Unit 8 Superposition of waves AS/A Level Physics Cambridge CAIE 9702

[Music]

hey everyone welcome to PL Academy hope

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it with your friends and maybe drop a

comment your support really helps me

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videos in this video I covered

everything on the syllabus like you can

see in the figure

principle of

superposition superposition is the

principle that when waves of the same

type meet at a point they individual

displacements combined to produce a

resultant

displacement as a blue wave travels to

the left and a red wave travels to the

right initially they are in Phase

meaning their phase difference is zero

when these waves meet their individual

displacements combine constructively

resulting in a maximum resultant

displacement that is twice the amplitude

of either individual wave this is called

constructive superposition which occurs

when waves are in Phase with phase

differences of 0 Dee creating the

maximum

amplitude when a blue wave travels One

Division to the left and a red wave

travels One Division to the right they

individual displacements combined to

create the resultant displacement as

shown

when a blue wave travels One Division to

the left and a red wave travels One

Division to the right the phase

difference between two waves is pi

radians or

180° they are said to be in

antiphase their individual displacements

combined destructively resulting in a

minimum resultant displacement which can

be zero if the waves have equal

amplitudes this is called destructive

superposition which occurs when waves

are in anti phase with phase differences

of Pi radians or

180° creating the zero

amplitude when a blue wave travels One

Division to the left and a red wave

travels One Division to the right their

individual displacements combined to

create the resultant displacement as

shown when a blue wave travels One

Division to the left and a red wave

travels One Division to the right the

phase difference between two waves is 2

pi radians or

360° they are said to be in Phase their

individual displacements combine

constructively resulting in a maximum

resultant displacement that is twice the

amplitude of either individual wave this

is called constructive superposition

which occurs when waves are in Phase

with phase differences of 2 pi radians

or

360° creating the maximum

amplitude therefore we can be concluded

that

the constructive superposition which

occurs when waves are in Phase with

phase differences of 0 2 pi 4 Pi 6 pi

and so on the destructive superposition

which occurs when waves are antiphase

with phase differences of Pi 3 Pi 5 Pi 7

pi and so

on exam style question one the three

waves shown in each diagram have the

same amplitude and frequency but

different phase they are added together

to give a resultant wave in which case

is the resultant wave zero at this

instant in a choice a there are three

waves as blue violet and green coolers

like this at this point displacements of

three waves combined to be zero like

this at this point displacements of

three waves combined to be zero like

this at at this point displacements of

three waves combined to be zero like

this and any positions the resultant

displacements of three waves combined

together to become zero like this so a

is

correct stationary wave stationary waves

or standing waves are formed by the

superposition or interference of two

Progressive waves traveling in opposite

directions with the the same frequency

wavelength and speed this interference

results in a pattern of nodes and

antinodes at nodes are the destructive

superposition their phase differences

are Pi 3 Pi 5 Pi 7 pi and so on or

180°

540°

900° 1,

1260° and so on their path differences

are Lambda over 2 3 Lambda over 2 5

Lambda over 2 7 Lambda over 2 and so on

at anodes are the constructive

superposition their phase differences

are 0 2 pi 4 Pi 6 pi and so on or 0°

360°

720° 1, 80° and so on their path

differences are 0 Lambda 2 lamb LDA 3

Lambda and so on when a blue wave

travels to the right and a red wave

travels to the left as shown they are

moving in opposite directions and they

have same frequency wavelength and speed

at the initial they are in phase and

phase difference is zero they superpose

or interfere with each other to create

the stationary wave shown in Black Wave

After 1/8 of a period each wave travels

1/8 of a wave length in opposite

directions resulting in a separation of

1/4 of a

wavelength they superpose to form the

stationary wave shown in Black Wave

After 1/4 of a period each wave travels

1/4 of a wavelength in opposite

directions resulting in a separation of

half the

wavelength they superpose to form the

stationary wave shown in Black Wave

After 38 of a period each wave travels

38 of a wavelength in opposite direction

CS resulting in a 34s wavelength in

separation they superpose to form the

stationary wave shown in Black Wave

after half of period each wave travels

half a wavelength in opposite directions

resulting in a full wavelength

separation they superpose to form the

stationary wave shown in Black Wave when

two waves traveling at high speeds in

opposite directions superpose they can

form stationary waves which appear as a

series of Loops as shown this creates a

pattern of nodes and antinodes as shown

at nodes the amplitude is at minimum in

this diagram it is zero because the

waves have equal

amplitudes at antinodes the amplitude is

maximum the distance between two Loops

is equal to wavelength Lambda so the

distance between consecutive nodes or

antinodes are half of the wavelength

exam style question two the diagram

shows a sketch of a wave pattern over a

short period of time which description

of this wave is correct in the diagram

shown the stationary wave the distance

for two Loops is a wavelength for one

meter has five Loops which is equivalent

to 2.5

wavelengths so we can determine the

wavelength of the wave is equal to 1 m /

2.5 which is 0.4 M or 40

cm the transverse waves and longitudinal

waves can be occurred the stationary

wave in the choice a the wave is

longitudinal has a wavelength of 20 cm

and is

stationary this is

incorrect in the choice B the wave is

transverse has a wavelength of 20 cm and

is

stationary this is incorrect

in the choice C the wave is transverse

has a wavelength of 40 cm and is

Progressive this is

incorrect in the choice d The Wave is

transverse has a wavelength of 40 cm and

is

