textbook sections 28-1 -- 28-3
DESCRIPTION
Physics 1161: Lecture 20 Interference. textbook sections 28-1 -- 28-3. +1. t. -1. +1. t. -1. +2. t. -2. Superposition. Constructive Interference. +. In Phase. +2. t. -2. Superposition. Destructive Interference. +1. t. -1. +. +1. Out of Phase 180 degrees. t. -1. - PowerPoint PPT PresentationTRANSCRIPT
• textbook sections 28-1 -- 28-3
Physics 1161: Lecture 20Interference
Superposition
t
+1
-1
t
+1
-1
t
+2
-2
+
Constructive Interference
In Phase
Superposition
t
+1
-1
t
+1
-1
t
+2
-2
+
Destructive Interference
Out of Phase180 degrees
Which type of interference results from the superposition of the two waveforms shown?
1 2 3
0% 0%0%
1. Constructive2. Destructive3. Neither+
Different f
-1.5
-1
-0.5
0
0.5
1
1.5
-1.5
-1
-0.5
0
0.5
1
1.5
Which type of interference results from the superposition of the two waveforms shown?
1 2 3
0% 0%0%
1. Constructive2. Destructive3. Neither+
Different f
-1.5
-1
-0.5
0
0.5
1
1.5
-1.5
-1
-0.5
0
0.5
1
1.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Interference for Light …• Can’t produce coherent light from separate
sources. (f 1014 Hz)
Single source
Two different paths
Interference possible here
• Need two waves from single source taking two different paths– Two slits– Reflection (thin films)– Diffraction*
Coherent & Incoherent Light
Double Slit Interference Applets
• http://www.walter-fendt.de/ph14e/doubleslit.htm
• http://vsg.quasihome.com/interfer.htm
Young’s Double Slit Applet
http://www.colorado.edu/UCB/AcademicAffairs/ArtsSciences/physics/PhysicsInitiative/Physics2000/applets/twoslitsa.html
Young’s Double Slit Layout
Interference - Wavelength
Light waves from a single source travel through 2 slits before meeting at the point shown on the screen. The interference will be:
1 2 3
0% 0%0%
1. Constructive2. Destructive3. It depends on L
Screen a distance L from slits
Single source of monochromatic light
d
2 slits-separated by d
L
Light waves from a single source travel through 2 slits before meeting at the point shown on the screen. The interference will be:
1 2 3
0% 0%0%
1. Constructive2. Destructive3. It depends on L
Screen a distance L from slits
Single source of monochromatic light
d
2 slits-separated by d
L
The rays start in phase, and travel the same distance, so they will arrive in phase.
Young’s Double SlitCheckpoint
Screen a distance L from slits
Single source of monochromatic light
d
2 slits-separated by d
1) The pattern of maxima and minima is the same for original and modified experiments.
2) Maxima and minima for the unmodified experiment now become minima and maxima for the modified experiment.
L
½ shift
The experiment is modified so that one of the waves has its phase shifted by ½ . Now, the interference will be:
Young’s Double SlitCheckpoint
Screen a distance L from slits
Single source of monochromatic light
d
2 slits-separated by d
1) The pattern of maxima and minima is the same for original and modified experiments.
2) Maxima and minima for the unmodified experiment now become minima and maxima for the modified experiment.
L
½ shift
The experiment is modified so that one of the waves has its phase shifted by ½ . Now, the interference will be:
For example at the point shown, he rays start out of phase and travel the same distance, so they will arrive out of phase.
Young’s Double Slit Concept
Screen a distance L from slits
Single source of monochromatic light
d
2 slits-separated by d
L
At points where the difference in path length is 0, ,2, …, the screen is bright. (constructive)
At points where the difference in path
length is
the screen is dark. (destructive)
2
5 ,2
3 ,2
Young’s Double Slit Key IdeaL
Two rays travel almost exactly the same distance. (screen must be very far away: L >> d)
Bottom ray travels a little further.
Key for interference is this small extra distance.
d
Path length difference =
d
Young’s Double Slit Quantitative
Destructive interference dsin (m
12)
Constructive interference dsin m
where m = 0, or 1, or 2, ...
d sin
Need < d
d
Destructive interference dsin (m
12)
Constructive interference dsin m
where m = 0, or 1, or 2, ...
Young’s Double Slit Quantitative
y
sin() tan() = y/L
dLmy
d
Lmy
21
L
A little geometry…
Young’s Double Slit Under WaterCheckpoint
d
L
y
When this Young’s double slit experiment is placed under water, how does the pattern of minima and maxima change?
