Physics 1B03summer-Lecture 10
Interference of Waves
BeatsDouble Slit
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
time
t
in phase 180o out of phase in phase
Physics 1B03summer-Lecture 10
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:
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
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.
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
Example:
Two sources, in phase; waves arrive by different paths:
At detector P:
)sin(
)sin(
22
11
tkrAy
tkrAy
detectorS1
S2
Pr1
r2
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
Solution:
Physics 1B03summer-Lecture 10
10 min rest
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
Infrared
Red 780 nm
Yellow 600 nm
Green 550 nm
Blue 450 nm
Violet 380 nm
Ultraviolet
Visible-LightVisible-Light SpectrumSpectrum
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
Our galaxy (Milky Way) at viewed at different wavelengths
Physics 1B03summer-Lecture 10
Radio lobes (jets) from a supermassive black hole at the center of the galaxy NGC 4261
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
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.
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
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, ………
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
Solution:
Physics 1B03summer-Lecture 10
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 ?
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
10 min rest
Physics 1B03summer-Lecture 10
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)
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
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 !
Physics 1B03summer-Lecture 10
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 ?
Physics 1B03summer-Lecture 10
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
Physics 1B03summer-Lecture 10
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?
Physics 1B03summer-Lecture 10
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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