stationary this is

correct stationary wave of the stretched

string a vibrator generates an incident

wave that travels to the left this wave

reflects at the pulley creating a

reflected wave traveling to the right

the incident and reflected waves have

the same frequency wavelength and speed

these two waves superpose to produce a

standing wave the simplest standing wave

pattern with a single Loop is called the

fundamental frequency or the first

harmonic the fixed ends at the vibrator

and pulley are always formed nodes

because the string cannot free to

vibrate

the vibrator varies the frequency until

a stationary wave with two Loops second

harmonic three Loops third harmonic four

Loops fourth harmonic and so on let L be

the length of the stretched string

recall that two Loops of stationary wave

equals one

wavelength for first harmonic has a

single Loop indicating that L equals

half of wavelength so the wavelength

equals 2 L the fundamental frequency F1

can be calculated by dividing the speed

V by wavelength Lambda substituting

Lambda equal 2 L so fub1 = V / 2L 4

second harmonic has two Loops indicating

that L equals wavelength Lambda the

second harmonic frequency FS2 is

calculated by dividing the speed V by

wavelength Lambda substituting Lambda

equals l so FS2 = V / l or 2 v/ 2 L and

V / 2 L =

fub1 so FS2 is twice as fub1 43d

harmonic has three Loops indicating that

L equals 3 Lambda / 2 so Lambda equal 2

L over 3 the third harmonic frequency F3

is calculated by dividing the speed V by

wavelength Lambda substituting Lambda =

2 L over 3 so F3 = 3 V / 2 L and V / 2 L

equals

FS1 so F3 is three times of fub1 for

fourth harmonic has four Loops

indicating that L equal 2 Lambda so

Lambda equal L / 2 the fourth harmonic

frequency F3 is calculated by dividing

the speed V by wavelength Lambda

substituting Lambda = L / 2 so F4 = 2 V

/ l or 4 V / 2 L and V / 2 L =

fub1 so F4 is fourth times of

fub1 exam style question three a

stationary wave is produced on a string

that is stretched between two fixed

points that are a distance of 1.35 M

apart as shown the speed of the waves on

the string is 450

m/s what is the frequency of oscillation

of the stationary wave the standing wave

has three Loops which corresponds to 1.5

wavelengths so 1.35 m equal 1.5

wavelengths this means one wavelength is

1.35 m / 1.5 is equal to 0.9 M the

frequency of the stationary wave can be

calculated by the equation FAL v/ Lambda

substituting V equal 450 m/s and Lambda

equal 0.9 M we get the frequency equals

500

htz so C is

correct the stationary wave of sound

wave in air column with close one end a

speaker generates an incident sound wave

that travels through a pipe closed at

one end this wave reflects at the closed

end creating a reflected wave traveling

in the opposite direction the incident

and reflected waves have the same

frequency wavelength and speed the

superposition of these two waves can

produce a standing wave the simplest

standing wave with a quarter Loop

pattern is called the fundamental wave

or the first

harmonic this corresponds to the lowest

frequency that can be heard hence the

term fundamental frequency a closed end

of the pipe forms a node because air

particles cannot vibrate freely an open

end forms an antinode because air

particles can vibrate freely the speaker

varies the frequency until a stationary

wave with second her second harmonic

three herd third harmonic and so on let

L be the length of the air column recall

that two Loops of stationary wave equals

one

wavelength for first harmonic has an

one4 Loop pattern indicating that L

equals 1/4 of a wavelength so the

wavelength equals 4 L the fundamental

frequency fub1 can be calculated by

dividing the speed V by wavelength

Lambda substituting Lambda equal 4 L so

fub1 = v/ 4 L 4C harmonic has a 3/4 Loop

pattern indicating that L equals 3/4 of

a wavelength so the wavelength equals 4

L over 3 the second harmonic frequency

FS2 can be calculated by dividing the

speed V by wavelength Lambda

substituting Lambda equals 4 L over 3 so

F2 = 3 V over 4 L and V over 4 L = fub1

so FS2 = 3 * of fub1 43 harmonic has a

54 Loop pattern indicating that L equals

5/4 of a wavelength so the wavelength

equals 4 L the third harmonic frequency

F3 can be calculated by dividing the

speed V by wavelength Lambda

substituting Lambda equals 4 L over 5 so

F3 = 5 V over 4 L and V over 4 L =

fub1 so F3 = 5 * of

fub1 exam style question four a long

glass tube is almost completely immersed

in a large tank of water a tuning fork

is struck and held just above the open

end of the tube as it is slowly raised a

louder sound is first heard when the

height h of the end of the tube above

the water is 18.8

CM a louder sound is next heard when H

is 56.4

cim the speed of sound in air is 330

m/s what is the frequency of the sound

produced by The Tuning Fork the first

herd when H is 18.8

CM the next herd when H is 56.4

CM so half of wave length equal

56.4 - 18.8 is equal to 37.6

CM therefore 1 wavelength equal 2 *

37.6 is equal to

0.752 M the frequency can be calculated

using the equation FAL V / Lambda

substituting V = 330 m/s and Lambda

equal 0.7 5 2

m we get the frequency F equal 440 Hertz

for two significant figures so B is

correct the stationary wave of sound

wave in air column with open both ends a

speaker generates an incident sound wave

that travels through an open pipe this

wave reflects at the open end creating a

reflected wave traveling in the opposite

direction

the incident and reflected waves have

the same frequency wavelength and speed

the superposition of these two waves can

produce a standing wave the simplest

standing wave with one Loop pattern is

called the fundamental wave or the first

harmonic this corresponds to the lowest

frequency that can be heard hence the

term fundamental

frequency both open ends forms an

antinode because air particles can

vibrate freely

the speaker varies the frequency until a

stationary wave with second her second

harmonic three herd third harmonic and

so on let L be the length of the air

column recall that two Loops of

stationary wave equals one

wavelength for first harmonic has an one

Loop pattern indicating that L equals

half of a wavelength so the wavelength

equals 2 L the fundamental frequency

fub1 can be calculated by dividing the

speed V by wavelength Lambda

substituting Lambda equals 2 L so fub1 =

V / 2 L 4 second harmonic has a two

Loops pattern indicating that L equals

one

wavelength the second harmonic frequency

