Download n - Eastchester High School

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AP Physics – Electromagnetic Waves, Reflection and Refraction of Light –
Chapter 23-24
Chapter 23 - Reading pp. 719-739 - text HW #17, 18, 24, 27
Chapter 24 - Reading pp. 748-771 – text HW #6, 7, 8, 9, 12, 13, 17, 18, 19,
20, 24, 28, 29, 32, 33, 36, 37, 38, 39, 42, 30(E.C.)
The story so far… In the last chapter, we studied mechanical waves such
as sound waves which required a matter medium for propagation.
And now…
23.1- Electromagnetic Waves- waves resulting from the
vibration of charged particles. Such waves are the result of:
1) the changing electric field around the oscillating particle
and;
2) the changing magnetic field induced by the changing
electric field.
Since these vibrations manifest themselves in the field, they
do not require a matter medium for transmission.
James Clerk Maxwell – 1865 – developed a
theoretical mathematical theory that showed the
above relationship between electrical and
mathematical phenomena. Maxwell’s four equations
allowed him to derive the speed of propagation for
EM Waves in a vacuum, which is:
c=2.99792458 x 108 m/s.
Wow. Maxwell’s theoretical EM waves would travel at the speed of light!
This means:
Light is an Electromagnetic wave!
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23.2 - This led to the discoveries of Heinrich Hertz in 1887. He used a variable AC
circuit that could alternate the charge on the two metal spheres on his transmitter
circuit (left), causing a spark to jump at different frequencies. As he adjusted the
frequency of the receiver by changing the spacing between the brass spheres, he
found that he could get a spark to jump across the gap in the receiver.
Hertz invented the first Radio.
Transmitter
Receiver
Note:
1) In air, light travels at approximately c=3 x 108 m/s
(for calculations in air, use speed of light in a vacuum)
2) In denser optical media (water, glass, diamond) Light and
EM waves move SLOWER than c.
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23.5 - The Electromagnetic Spectrum
Which has the highest frequency? Gamma rays
Which has the lowest frequency? Radio waves
Which has the largest wavelength? Since c=fλ, radio waves
Which has the smallest wavelength? Gamma rays
Which has the most energy? Gamma rays Why? E = hf
Which has the least energy? Radio waves
Note: for mechanical waves (and harmonic oscillators like the pendulum and mass
spring system), energy affects the amplitude of the vibration, only. Whereas a
higher frequency for a mechanical wave yields the same energy per oscillation, a
higher frequency for EM waves means MORE ENERGY per oscillation, if you think
of light as a photon of course, but I digress…
Back to light as a wave…
Fluoroscope Videos
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The Visible Light Spectrum- Some important facts:
1) ROYGBIV
2) Wavelength range of visible light in air/vacuum:
In nanometers (1x10-9m): 400-700 nm (approximate)
Note: 400nm = Violet, 700 nm=Red
In Angstroms (1x10-10m): 4000-7000 Å
3) Frequency range of visible light in air/vacuum:
4.3 x 1014 - 7.5 x 1014 Hz (approximate)
Note: What is the wavelength and frequency range of
visible light in water?
300-500 nm (approximate)
4.3 x 1014 - 7.5 x 1014 Hz (approximate)
Therefore, the color/characteristic of a wave is determined
by the wave’s FREQUENCY.
Now go home and do HW Chapter 23 -- #17, 18, 24, 27
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AP Physics - Reflection and Refraction of light - Chapter 24
Chapter 24 - Reading pp. 748-771 – text HW #6, 7, 8, 9, 12, 13, 17, 18, 19,
20, 24, 28, 29, 32, 33, 36, 37, 38, 39, 42, 30(E.C.)
24.2 Measuring the Speed of Light in Air –
Precisely measured in 1926 by Albert Michelson, after the predecessor to
the US Geologic Survey painstakingly measured the baseline distance
from the Mount Wilson Observatory in Los Angeles to Old Baldy, 22 mi
away (35 km) and he was still off by about 48000 m/s.
How did Michelson use this apparatus? The rotating mirror had to be in
the exact orientation above to properly reflect light from the light bulb to
the telescope. The frequency of rotation was increased until the point
when he could first see the reflection of the light through the telescope.
