Interference of Waves Beats Double Slit

Physics 1B03summer-Lecture 10

Interference of Waves

BeatsDouble Slit


BeatsBeats

Two waves of different frequencies arriving together produce a fluctuation in power or amplitude.

Since the frequencies are different, the two vibrations drift in and out of phase with each other, causing the total amplitude to vary with time.

y

time

1 beat


time

t

in phase 180o out of phase in phase


Same amplitudes, different frequencies:

)cos( 11 tAy )cos( 22 tAy

Trigonometry: cos a + cos b = 2 cos [(a-b)/2] cos [(a+b)/2]

Result:

ttA

yyy

2cos

2cos2 2121

21

slowly-varyingamplitude

SHM at averagefrequency

The math:


Note:

2 beats per cycle of

t2

cos 21

# beats/second =

22 21 ff

The beat frequency (number of beats per second) is equal to the difference between the frequencies:

21 fffb


Quiz

Two guitar strings originally vibrate at the same 400-Hz frequency. If you hear a beat of 5Hz, what is/are the other possible frequencie(s) ?

a) 10 Hzb) 395 Hzc) 405 Hzd) 395 Hz and 405 Hz


InterferenceInterference

2 waves, of the same frequency; arrive out of phase.

Eg: y1=Asin t y2=Asin (t+)

Then yR= y1 + y2 = AR sin(t+R),

and the resultant amplitude is AR=2Acos(½.

Identical waves which travel different distances will arrive out of phase and will interfere, so that the resultant amplitude varies with location.


Phase difference :

Define

Then, at detector:

rkrrktkrtkr )()()( 2121

(pick starting timeso initial phase is zero here)

cycles) (or

radians2

r

rrk

)2

sin()2

cos2(

)sin(

)sin(

2

1

tAy

tAy

tAy

R


Example:

Two sources, in phase; waves arrive by different paths:

At detector P:

)sin(

)sin(

22

11

tkrAy

tkrAy

detectorS1

S2

Pr1

r2


8 m

x

2 speakers, in phase; f = 170 Hz (so = 2.0 m; the speed of sound is about 340 m/s)

As you move along the x axis, where is the sound:

a) a minimum (compared to nearby points)?b) a maximum (compared to nearby points)?

detector


Solution:


10 min rest


Interference of LightInterference of Light

Light is an electromagnetic (EM) wave.

Wave properties:

Diffraction – bends around corners, spreads out from narrow slits

Interference – waves from two or more coherent sources

interfere


Electromagnetic WavesElectromagnetic Waves

Usually we keep track of the electric field E :

Electric field amplitude

all vBE

,,

E B (magnetic field)Eo

)sin(),( tkxtx AE

v

Electromagnetic waves are transverse waves


Infrared

Red 780 nm

Yellow 600 nm

Green 550 nm

Blue 450 nm

Violet 380 nm

Ultraviolet

Visible-LightVisible-Light SpectrumSpectrum


The Electromagnetic SpectrumThe Electromagnetic Spectrum

(m) f (Hz) 300 106

3 108

3 x 10-3 1011

3 x 10-6 1014

7 x 10-7 5x1014

4 x 10-7

3 x 10-9 1017

3 x 10-12 1020

RadioTV

Microwave

Infrared

Visible

UltravioletX rays rays


Our galaxy (Milky Way) at viewed at different wavelengths


Radio lobes (jets) from a supermassive black hole at the center of the galaxy NGC 4261


Double SlitDouble Slit(Thomas Young, 1801)

Result: Many bright “fringes” on screen, with dark lines in between.

screen

double slit

separation d

θ

m=2m=1

m=0 (center) m=-1 m=-2

incident light


The slits act as two sources in phase. Due to diffraction, the light spreads out after it passes through each slit. When the two waves arrive at some point P on the screen, they can be in or out of phase, depending on the difference in the length of the paths.

The path difference varies from place to place on the screen. Pr1

r2

Δr = r

1-r2

d

To determine the locations of the bright fringes (interference maxima), we need to find the points for which the path difference r is equal to an integer number of wavelengths.