F2 can be calculated by dividing the

speed V by wavelength Lambda

substituting Lambda equals l so F2 = V /

l or 2 V / 2 L and v/ 2 L =

fub1 so FS2 = 2 * of fub1 43 harmonic

has a three Hales Loops pattern

indicating that L equals 3 halves of a

wavelength so the wavelength equals 2 L

over 3 the third harmonic frequency F3

can be calculated by dividing the speed

V by wavelength Lambda substitute Lambda

= 2 l/ 3 so F3 = 3 V / 2 L and v/ 2 L =

fub1 so F3 = 3 * of

fub1 exam style question five stationary

sound waves can be formed in the air

Columns of pipes one type of pipe is

closed at one end and open at the other

end

another of pipe is open at both ends

which pipe can form a stationary sound

wave with the lowest frequency let V

represent the speed of sound in air

Choice a consider a pipe open at both

ends the lowest frequency or fundamental

frequency standing wave has a half

wavelength pattern one Loop is shown one

Loop is equivalent to half

wavelength so 2 L equals a half of

Lambda and Lambda equal 4 L using the

equation the frequency F equal speed V

over wavelength Lambda substituting

Lambda equals 4 L so F = 0.25 V over L

Choice B consider a pipe is closed at

one end the lowest frequency or

fundamental frequency standing wave has

a quarter wavelength pattern half Loop

is shown a half Loop is equivalent a

quarter

wavelength so 2 L = 1/4 Lambda and

Lambda equal 8 L using the equation the

frequency F equal speed V over

wavelength Lambda substituting Lambda

equal 8 l so F =

0.125 V / L Choice C consider a pipe

open at both ends the lowest frequency

or thund mental frequency standing wave

has a half wavelength pattern one Loop

is shown one Loop is equivalent to half

wavelength so L equals a half of Lambda

and Lambda equal 2 L using the equation

the frequency FAL speed V over

wavelength Lambda substituting Lambda

equal 2 L so F equal 0.5 V / L Choice d

consider a pipe is closed at one end the

lowest frequency or fundamental

frequency standing wave has a quarter

wavelength pattern half Loop is shown a

half Loop is equivalent to quarter

wavelength so L equal 1/4 Lambda and

Lambda equal 4 L using the equation the

frequency F equal speed V over

wavelength Lambda substituting Lambda

equals 4 L

so F equal 0.25 V / l so B has the

lowest

frequency stationary wave of

microwave an incident microwave

traveling toward a metal reflector is

reflected and travels in the opposite

direction the incident and reflected

waves have the same frequency wavelength

and speed the superposition of the

incident and reflected microwaves

creates a stationary wave pattern with

nodes and

antinodes a detector measures the

microwave signal intensity of

microwave this detector can be used to

locate the nodes and anten noes of this

standing wave by moving the detector

between the microwave source and the

reflector the alternating points of

minimum and maximum signal intensity can

be observed the points of minimum signal

intensity correspond to the nodes the

points of Maximum signal intensity

correspond to the ant noes the distance

between successive nodes or antinodes is

equal to half of a

wavelength exam style question six in an

experiment to demonstrate a stationary

wave two microwave transmitters emitting

waves of wavelength for cmers are set

facing each other as shown a detector is

moved along a straight line between the

transmitters it detects positions of

Maximum and minimum signal the detector

is a distance D from the leftand

transmitter assume that both

transmitters are at antinodes of the

stationary wave which row gives a value

of D for a maximum and for a minimum at

the maximum signals is the

antinodes at the minimum signals is the

nodes the the question given the

wavelength of microwave is 4 cm so half

of a wavelength is 2

cm the distance between consecutive

antinodes or nodes is 2

cm and the distance between consecutive

antinodes and nodes is 1

cenm from the diagram shows that the

value of D for a maximum signals or

antinodes occur to be 0 2 4 6 8 and so

on which are even numbers the value of D

for minimum signals or nodes occur to be

1 3 5 7 and so on which are odd numbers

so the value of D for a maximum can be

46 and 48

CM the value of D for a minimum can be

47

CM so C is

correct difference between the

stationary wave and the progressive wave

the progressive wave can be transferred

the energy from one place to another the

progressive wave transfers the energy in

this direction the stationary wave

cannot be transferred the energy from

one place to

another however the energy is stored in

the vibrating particles like this all

the points along the progressive wave

have the same

amplitude these amplitude is the maximum

displacement that all point in wave can

be oscillated

the stationary wave has the amplitude of

different point to vary from a maximum

to zero the zero amplitude is at nodes

the maximum amplitude is at

antinodes all the points over one

wavelength in Progressive wave have

different phases all the points between

successive nodes are in phase for

example all points in same color section

her phase difference is zero all points

in pink and yellow region have phase

difference equals

360° which is also in Phase all points

in red and yellow regions have phase

difference equals

720° which is also in Phase all points

in blue and yellow regions have phase

difference equals

180° which is

antiphase all points in green and yellow

regions have phase difference equals

540° which is antiphase

exam style question seven which

statement concerning a stationary wave

is correct Choice a all the particles

between two adjacent nodes oscillate in

Phase this is correct Choice B the

amplitude of the stationary wave is

equal to the amplitude of one of the

Waves creating it this is

incorrect the amplitude of a standing

wave varies from0 Z at the nodes to a

maximum which can be twice the amplitude

of the individual waves at the anten

noes Choice C the wavelength of the

stationary wave is equal to the

separation of two adjacent

nodes this is

incorrect the distance between two

adjacent nodes is half a

wavelength Choice D there is no

displacement of a particle at an

antinode at any time this is

incorrect at antinodes it's displacement

is maximum while no displacement at

[Music]

nodes hey everyone welcome to PL Academy

hope you found this video

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

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in this video I covered everything on