(That frequency was 32000 rpm, or 533.33 Hz, a period of .001875 sec/rotation, or a time of
.001875/8=.000234 sec for 1/8 a rotation. This yields a v=70000m/.000234sec=2.9866x108 m/s)
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(using 3x108m/s, t=2.3933x10-4s,
ω=3281.65Rad/s or 522Hz or 31320rpm)
(this would be a quarter turn in same
time, or double the frequency)
Also, read about the first successful terrestrial (non astronomical) method (1849) FIZEAU’S
technique for calculating the speed of light in air on page 752.
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24.4-24.5 Reflection of Light
When a light ray is incident upon a boundary leading into a
second medium, it is reflected back into the first medium,
governed by the law of reflection.
θi=θr (measured to a line normal to the boundary)
Regular reflection and diffuse reflection?
DEMO EACH (Laser on paper and on
mirror and smoke machine)
#1) Find the angle of reflection off of mirror M2
Now go home and do HW Chapter 24 #6,7,8
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24.5 REFRACTION
I. What happens when a ray of light traveling through a transparent
medium encounters the boundary leading into another transparent
medium?
1. Partial Reflection
2. Partial Absorption
3. Partial Transmission – this part may be refracted.
Why do materials look darker when they are wet?
If your shirt gets wet, the light that is illuminating the shirt now has to pass through a layer of water.
Since some of the light is reflected and some gets absorbed, less light illuminates the shirt, less light reflects
off the shirt, and therefore the shirt looks darker.
Refraction – the bending of a wave as it travels from one
medium to a different medium density at an angle. (as it
bends, its speed also changes)
DEMO: Blackboard optics or beaker of Water / Laser
The Index of Refraction of a transparent medium is defined
as the ratio of the speed of light in a vacuum to the speed of
light in the medium
c
n
v
Indices of refraction measured with λ = 589 nm (Yellow line in
bright-line spectrum of sodium)
(Text Ref. Table 24-1)
Air
n = 1.000293
Water
n = 1.333
Crown Glass n = 1.52
Flint Glass
n = 1.66
Diamond
n = 2.419
Zircon
n = 1.923
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#2) What is the velocity of light (λ = 589 nm) in
A) air?
B) crown glass?
C) diamond?
#3) The speed of light in a material is 1.807 x 108 m/s. That material may
be ___.
#4) What material may have an index of refraction of 0.6?
Therefore, how is the index of refraction “n” related to the speed in that
medium?
#5) In the table of indices of refraction, why was in necessary to mention
that the values were those measured with λ = 589 nm? Some media
are DISPERSIVE, which means waves of different
wavelengths travel at different speeds in that medium.
This means that:
Index of refraction is a function of the wavelength.
Hmmm…. Why is index of refraction not a function of frequency? Try to
derive an equation with n and f in it.
v1
n
c
v
f
f
o
o
1
or
2
n1
1
n2
2
v2
f
f
v1
v2
c
c
n1
n2
n2
n1
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#6)Draw the path as the ray enters the water
Compare the following:
v1 > v2
f1 = f2
λ1 > λ2
*Remember: when a ray is refracted, the wave’s frequency
and phase remain the same.
Refraction – Which way does it bend?
When a light ray goes from
low n to high n, its speed
________ and the ray bends
TOWARDS the normal.
When a light ray goes from
high n to low n, its speed
________ and the ray bends
AWAY from the normal.
Gruber says: remember these two important memory tricks:
HIGH to LOW,
AWAY we go!
LOW to HIGH,
Stay close BY!
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Recall, Willebrord Snellius’ Law (1621)
n1sinθ1 = n2sinθ2
#7) For each light ray (λ=589 nm), draw the path and find the angle of
refraction.
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#8) Trace the path
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#9) A monochromatic ray with wavelength of 6.5 x 10-7 m in air is incident upon a
flint glass boundary. The angle of incidence is 40o.
A) Sketch the path and find the angle of refraction.
B) Find the speed of light in the flint glass (n=1.66).
C) Find the frequency in the flint glass.
