence of a sound wave this represents

the displacement of air particles as a

sound wave propagates to the right this

creates the compression and rif action

like this this graph plots a particle

displacement against distance this graph

plots pressure against

distance exam style question one

the graph shows the variation of the

displacement of particles with distance

along a transverse wave at an instant in

time the wave is moving to the right

which position along the wave

corresponds to a point where particles

in the wave are traveling the fastest

upwards when the wave moves to the right

as shown point a is at the equilibrium

position moving

upwards this is where the particles have

maximum upward velocity point B is at a

maximum negative displacement a trough

at this instant its Vertical Velocity is

zero it is about to move

upwards Point C is at the equilibrium

position moving

downwards this is where the particles

have maximum downward velocity Point D

is at a maximum positive displacement a

Crest at this instant its Vertical

Velocity is zero it is about to move

downwards

so a is

correct exam style question two the

diagram illustrates the position of

particles in a progressive sound wave at

one instant in time the speed of the

wave is v p and Q are two points in the

wave a distance L apart what is an

expression for the frequency of the wave

the distance PQ equals half of

wavelength Lambda

so the distance l equals half of Lambda

and Lambda equals 2 L the frequency F

can be calculated using F equal speed V

/ wavelength Lambda substituting Lambda

equal 2 L we get the frequency f = v / 2

L so a is

correct exam style question three a wave

moves along the surface of water the

diagram shows the variation of

displacement s with distance along the

wave at time T equals z which graph best

shows the variation with time T of the

displacement s of the point p on the

wave when the wave moves to the right as

shown the point P will initially move

upwards towards a maximum displacement

then downwards through its equilibrium

position to a minimum displ M and

finally return to its starting position

as shown so the displacement time graph

for Point P should show this

characteristic

oscillation option D correctly

represents this pattern of

[Music]