the syllabus like you can see in the

figure defraction of wave defraction is

the spreading out of a wave as it passes

through a narrow Gap aperture or around

an obstacle the amount of defraction

depends on the wavelength of the wave

compared with the size of the Gap

aperture or the size of the obstacle

when straight wave fronts in encounter a

gap or aperture in a barrier they pass

through and defract spreading out into

the region Beyond The Wave fronts curve

at the edges of the Gap while the

central portion remains relatively

straight this shown that these area are

diffracted this area is not

defract as the size Gap decreases to

approach the wavelength defraction

becomes more resulting in Greater

spreading and curvature of the wave

fronts as shown as the size Gap

decreases to very smaller than the

wavelength defraction increases

resulting in Greater spreading and

curvature of the wave fronts but the

defraction pattern may be become less

distinct as shown so we can conclude as

follows when the aperture is much larger

than the wavelength defraction is

minimal Little Wave forms curvature when

the aperture size is comparable to the

wavelength defraction increases greater

curvature of wave fronts

when the aperture is much smaller than

the wavelength defraction is significant

but the defraction pattern may become

less

distinct when a wave encounters an

obstacle defraction occurs around the

edges when straight wave fronts with a

wavelength much smaller than the width

of the obstacle defraction is minimal

resulting in slight curvature of the

wave fronts at the edges as shown

this creates a shadow region behind the

obstacle where the wave is blocked as

wavelength increases defraction

increases leading to Greater curvature

of the wave fronts at the edges as shown

this results in a less distinct or no

Shadow region behind the obstacle as the

wave bends around it when a sound wave

passes through a doorway the size of the

doorway is comparable to the wavelength

of the sound causing the defraction to

occur when light passes through a

doorway the size of the doorway is

typically much larger than the

wavelength of light resulting in no

defraction to investigate the defraction

of the visible light using a single slit

the slit width should be less than

approximately 10 power -4 M or 0.1

mm this is because the wavelength of

visible light ranges from about 10^ -6

to 10 power -7 M when monochromatic

light from a laser Source passes through

a single slit it diffracts as shown the

central Fringe of the defraction pattern

has the highest intensity and greatest

width decreasing the slit width

increases defraction leading to Greater

spreading of the light and a decrease in

intensity this increased defraction also

results in a wider Central Fringe the

defraction pattern of white light

through a single slit is shown the

central Fringe is white while the other

fringes display a spectrum of colors

with blue light closer to the center and

red light further away this occurs

because red light has a longer

wavelength than blue light and longer

wavelengths defract more spread out more

the defraction patterns of red green and

blue light through a single slit are

also shown the red lights Central Fringe

is the widest while the blue lights

Central Fringe is the

narrowest this is consistent with red

lights longer wavelength which causes

greater defraction than green or blue

light to investigate the defraction of

water waves using a single slit in a

ripple tank the wave fronts of waves are

generated using a vibrating bar and move

towards a gap in a barrier where the

Ripple stri strike the barrier they are

reflected back where they arrive at the

Gap however they pass through and spread

out into the space beyond it is this

spreading out of waves as they travel

through a gap or past the edge of a

barrier a ripple tank is a shallow tray

of water with a light source shining

down through it the light illuminates

the wave fronts making them visible a

straight Dipper can be used to create

straight wave fronts in a ripple tank

when the Dipper is vibrated up and down

it creates a series of parallel wave

fronts passing through a gap to observe

the defraction of the water wave on a

screen the stroboscope or video camera

is used to make the slow motion of the

wave fronts of water wave on the

screen exam style question one the speed

of sound in air is 330

m/s which size of architectural features

in a large Concert Hall would best

defract sound waves of frequency 0.44

khz the size of architecture features

should be comparable to wavelength of

the sound waves resulting in the best

defraction so we can calculate the

wavelength of the sound first using the

equation wavelength Lambda equals speed

V over frequency F substituting V equal

330 m/ second and f = 0.44 * 10^ 3 Hertz

we get the wavelength Lambda equal 0.75

m or 750

mm so the size of architecture features

should be 750

mm exam style question two in an

experiment water waves in a ripple tank

are incident on a gap as shown some

defraction of the water waves is

observed which change to the experiment

would provide a better demonstration of

defraction the best defraction occurs

when the size of Gap is comparable to

the

wavelength Choice a increase the

amplitude of the

Waves this is incorrect because the

amplitude does not affect

defraction Choice B increase the

frequency of the

Waves this is incorrect because

increasing the frequency decreases the

wavelength which would reduce

defraction Choice C increase the

wavelength of the Waves this is correct

because increasing the wavelength makes

it closer to the gap size leading to

more

defraction Choice D increase the width

of the Gap this is incorrect because

increasing the Gap width decreases the

amount of defraction

[Music]