D) Find the wavelength in the flint glass.
E) What color is the light in air?
F) What color is the light in the flint glass?
Note: We know that
n
c
v
n
o
n
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What other ratio is n equal to?
Now go home and do HW Ch. 24 # 9,12,13,17,18,19,20,24,28,29
#10) A ray of light enters and exits a glass of water.
θ = 20°
A) Trace the path as light enters and exits
the glass of water. Show all angles.
B) What is the time it takes for the light to
pass through the entire glass of water?
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24.6 Dispersion and Prisms
What is white light?
White light is what the human eye
perceives when all three cones in
the eye are hit by a combination of
all visible wavelengths.
Review: Which color in the visible spectrum has the
Longest wavelength ______
Shortest wavelength ______
Recall that the index of refraction is a function of wavelength. This
means that light of different wavelengths will be bent at different angles
when entering a dispersive refracting material. So…
What happens when white light enters a glass prism?
It undergoes DISPERSION.
Dispersion – The breaking up of polychromatic light into
separate monochromatic rays as it passes into a dispersive
medium or prism.
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According to the above graph, Index of refraction decreases
with increasing wavelength. This means that
VIOLET light bends the MOST VIOLENTLY, and
Red light bends the least.
Trace the path as the white light enters and exits the prism.
In the prism, which color travels the
fastest?
slowest?
24.7 How are rainbows made?
Double Rainbows?
Video – how rainbows are made.
Video – Whoa… Double Rainbow… Why?
Video – How double rainbows are made
http://www.youtube.com/watch?v=FJVvtOy-ukE 22:50-40 min
Now go home and do HW ch. 24 #32,33
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24.9 Total Internal Reflection
Total internal reflection occurs when a ray, moving from a
medium of high n to low n, is incident at an angle greater
than the critical angle.
The critical angle, θc, of a medium is the angle that yields
an angle of refraction = 90º
#11) Monochromatic light travels from air into water (n=1.33). Complete
the chart for the following angles and draw each refracted ray.
θincident
0°
25°
48.7534664°
89°
Θrefracted
Hence, when light is traveling from n1 to n2 where n1 < n2
(low to high), refraction occurs for 0 < θi < 90.
Monochromatic light travels from water into air. Complete the chart for
the following angles and draw each refracted ray.
θincident
0°
25°
48.7534664°
89°
Θrefracted
Hence, when light is traveling from n1 to n2 where n1 > n2
(high to low), refraction occurs only for 0 < θi < θc.
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Critical angle (θc): When a ray moves from high n to low n
at θi = θc, the refracted ray moves parallel to the boundary.
Thus:
If θi < θc then refraction occurs.
If θi = θc then refraction occurs and θr = 90°.
If θi > θc then total internal reflection occurs.
Using Willebrord Snellius’ Law, derive the formula for θc in terms of n1
and n2, and the special case equation when n2 is air.
http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/totintrefl/index.html
DEMO: Show total internal reflection with laser and block.
DEMO: Fiber-optic cables
DEMO: Water/laser through hole in cylinder (Use Torricelli’s Law apparatus)
DEMO: Beer mug filled with beer in air and water. Discussion on next page.
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Explain this one: a frosty mug (soda-lime glass, n = 1.5) filled with your favorite
beer, when viewed in air, looks fuller than when viewed in a medium such as
water. In water, the very thick glass of the beer mug is noticeable.
View from above:
Hints:
#12(a) For a θi =40°, find θr going from air into glass and from water into
glass.
(b) Find the critical angle for the glass-air boundary. Repeat for the glasswater boundary.
Putting the mug in water changes the outer glass interface from an air-glass
interface to a water-glass interface. This means there is ________ of a bend as
light rays travel from water into the glass and travels through the wall. This also
means that there is a _________ critical angle at the glass-water interface. When
the ray hits the other side, there is a ________ chance of the incident light ray
undergoing total internal reflection, so refraction ________, and then light that
passed through the glass hits your eye.
DEMO: Fiber-optic cables in water?
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#13) Find the critical angle for a diamond–air boundary.
Diamond ring and laser in and out of water. Glycerine?