motion hey everyone welcome to PL

Academy hope you found this video use

useful if you did it would be awesome if

you could subscribe like the video share

it with your friends and maybe drop a

comment your support really helps me

make more videos in this video I covered

everything on the syllabus like you can

see in the

figure Doppler effect for sound waves

when a stationary ambulance car emits

sound waves the wave front spread out

symmetrically forming circles around the

source if the wave Source moves the

waves can become compressed in front of

it and stretched behind it this movement

changes the wavelength and frequency of

the waves in front of the source the

wavelength decreases and the frequency

increases person a hears sound that has

a higher frequency than when the car was

stationary this observed frequency can

be calculated this equation

behind the source the wavelength

increases and the frequency

decreases so person B hears a sound with

a lower frequency than when the car was

stationary this observed frequency can

be calculated this

equation these apparent changes in

frequency which occur when a source of

waves is moving is called the Doppler

effect and is a property of all waves we

can conclude that the Doppler effect

happens when a source of waves moves

relative to a stationary Observer The

observed frequency is different from the

source

frequency if the source of waves moves

towards the Observer The observed

frequency F0 increases and The observed

wavelength Lambda 0

decreases if the source of waves moves

away the Observer The observed frequency

F0 decreases and The observed wavelength

Lambda 0

increases the observed frequency F0 can

be calculated using the following

formula F0 = FS * V and dividing by v+

or minus v s where F0 is the observed

frequency in hertz FS is the sound

frequency from The Source in hertz V is

the speed of the sound wave in me/

second and vs is the speed of the source

in me/ second we use addition in the

denominator when the source moves away

from The Observer and subtracts

when the source moves towards the

Observer exam style question one a toy

motorboat moving with constant velocity

V vibrates up and down on the surface of

a pond this causes the boat to act as a

source of circular water waves of

frequency 2 Hertz the speed of the waves

is 1.5

m/s a man standing at the edge of the

pond obs observes that the waves from

the boat approach him with a frequency

of 3 Hertz the formula for Doppler

effect calculations with sound waves may

also be used for water waves what is a

possible value of V The observed

frequency to increase from 2 Herz to 3

Herz indicating that the source of wave

moves towards the man we can calculate

the speed of the source of the wave

using the formula for Doppler effect F0

equal FS time V and dividing by V minus

v s rearranging the extension as v s

equal V minus fsv dividing by

F0 the frequency of wave FS equal

2z the speed of wave V equals 1.5

m/s The observed frequency F 0 = 3

Herz substituting V = 1.5 m/ second FS =

2 Herz F0 = 3 Herz we get the speed of

source of wave equals 0.5

m/s so B is

correct exam style question two a train

travels in a straight line at a constant

speed of 30

m/s the Train's horn continuously emits

sound of frequency 2,000 400 Hertz a

stationary Observer stands next to the

train track the train approaches the

stationary Observer passes him and then

moves away the speed of sound is 340

m/s what is the maximum difference in

the frequencies of the sound heard by

the stationary Observer The observed

frequency when a train moved towards the

Observer can be calculated using this

formula the speed of The Source v s

equal 30

m/s the speed of sound V equal 340

m/s the frequency of sound FS equals

2,400

htz substituting the all values in the

formula we get the observed frequency

when a train moved towards the Observer

equals

26324 6 Herz The observed frequency when

a train mover way the Observer can be

calculated using this

formula substituting the all values of

FS V and vs in the formula like this we

get the observed frequency when a train

move away the Observer equals 2,

25.4 1 Hertz the maximum difference in

the frequencies of the sound heard by

the stationary Observer equals 430 htz

for two significant

figures so C is

[Music]

correct hey everyone welcome to PL

Academy hope you found this video

useful if you did it would be awesome if

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it with your friends and maybe drop a

comment your support really helps me

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in this video I covered everything on

the syllabus like you can see in the

figure electromagnetic spectrum the

properties of the electromagnetic

spectrum

include they are all transverse waves

they can all travel through the vacuum

they all travel at the same speed in a

vacuum which is approximately 3 * 10^ 8

m/s

this speed is also a good approximation

for their speed in air therefore if an

electromagnetic waves wavelength is

known its frequency in a vacuum can be

calculated using the wave equation v = f

* Lambda electromagnetic waves comes at

many different frequencies and

wavelengths in a table lists different

types of electromagnetic spectrum and

their approximate wavelengths in a

vacuum as gamma rays have an

approximately wavelength range of 10

power -14 to 10 power1 m x-rays have an

approximately wavelength range of 10

power -13 to 10 power1 m ultraviolet

have an approximately wavelength range

of 10 power -14 to 10 power 9 M visible

light have an approximately wavelength

range of 400 nanom to 700 nanom

infrared have an approximately

wavelength range of 10 power -7 to 10

power -3 M microwave have an

approximately wavelength range of 10

power -3 to 10 power -1 m radio waves

have an approximately wavelength range

of 10 power -1 to 10 power 5 m these

electromagnetic waves are ordered

according to decreasing frequency and

increasing wavelength

it is important to recognize that there

are no sharp boundaries between these

types of

radiation their properties gradually

change with

wavelength for example there is no

precise wavelength at which radiation

ceases to be ultraviolet and becomes X

radiation we can calculate the

approximate frequency range of gamma

rays to be 10 power 19 to 10^ 22 Herz

the approximate frequency range of

x-rays is is 10^ 17 to 10 power 21 Hertz

the approximate frequency range of

ultraviolet is 10 power 15 to 10^ 17

Hertz the approximate frequency range of

visible light is 10 power 14 to 10 power

15 Hertz the approximate frequency range

of infrared is 10 power 11 to 10 power

14 Hertz the approximate frequency range

of microwave is 10 10^ 9 to 10 power 11

Hertz the approximate frequency range of

radio waves is 10^ 3 to 10^ 9 Hertz you

should remember the approximate

wavelength ranges of electromagnetic

spectrum exam style question one the

graph shows how the intensity of

electromagnetic radiation emitted from a

distant star varies with wavelength

in which region of the electromagnetic

spectrum is the radiation of greatest

intensity the peak of the graph

corresponds to a wavelength of

approximately 150 nanom equivalent to

1.5 * 10 power7

m which is in the ultraviolet region so

C is

correct exam style question two the

diagram shows the principal regions of

the electromagnetic spectrum with some

details labeled the diagram is not to

scale what is a typical order of

magnitude of the wavelength of the

radiation in region Q at the region Q is

the microwave region the microwave has

the approximate wavelength range to be

10^ -3 to 10^ -1

M so C is

correct exam style question three an

electromagnetic wave has a wavelength of

138 pomet in a vacuum to which region of

the electromagnetic spectrum does this

wave belong the wavelength 138 pom equal

138 * 10^ -12 M or equivalent to 1.38 *

10^ -10

M the x-rays has the approximate

wavelength range to be 10^ -33 to 10^ 9

M so this wavelength is

[Music]