hey everyone welcome to PL Academy hope

you found this video

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 interference of wave inter

interference occurs when two coherent

waves meet at a point they superpose

producing a pattern of nodes and

antinodes at the antinodes have the

phase differences to be 0er 2 pi 4 Pi 6

pi and so on or 0

360°

720° and so on and their patch

differences are zero Lambda 2 Lambda 3

Lambda and so on at the nodes have the

phase differences to be Pi 3 Pi 5 Pi 7

pi and so on or

180°

540° 900° and so on and their patch

differences are half a Lambda 3 half

Lambda 5 half Lambda and so on waves are

considered coherent when they satisfy

the following conditions they have a

constant phase

difference this is the most important

condition they are the same type of wave

they have the same frequency and

wavelength for example green and blue

waves are

coherent this is because both waves

maintain a constant phase difference of

180° and they have the same frequency

and

wavelength their amplitude are not equal

but do not affect

coherence pink and blue waves are

coherent this is because both waves

maintain a constant phase difference of

90° and they have the same frequency and

wavelength their amplitude are not equal

but do not affect

coherence red and green waves are not

coherent this is because they do not

maintain a constant phase difference and

do not have the same frequency or

wavelength to investigate the

interference of water waves in the

Ripple tank two Dippers connected to the

same oscillator vibrate with the same

frequency generating two coherent water

waves when the bar vibrates each Dipper

acts as a source of circular ripples

spreading

outwards where these sets of ripples

overlap an interference pattern is

projected onto the screen below this

interference pattern is visible on the

screen the pattern consists of

alternating regions of constructive and

destructive interference

along this line are locations of

destructive interference resulting in

nodes along this line are locations of

constructive interference resulting in

antinodes along this line are locations

of destructive interference resulting in

nodes along this line are locations of

constructive interference resulting in

antinodes along this line are locations

of destructive interference resulting in

nodes

along this line are locations of

constructive interference resulting in

antinodes along this line are locations

of destructive interference resulting in

nodes another way to observe

interference in a ripple tank is to use

plain waves passing through two gaps in

a barrier the water waves are defract at

the gaps and then interfere Beyond

creating a similar pattern

the interference pattern in the Ripple

tank two sources S1 and S2 produce

coherent waves with a phase difference

of zero the waves have the same

wavelength Lambda given the red line

represents the crest and green line

represents the trough along the yellow

lines are locations of constructive

interference resulting in

antinodes these points are formed the

superposition of wave crests with crests

or wave trough with troughs along this

black lines are locations of destructive

interference resulting in

nodes these points are formed the

superposition of wave crests with

troughs recall that at the nodes have

the phase differences to be Pi 3 Pi 5 Pi

7 pi and so on or

180°

540° 900° and so on and their patch

differences are half Lambda 3 half

Lambda 5 half Lambda and so on at the

antinodes have the phase differences to

be zero 2 pi 4 Pi 6 pi and so on or zero

360°

720° and so on and their patch

differences are zero Lambda 2 Lambda 3

Lambda and so on the difference in

distance from any point along black and

yellow lines to sources S1 and S2

respectively is the path difference for

that line the central yellow line marks

the zero order antinode

a0 at this antinode both the path

difference and the phase difference are

zero this means that the distance from

any point on this line to the two

sources S1 and S2 is equal for example

at point a on this line the distance S1

to a is 5 Lambda and the distance S2 to

a is also 5 Lambda at the left next

black line marks the first order node

N1 at this node the path difference is

half a wavelength and the phase

difference are

180° this means that the difference in

distance from any point on this line to

the two sources S1 and S2 is half a

wavelength for example at point B on

this line the distance S1 to B is 5

Lambda and the distance S2 to B is also

5.5 Lambda at the next yellow line marks

the first order anode

A1 at this anode the path difference is

one wavelength and the phase difference

are

360° this means that the difference in

distance from any point on this line to

the two sources S1 and S2 is one

wavelength for example at Point C on

this line the distance S1 to C is 4.5

Lambda and the distance S2 to C is also

5.5 Lambda at the next black line marks

the second order node

N2 at this node the path difference is

three half wavelength and the phase

difference are

540° this means that the difference in

distance from any point on this line to

the two sources S1 and S2 is three half

wavelength for example at Point D on

this line the distance S1 to D is 3

Lambda and the distance S 2 to D is also

4.5 Lambda at the next yellow line marks