Now go home and do HW Ch. 24 #36,37,38,39,42,30(E.C.)
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#14 (A) Trace the path as a beam of monochromatic light in air enters and
exits the glass prism. (Show all angles)
B) Repeat when the prism is submerged in water (n = 1.33)
C) What would be the index of refraction of an unknown liquid that the
prism would need to be dropped into such that the ray would refract at
900 at the hypotenuse boundary?
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Here’s another interesting question: If a fish is underwater, and I see it,
does that mean it can see me? Conversely, if a fish is underwater, and it
sees me, does that mean I can see it?
#15) (A) A fish is 6 m below the water surface. Where does a person in
the air have to look to see the fish?
The critical angle makes a “cone of visibility” for seeing
the fish. The person in the boat has to be in the area of
radius R above the fish.
(B) Determine R.
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So what does the fish see?
Swimming pool vid 1– Scuba diver surfacing vid 2
#16) motivated by his neighbor’s fake rock hide-a-key, Mr. Gruber decides he is
going to build one using a coaster of diameter _____, hang the key from the
center of it, and let it float around the pool. What is the maximum length that
the key can hang from the coaster so that no one can see it from above the
surface of the pool?
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Inverse Square Law of Illumination
Luminous Intensity – The Luminous strength of a light
source in a particular direction. Roughly speaking, it’s how
bright the source is.
(Units: candela (Cd) or CandlePower).
A 40-watt tungsten filament ≈ 35 Cd
A 40-Watt fluorescent lamp ≈ 200 Cd
Why? Fluorescent lamps are more efficient (heating a low
pressure gas requires less thermal energy than a solid
tungsten filament
Note: Luminous intensity is not to be confused with
intensity of illumination.
Intensity of illumination (Unit – lumen (lm))
1 lm is defined as the amount of light falling upon a 1-m2
area from a 1-cd light source a distance 1 m away.
Note:
I
1
2
r
#17)
(A)
(B)
At a distance of 2 m from the 1-cd light source, what is the intensity of
illumination on a 1-m2 area? 0.25 lumen
The intensity of radiation from the Sun is 9126 W/m2 at the distance of
Mercury (0.387 Astronomical Units); Determine the approximate intensity
of radiation at the distance of Earth (1 AU)? 1367 W/m2
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(C) Graph the intensity of a light source vs. the distance from the source.
INTENSITY = POWER/AREA
(SI units: Watt/m2)
In the Modern Physics section, we will talk about Intensity
as photons falling upon a unit area per unit time.
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AP Physics – Wave Optics – Diffraction, Interference and Polarization –
Chapter 26
Chapter 26 - Reading pp. 809-833- text HW #1,5,6,9,17,18,27,28,38,40
26.2 Diffraction and Young’s Double Slit Interference
What is Diffraction?
Diffraction is the spreading out of a wave when it passes
through an opening (slit) or around a barrier
Draw what happens to the parallel wave fronts when they encounter the
break in the sea wall as shown:
Draw what happens to the parallel wave fronts when they encounter two
breaks in the sea wall as shown:
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Remember interference?
Interference: The resulting wave pattern that occurs when
two waves pass through each other in the same medium
What are the conditions in which 2 waves have maximum
constructive interference?
In phase, same frequency (and same amplitude helps too.)
What are the conditions in which 2 waves have maximum
destructive interference?
180° phase difference, same frequency, same amplitude.
For points A, B and C in the interference pattern below, determine the
following:
Type of Interference
Point A
Constructive
Point B
Destructive
Point C
Constructive
Distance from S1
3λ
3.5λ
4λ
Distance from S2
4λ
4λ
4λ
Path difference δ
1λ
½λ
0λ
DEMO: Ripple Tank Simulator http://www.falstad.com/ripple/
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And now…
Thomas Young’s Double Slit interference experiment (1801)
There was much debate over the nature of light in the 1600s. Hooke, Huygens, and Euler proposed a wave
theory of light, while Newton rejected wave theory and proposed a corpuscular (particle) theory of light
(interesting). Of course, everyone accepted Newton’s Theory – duh – it was Newton – SMH. Anyway, when
Young revisited the debate in the late 1700s, going against Newton, this was akin to heresy. But his double slit
experiment shut everyone up, at least until 1901 when Max Planck resurrected Newton’s particle theory
under the name “Quantum Theory”.