x-rays hey everyone welcome to PL

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figure

polarization electromagnetic waves are

transverse waves meaning their

oscillations occur perpendicular to the

direction of energy transfer or wave

propagation in general the oscillations

of an unpolarized electromagnetic wave

occur in all directions perpendicular to

this direction of

propagation therefore the unpolarized

electromagnetic wave vibrate in all

directions perpendicular to the

direction of wave

propagation an unpolarized

electromagnetic wave has a mixture of

these different vibration

directions this can be represented by a

vector diagram showing vibrations in

multiple planes perpendicular to the

direction of

propagation polarization is the process

of restricting the oscillations of an

electromagnetic wave to a single plane a

polarized electromagnetic wave vibrates

in Only One Direction perpendicular to

the direction of wave

propagation this can also be represented

by a vector diagram showing oscillations

confined to a single plane

as vertically polarized electromagnetic

wave and horizontally polarized wave as

shown therefore we can conclude that the

longitudinal waves cannot be polarized

because the vibrations are parallel to

the direction of travel of the wave

energy thus if a wave can be polarized

it means that it must be

transverse the unpolarized

electromagnetic wave can be polarized by

using a polarizing filter this allows

oscillations only in one direction to

pass through and parallel to

transmission axis when an unpolarized

electromagnetic wave with the intensity

one passes through a polarizing filter

with a vertical transmission axis the

transmitted wave is vertically polarized

and its intensity is reduced by half if

a second polarizing filter is added with

its transmission axis aligned with the

first filter all of the polarized wave

passes through and the intensity Remains

the Same as the second filter is rotated

the intensity of the transmitted light

decreases this decrease follows malo's

law if the second polarizing filters

transmission axis is at right angles to

the first no light is

transmitted the intensity is

zero malus is law used to calculate the

intensity I have a plain polarized wave

after transmission through a polarizing

filter the malo's formula is I equal i0

cos Theta 2 where I is the transmitted

wave intensity from the filter i0 is the

incident wave intensity on the filter

and the angle Theta is the angle between

the transmission axis of the two

filters this graph shows that the

transmitted wave intensity varies with

the angle Theta according to a cosine

squar

relationship the intensity is zero when

the transmission axis of the filter is

perpendicular to the plane of

polarization of the incident wave at 90°

270° the intensity is maximum when the

transmission axis of the filter is

parallel to the plane of polarization of

the incident wave at 0°

180° and 360°

exam style question one when plain

polarized light of amplitude a is passed

through a polarizing filter as shown the

amplitude of the light emerging is a COS

Theta the intensity of the initial beam

is I what is the intensity of the

emerging light when Theta is

60° the intensity 1 is directly

proportional to the amplitude squared

the initial intensity 1 is Ka a^ 2 where

K is constant value and an a is the

initial

amplitude the intensity of emerging

light I2 equal k a COS Theta 2 where K

is constant value and Theta equal

60° ka^ 2 equal intensity 1 and cos 60

is equal to

0.5 I2 equal I

0.52 so the intensity of emerging light

I2 equal 0.25

I exam style question two a vertically

polarized electromagnetic wave of

intensity 1 is incident normally on a

polarizing filter the transmission axis

of the filter is at an angle of 30° to

the vertical the transmitted wave from

the first filter is then incident

normally on a second polarizing filter

the transmission axis of this filter is

at an angle of 90° to the vertical what

is the intensity of the wave after

passing through the second filter the

intensity of wave after passing through

the second filter can be calculated

using the malus's Law formula when the

vertically polarized wave passes through

the first filter the angle Theta is the

angle between vertically plane and

transmission axis of first filter which

is

30° so the intensity i1 of wave passing

the first filter equals i0 cos 30 s

which is 0.75

i0 when the polarized wave with

intensity i1 passes through the second

filter the angle between the plane

polarized wave i1 and transmission axis

is 90 + 30 is equal to

120° so the intensity I2 = i1 cos

1202 which is 0.19 i0 for two

significant figures so C is

correct exam style question three a

student investigates the polarization of

microwaves the microwaves from the

transmitter are vertically

polarized a metal Grill acts as a

polarizing filter when placed between

the microwave transmitter and the

receiver the reading on the voltmeter is

proportional to the intensity of

microwaves transmitted through the grill

when the transmission axis of the grill

is vertical the voltmeter reads 3.5

volts the grill is then rotated through

an angle Theta the voltmeter now reads

2.20 volts what is Theta the reading on

the voltmeter is direct L proportional

to the intensity one of wave passing

through the grill so I equals KV where K

is constant value and V is the reading

on the

voltmeter the intensity 1 Z of wave

before passing the grill equals

3.5k the intensity one of wave after

passing the grill equals

2.2k the intensity of wave after passing

through the metal Grill can be

calculated using the malo's Law formula

substituting I =

2.2k and i0 =

3.5k we solve the angle theta equals

37.5 de for three significant

figures I hope you found this video

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