the second order antinode

A2 at this antinode the path difference

is two wavelength and the phase

difference are

720° this means that the difference in

distance from any point on this line to

the two sources S1 and S2 is two

wavelength for example at Point e on

this line the distance S1 to e is 4

Lambda and the distance S 2 to e is also

6 Lambda at the next black line marks

the third order node

N3 at this node the path difference is

five half wavelength and the phase

difference are

900° this means that the difference in

distance from any point on this line to

the two sources S1 and S2 is five half

wve length for example at Point F on

this line the distance S1 to F is 3 l

Lambda and the distance S2 to F is also

5.5 Lambda at the next yellow line marks

the third order antinode

A3 at this anode the path difference is

three wavelength and the phase

difference are 1,

180° this means that the difference in

distance from any point on this line to

the two sources S1 and S2 is three

wavelength for example at Point G on

this line the distance S1 1 to G is 2.5

Lambda and the distance S2 to G is also

5.5 Lambda this pattern is symmetric

about the central yellow line like

this exam style question one two sources

of microwaves p and Q produce coherent

waves with a phase difference of

180° the waves have the same wavelength

Lambda and at the point s there is a

minimum in the interference pattern

produced by waves from the two sources

the distance Qs minus PS is called the

path difference which expression could

represent the path difference at the

initial the waves from sources p and Q

have phase difference of

180° so the path difference between two

waves is half a

wavelength the path difference for

minimum interferences at Point s can be

one half Lambda 3 halfes Lambda 5 Hales

Lambda and so on the half Lambda path

difference from the sources adds to any

path difference to determine the overall

path difference at Point s Choice a the

overall path difference = 1/4 Lambda +

12 Lambda = 3/4 Lambda this is

incorrect Choice B the overall path

difference equal 1 12 Lambda + 1 12

Lambda = 1 lamb Lambda this is

incorrect Choice C the overall path

difference equals 1 Lambda + 1 12 Lambda

equal 3es Lambda this is correct Choice

D the overall path difference equals 3es

Lambda + 1 12 Lambda = 2 Lambda this is

incorrect exam style question two an

outdoor concert has two large speakers

beside the stage for broadcasting

music in order to test the speakers they

are made to emit sound of the same

wavelength and the same

amplitude the curved lines in the

diagram represent wave fronts where is

the loudest sound heard the wave fronts

connect points of equal phase in the

sound wave such as

compressions the loudest sound occurs

where constructive interference takes

place constructive interference happens

when compressions from both speakers

overlap or when rare factions from both

speakers

overlap this occurs where the wave

fronts from the two speakers

intersect so D is

correct investigating the interference

of sound waves two identical

loudspeakers connected to a single

generator produce coherent sound waves

with a phase difference of zero as a

microphone is moved along line AB in

front of the loudspeakers a series of

alternating loud and quiet regions is

observed at the loud regions

constructive interference occurs

resulting in a sound intensity that is

twice that of a single

loudspeaker these regions are called

antinodes at the quiet regions

destructive interference occurs

resulting in a sound intensity that is

lower than that of a single

loudspeaker these regions are called

nodes the the central loud region marks

the zero order

antinode at this antinode both the path

difference and the phase difference are

zero at the next quiet region marks the

first order node at this node the path

difference is half a wavelength and the

phase difference are Pi

radians at the next loud region marks

the first order

antinode at this antinode the path

difference is one wavelength and the

phase difference are 2 pi radi Ian at

the next quiet region marks the second

order node at this node the path

difference is three half wavelength and

the phase difference are 3 Pi

radians this pattern is symmetric about

the central quiet like this the graph

shows the variation of sound intensity

detected by the microphone with distance

from point A as displayed on the

C the intensity at nodes may not be

exactly zero due to the presence of

background

noise exam style question three two

loudspeakers are placed near to each

other and facing in the same direction a

microphone connected to an oscilloscope

is moved along a line some distance away

from the loudspeakers as shown which

statement about the waves emitted by the

loudspeakers is not a necessary

condition for the microphone to detect a

fixed point along the line where there

is no sound Choice a the waves must be

emitted in Phase this is not a necessary

because the interference can be occurred

if the waves emitted in out of phase

Choice B the waves must be emitted with

a similar

amplitude this is a necessary because we

need to detect where no sound which

means that completely destructive

interference Choice C the waves must

have the same

frequency this is necessary because it

is condition of coherent waves Choice D

the waves must have the same