For dark fringes (minima), the path difference is integer multiples of half of a wavelength.


For light, the slits will usually be very close together compared to the distance to the screen. So we will place the screen “at infinity” to simplify the calculation.

Pr1

r2

Δr = r

1-r2

d

r >> d, r1 & r2 nearly parallel

d

Δr

θ

θ

Δr = d sin θ

move P to infinity


Interference:Interference: 2 coherent waves, out of phase 2 coherent waves, out of phase due to a path difference due to a path difference r:r:

Constructive Interference (maximum intensity)

for = 0, ±2π, ±4π, ±6π, ……… -> Δr =0, ±, ±2 , ±3 , ………

cycles

radians 2difference phase

r

r

Destructive Interference (minimum intensity)

for = ±π, ±3π, ±5π, ……… -> Δr =±λ/2, ±3λ/2, ±5λ/2, ………


Constructive Interference: (bright)

Δr = mλ ord sin θ = mλ, m = 0, ±1, ±2, …

But, if the slit-screen distance (L) is large, then sinθ~θand so sinθ=θ=y/L (in radians):

L

yθ

So we have: mL

dy

d


Destructive Interference: (no light)

Δr = (m + ½)λ ord sin θ = (m + ½) λ, m = 0, ±1, ±2, …

So, we have: )2

1( m

L

dy


Two slits are illuminated with red light to produce an interference patter on a distant screen. If the red light is replaces with blue light, how does the pattern change?

A) The bright spots move closer togetherB) The bright spots move farther apartC) The pattern does not changeD) The patter doesn’t chance, but the width of the

spots changes

Quiz


ExamplExamplee

Where are a) the bright fringes?b) the dark lines?

(give values of y)

3 m

y

0

2 slits, 0.20 mm apart; red light ( = 667 nm)

screen


Solution:


Example

A double slit interference patter is observed on a screen 1.0m behind two slits spaced 0.3mm apart. Ten bright fringes span a distance of 1.65 cm.

What is the wavelength of light used ?


Which of the following would cause the separation between the fringes to decrease?

A) Increasing the wavelength B) Decreasing the wavelength C) Moving the slits closer together D) Moving the slits farther apart E) None of the above

Quiz


10 min rest


Refractive IndexRefractive Index

material refractive index speed of light

vacuum 1 c 300,000 km/sair 1.0003glass about 1.5 200,000 km/swater 1.333 225,000 km/sdiamond 2.4 125,000 km/s

The speed of light depends on the material. We define the refractive index “n” as

n = (speed of light in vacuum)/(speed of light in a material)


Question:

A beam of yellow light (wavelength 600 nm), travelling in air, passes into a pool of water. By what factor do the following quantities change as the beam goes from air into water?

A) speed B) frequency C) wavelength


Reflection and Phase Reflection and Phase ChangeChange

Light waves may have a 180° phase change when they reflect from a boundary:

“optically dense” medium (larger refractive index)

no phase change at this reflection180° phase change

when reflecting from a denser medium

Just remember this : low to high, phase shift of pi !


Example: Thin filmExample: Thin film

What is the minimum thickness of a soap film (n 1.33) needed to produce constructive interference for light with a 500nm wavelength ? (air : n 1.00).

What about destructive interference ?


Example: Antireflection Example: Antireflection coatingscoatings

To reduce reflections from glass lenses (n 1.5), the glass surfaces are coated with a thin layer of magnesium fluoride (n 1.38). What is the correct thickness of the coating for green light (550 nm vacuum wavelength)?

MgF2air glass


ExampleExample

A beam of 580 nm light passes through two closely spaced glass plates (nglass=1.6), as shown in the figure below. For what minimum nonzero value of the plate separation d is the transmitted light dark?


QuizQuiz

Why do we see many colours on a soap bubble?

A) because white light is made up of different wavelengths

B) because the bubble has different thickness

C) both A and B

D) because the bubble is round and light reflects from the other side

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Interference of Waves Beats Double Slit