1) the pattern stays the same
2) the maxima and minima occur at smaller angles
3) the maxima and minima occur at larger angles
Young’s Double Slit Under WaterCheckpoint
d
L
y
When this Young’s double slit experiment is placed under water, how does the pattern of minima and maxima change?
1) the pattern stays the same
2) the maxima and minima occur at smaller angles
3) the maxima and minima occur at larger angles…wavelength is shorter under water.
Young’s Double SlitCheckpoint
In Young’s double slit experiment, is it possible to see interference maxima when the distance between slits is smaller than the wavelength of light?
1) Yes 2) No
Young’s Double Slit CheckpointIn Young’s double slit experiment, is it possible to see interference maxima when the distance between slits is smaller than the wavelength of light?
1) Yes 2) No
Need: d sin = m => sin = m / d
If > d then / d > 1
so sin > 1
Not possible!
Reflections at Boundaries
Free End ReflectionNo phase change
Slow Mediumto
Fast Medium
Fast Mediumto
Slow Medium
Fixed End Reflection180o phase change
Newton’s Rings
Iridescence
Iridescence
Soap Film Interference• This soap film varies in
thickness and produces a rainbow of colors.
• The top part is so thin it looks black.
• All colors destructively interfere there.
Thin Film Interference
n1 (thin film)
n2
n0=1.0 (air)
t
1 2
Get two waves by reflection from the two different interfaces.
Ray 2 travels approximately 2t further than ray 1.
Reflection + Phase Shifts
n1
n2
Upon reflection from a boundary between two transparent materials, the phase of the reflected light may change.
• If n1 > n2 - no phase change upon reflection.
• If n1 < n2 - phase change of 180º upon reflection. (equivalent to the wave shifting by /2.)
Incident wave Reflected wave
Thin Film Summary
n1 (thin film)
n2
n = 1.0 (air)
t
1 2
Ray 1: d1 = 0 or ½
Determine d, number of extra wavelengths for each ray.
If |(d2 – d1)| = ½ , 1 ½, 2 ½ …. (m + ½) destructiveIf |(d2 – d1)| = 0, 1, 2, 3 …. (m) constructive
Note: this is wavelength in film!
(film= o/n1)+ 2 t/ film
Reflection Distance
Ray 2: d2 = 0 or ½
This is important!
Thin Film Practice
nglass = 1.5
nwater= 1.3
n = 1.0 (air)
t
1 2
d1 =
d2 =
Blue light (o = 500 nm) incident on a glass (nglass = 1.5) cover slip (t = 167 nm) floating on top of water (nwater = 1.3).
Is the interference constructive or destructive or neither?
Phase shift = d2 – d1 =
Thin Film Practice
nglass = 1.5
nwater= 1.3
n = 1.0 (air)
t
1 2
d1 = ½
d2 = 0 + 2t / glass = 2t nglass/ 0= 1
Blue light (o = 500 nm) incident on a glass (nglass = 1.5) cover slip (t = 167 nm) floating on top of water (nwater = 1.3).
Is the interference constructive or destructive or neither?
Phase shift = d2 – d1 = ½ wavelength
Reflection at air-film interface only
Blue light = 500 nm incident on a thin film (t = 167 nm) of glass on top of plastic. The interference is:
1 2 3
0% 0%0%
nglass =1.5
nplastic=1.8
n=1 (air)
t
21
1. Constructive2. Destructive3. Neither
Blue light = 500 nm incident on a thin film (t = 167 nm) of glass on top of plastic. The interference is:
1 2 3
0% 0%0%
nglass =1.5
nplastic=1.8
n=1 (air)
t
21
1. Constructive2. Destructive3. Neither
d1 = ½ d2 = ½ + 2t / glass = ½ + 2t nglass/ 0= ½ + 1
Phase shift = d2 – d1 = 1 wavelength
Thin FilmsCheckpoint
The gas looks: • bright • dark
A thin film of gasoline (ngas=1.20) and a thin film of oil (noil=1.45) are floating on water (nwater=1.33). When the thickness of the two films is exactly one wavelength…
t =
nwater=1.3
ngas=1.20
nair=1.0
noil=1.45
The oil looks: • bright • dark
Thin FilmsCheckpoint
The gas looks: • bright• dark
A thin film of gasoline (ngas=1.20) and a thin film of oil (noil=1.45) are floating on water (nwater=1.33). When the thickness of the two films is exactly one wavelength…
t =
nwater=1.3
ngas=1.20
nair=1.0
noil=1.45
d1,gas = ½
The oil looks: • bright• dark
d2,gas = ½ + 2 d1,oil = ½ d2,oil = 2| d2,gas – d1,gas | = 2 | d2,oil – d1,oil | = 3/2
constructive destructive