wavelength this is necessary because it

is condition of coherent

waves investigating the interference of

light waves a white light source is used

rather than a laser a monochromatic

filter allows only one wavelength of

light to pass through a single slit

diffracts the light this defract light

arrives in phase at the double slit

which ensures that the two parts of the

double slit behave as coherent sources

of light the double slit is placed a

centimeter or two beyond the single slit

acts as two coherent sources the

interference pattern is observed on a

screen about a meter away the bright

fringes have equal intensity and are

equally fringes separation as shown the

experiment has to be carried out in a

darkened room as the intensity of the

light is low and the fringes are hard to

see when a red filter is used a

interference pattern is shown when a

green filter is used a interference

pattern is shown showing that the Fringe

separation is smaller than with red

light when a blue filter is used a

interference pattern is shown showing

that the Fringe separation is smaller

than with green light when no filter the

interference pattern of white light is

shown this pattern shows a spectrum

cooler with blue light fringes closer to

the center and red light fringes further

away this is because red light has a

longer wavelength than blue light and

longer wavelengths defract more spread

out more three key factors influence the

experimental setup slit width slit

widths are typically a fraction of a

millimeter since visible light

wavelengths are much smaller minimal

defraction occurs allowing for

sufficient light intensity to observe

fringes narrower slits would

significantly reduce light intensity

slit separation double slits are usually

separated by about a

millimeter larger separations would

result in closely spaced fringes making

them difficult to

distinguish screen distance the screen

is placed approximately 1 M from the

slits

this distance provides clearly separated

fringes without excessive dimming of the

light exam style question four light of

wavelength Lambda is incident normally

on two narrow slits S1 and S2 a small

distance apart bright and dark fringes

are observed on a screen a long distance

away from the slits the N dark Fringe

from the central bright Fringe is

observed at Point p on the screen which

equation is correct for all positive

values of n all positive values of n

equal 1 2 3 4 and so on the dark fringes

occur where destructive interference

take place which happens when the path

differences are 1 half Lambda 3 Hal

Lambda 5 Hales Lambda and so on Choice a

the path differences are 1 half Lambda

Lambda 3es Lambda and and so on this is

incorrect Choice B the path differences

are Lambda 2 Lambda 3 Lambda and so on

this is

incorrect Choice C the path differences

are 1 half Lambda 3 half Lambda 5 Hales

Lambda and so on this is correct Choice

D the path differences are three halfes

Lambda 5 Hales Lambda 7 Hales Lambda and

so on this is

incorrect Young's double slit experiment

a laser produces intense coherent light

as the light passes through the double

slits it is diffracted spreading out

into the space beyond this creates two

overlapping sets of waves and an

interference pattern is observed on a

screen placed about a meter away we will

focus on the interference pattern with

within this specific region within this

region the bright fringes exhibit equal

intensity and consistent Fringe

separation as shown the double slit

experiment can be used to determine the

wavelength Lambda of light the following

three quantities have to be measured

slit separation a in meters Fringe

separation X in meters this is the

distance between the centers of adjacent

bright or dark fringes s flip to screen

distance D in meters the formula for

calculating the wavelength is Lambda

equals ax / D this formula can explain

why the Fringe separation for red light

is greater than that for green and blue

light as shown in the diagram given a

constant slit separation a and slit to

screen distance D the wavelength Lambda

is directly proportional to The Fringe

separation X since red light has the

longest wavelength The Fringe separation

X for red light is greatest blue light

with the shortest wavelength exhibits

the smallest Fringe

separation exam style question five a

double slit interference experiment is

set up as shown fringes are formed on

the screen the distance between

successive bright fringes is found to be

4 m

M given the wavelength of red light

equals Lambda the slit separation equals

a the slit to screen distance equals D

and Fringe separation x = 4 mm as shown

in the diagram two changes are then made

to the experimental

Arrangement the double slit is replaced

by another double slit which has half

the spacing the screen is moved so that

its distance from the double slit is

twice as great what is now the distance

between successive bright

fringes from the equation Lambda equal

ax / D rearranging the equation as xal

Lambda D / a from the diagram the old

Fringe separation x = 4

mm substituting new D = 2D and new a = a

/ 2 so mux = 4 Lambda D / a substitute

Lambda d/ a = 4

mm so newx = 16

mm exam style question six the diagram

shows an arrangement for demonstrating

two Source interference using coherent

light of a single wavelength Lambda an