(DEMO: Laser/Diffraction Grating)
λ = wavelength
d = distance between the 2 slits
L = distance between slits and screen
Y = distance from central maximum to first
maximum
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Derivation of the equation for path difference:
δ = r2 – r1 = d sinθ = mλ
For constructive interference m = 0, +/-1, +/-2, +/- 3…
For destructive interference m = +/- (0+ ½), +/-(1 + ½)…
Determine the m values:
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Note: The textbook shows 2 formulas for the path difference; one
formula for the constructive maxima, and another for the destructive
maxima. It is easier to combine the cases using one formula.
Book:
constructive
δ = dsinθ = mλ
for m = 1,2,3,4...
destructive
δ = dsinθ = (m + ½)λ
for n = 0,1,2,3,4…
What does the intensity graph look like?
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Derivation of the APPROXIMATION equation relating λ, d, x, L:
This equation works as long as L>>d and d>>λ, because this
keeps θ really small, and thus:
sin θ ≈ tan θ
m
d
xm
L
So, assuming all else stays constant:
1) As L decreases, x _____________ (move the screen, move the slits move laser)
2) As d decreases, x _____________ CD-R vs DVD-R reflection interference pattern.
3) As λ decreases, x _____________ Demo: Green Laser, Red Laser with Diffraction Grating
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#18) Two slits are illuminated with coherent light of wavelength 560 nm.
The first order interference maximum appears at 6o.
(A)What is the distance separating the slits?
(B)Also, what is the path difference?
(C) If a diffraction grating were used, what is the number of lines per
mm?
#19) (A) Two slits separated by 1 x 10-5 m. If the distance between the
slits and the screen is 1.2 m, and the distance from the central maximum
to the first order maximum is 2.9 x 10-2 m, what is the wavelength of the
light?
(B) What is the distance from the central maximum to the 3rd order
minimum?
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#20) Two loudspeakers separated by 0.5 m and operating in phase send
out identical sound waves. An observer who is 10 m from the speakers
finds out the first sound maximum at a distance of 12.5 m as shown.
(A) What is the wavelength of the sound using path difference equation?
(.39 m)
(B) What is the wavelength of the sound using the approximation
equation. (.625 m)
(C) Give justification for the discrepancy.
For sin θ ≈ tan θ, both opposite sides must be small
compared to their respective adjacent side/hypotenuse
(X<<L, λ << d). Since both x and λ are relatively large, the
small angle approximation breaks down.
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#21) Light of wavelength 440 nm falls upon two slits spaced 0.31 mm
apart. What is the angle between the first and second dark fringes?
#22) Light of wavelength 700 nm is incident upon a diffraction grating
that has 1000 lines/cm. If the diffraction grating is 3 m from the screen,
find the distance from the central maximum to the 5th dark fringe.
Now go home and do HW #1,5,6,9
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26.4 Interference in Thin Films
I remember as a kid growing up in the 70s, walking around the
neighborhood after a rain storm, puddles and wet ground
everywhere looked like the picture to the right. Only now do I
realize it was a sign of the times: old cars back then were
notorious for leaking oil. In fact my 1973 Buick Le Sabre (in 1988)
had three leaks I had to keep up with at once: motor oil,
transmission fluid power steering fluid. Today I see less oil films
on the road, but is it due to cars leaking less, or to New Yorkers
buying new and keeping them for fewer years?
Now, to finally get to the point: what’s up with those rainbows in the
street, and in soap bubbles?
Reflection Rule:
i. When a ray of light reflects off of a surface, the reflected
ray undergoes a 180o phase change at the boundary if
n1 < n2 such as when n1 = air and n2 = water.
ii. If n1 > n2 no phase change occurs
Analogy: show reflected pulses Thick to thin string/Thin to thick string
n1 n2
n1 n2
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THIN FILMS CASE I – phase inversion occurs at BOTH surfaces
(So think of it as NO PHASE SHIFT):
(A) Derive a general expression for the possible thicknesses t of the oil film
such that the resultant intensity of the light reflected back into the air is a
maximum (constructive interference)
At minimum, to be in phase, make t= λfilm/2, then the wave in the oil
has to travel 2t (1λfilm) to meet the wave reflected off the top surface.