interference pattern is observed on a

screen 3.0 M away from the slits X and Y

which have a separation of 1.0

mm the central bright Fringe is at Q and

the second bright Fringe from the center

is at P what is the distance between q

and P the distance PQ equals

2x from the equation Lambda equal ax / D

rearranging the equation as xal Lambda D

/ a substituting D equal 3 m and a = 1.0

* 10^ -3 M we get The Fringe separation

x = 3 * 10^ 3 Lambda so 2x = 6 * 10^ 3

[Music]

Lambda hey everyone welcome to PL

Academy hope you found this video useful

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everything on the syllabus like you can

see in the

figure defraction grating a defraction

grating consists of a large number of

equally spaced lines ruled onto a glass

or plastic slide each line acts as a

source of defraction for the incident

light the gratings can have up to 10,000

lines per

cimer when light passes through the

grating an interference pattern of

fringes is

observed in the diagram monochromatic

light from a laser is incident normally

on a transmission defraction grating

beyond the grating interference fringes

are formed and can be observed on a

screen Sim similar to the double slit

experiment the fringes are also referred

to as Maxima the central Fringe is

called the zeroth order maximum

a0 the next Fringe is the first order

maximum

A1 the next Fringe is the second order

maximum A2 and so on the pattern is

symmetrical so there are two first order

Maxima A1 2 second order Maxima A2 and

so on as shown to analyze the pattern we

measure the angles Theta and beta at

which the Maxima are formed rather than

measuring their

separation with double slits the fringes

are equally spaced and the angles are

very small with a defraction grating the

angles are much greater and the fringes

are not equally

spaced the differences in the

interference patterns between defraction

gratings and double slits are as follows

brighter and sharper fringes defraction

gratings typically produce brighter and

sharper fringes than double slits

because more light passes through the

numerous slits in the grating larger

Fringe separation defraction gratings

generally exhibit larger Fringe

separations than double slits because

the slit spacing in gratings is much

smaller than the slit separation in

typical double slit

experiments the wavelength Lambda can be

determined with a defraction grating by

measuring the angles at which the Maxima

occur we can determine the wavelength of

the incident light and calculating using

the equation D sin theta equals n Lambda

where D is the grating spacing in meters

Theta is the angle between the zero

order maximum and N order maximum n is

the order of Maximum as 1 2 3 and so on

and Lambda is the wavelength of light in

meters if the grating has K line per

meter then the grating spacing D can be

calculated using d = 1 /

K exam style question one a beam of red

laser light of wavelength 633 nanom is

incident normally on a defraction

grating with 600 lines per

millimet the beam of red light is now

replaced by a beam of blue laser light

of wavelength

445

nanom a replacement defraction grating

is used so that the first order maximum

of the blue light appears at the same

position on the screen as the first

order maximum of the red light from the

original

laser this means that the angle Theta is

the same for both and the first order

corresponds n equal 1 how many lines per

millimeter are there in the replacement

defraction grating using the equation D

sin Theta equal n Lambda for the blue

light and the replacement grating we

have D2 sin Theta = 1 time blue Lambda

for the red light and the original

grating we have D1 sin Theta = 1 * red

Lambda dividing the first equation by

the second equation we get D2 sin Theta

over D1 sin Theta equal blue Lambda over

red Lambda so sin Theta is the same for

both it cancels out leaving D2 / D1

equal blue Lambda over red Lambda

substituting blue Lambda 445 nanom and

red Lambda equal

633

nanom the nanometers cancel out so we

get D2 =

445 over 633 *

D1 substituting D2 = 1 / K2 and D1 = 1 /

K1 where K2 and K1 are the number of

line per millimeters for the replacement

and original gratings

respectively substituting K1 equals 600

per

MIM we solve K2 = 850 per millim for two

significant

figures exam style question question two

monochromatic light of wavelength 690

nanom passes through a defraction

grating with 300 lines per millimet

producing a series of Maxima bright

spots on a screen what is the greatest

number of Maxima that can be observed

the defraction grating with 300 lines

per millimet so grating spacing D equals

1 over 300 mm or 1 over 300 * 10 power

-3

M using the equation D sin Theta equal n

Lambda substituting d = 1 300 power -3 M

Theta = 90° and Lambda = 690 * 10^ -9

M we solve for n order Maxima equal

4.83 but n must be a whole numbers so

the maximum order is four so the

greatest number number of Maxima that

can be observed is 4 + 1 + 4 is equal to

9 Maxima which includes the zeroth order

maximum and four Maxima on left and

right hand sides as shown we substitute

the angle theta equals 90° because this

represents the maximum possible angle

from the zero order maximum to the

highest order

maximum I hope you found this video

helpful

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