Thus:
c
f 0
m 0
m film
0
n
t
t
film
m=1,
n
2n
v
f
2
film
2…
(B) Derive a general expression for the possible thicknesses t of the oil film such
that the resultant intensity of the light reflected back into the air is a minimum.
(destructive interference)
At minimum, to be 180° out of phase, make t= λfilm/4, so the wave in oil has
to travel 2t (1/2 λfilm) to meet the wave reflected off the top surface.
Thus:
t
m
film
4
t
m 0
4n
m= 1, 3, 5…
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(C) Calculate the minimum thickness t for the example above such that
the resultant intensity of the light reflected back into the air produces:
i. constructive interference
ii. destructive interference
THIN FILMS CASE II (Soap Bubble)-phase inversion occurs at
one surface (So think of it as PHASE SHIFT)
(A) Derive a general expression for the possible thicknesses t of the soap
bubble such that the resultant intensity of the light reflected back into the air is
a maximum (constructive interference)
Because of the phase shift, to be in phase, make t= λfilm/4, then the
wave in the soap bubble has to travel 2t (1/2 λfilm) to meet the wave
reflected off the top surface.
Thus:
m 0
m film
t
t
4n m=1, 3, 5…
4
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(B) Derive a general expression for the possible thicknesses t of the soap
bubble such that the resultant intensity of the light reflected back into the air is
a minimum. (destructive interference)
Because of the phase shift, to be 180° out of phase, make t= λfilm/2, so the
wave in oil has to travel 2t (1λfilm) to meet the wave reflected off the top
surface.
Thus:
t
m
film
t
m 0
2n
m= 1, 3, 5…
2
(C) Calculate the minimum thickness t for the example above such that
the resultant intensity of the light reflected back into the air produces:
i. constructive interference
ii. destructive interference
#23) A thin film of oil (n = 1.26) coats the water in a marina (n = 1.33). If
light in air with wavelength 500 nm strikes the oil, what are the three
smallest thicknesses of the oil film such that the light reflected back in
the air interferes constructively.
Now go home and do Ch 26 HW # 17,18,27,28
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#24) 1984 AP Exam – Free Response #5
The surface of a glass plate is coated with a transparent thin film. A beam of
monochromatic light of wavelength 6 x 10-7 m in air is incident normally on surface
S1. The beam is partially transmitted and partially reflected.
(A) Calculate the frequency of the light in the air.
(B) Calculate the wavelength of the light in the thin film.
(C) Calculate the minimum thickness d1 of the film such that the light reflected
into the air is a minimum.
(D) Calculate the minimum thickness d2 of the film such that the light reflected
into the air is a maximum.
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Application of thin films: Antireflective (AR) coating on eyeglasses.
AR coatings serve two purposes:
i. They reduce the amount of reflected glare, making the glasses look better
on the wearer.
ii. They increase the amount of light that passes through the lens. This is
because the amount of reflection is related to the difference between indices of
refraction. Going from air (n=1.00) into crown glass (n=1.52) is a big jump and
leads to more reflection than going from air into a coating of MgF2 (n=1.38), or a
fluoropolymer (n=1.30). MgF2 on a crown glass surface gives a reflectance of
about 1%, compared to 4% for bare glass. MgF2 coatings perform much better on
higher-index glasses, especially those with index of refraction close to 1.9. MgF2
coatings are commonly used because they are cheap, and when they are designed
for a wavelength in the middle of the visible band they give reasonably good antireflection over the entire band.
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26.6 Single Slit Diffraction of light
When light is sent through a small single slit and the
transmitted wave is projected on a screen, a diffraction
pattern is established that is similar to the double slit
interference pattern with one big difference:
The single slit pattern has a larger central maximum (about
twice the size of the next weaker bright band).
This can be done in two ways:
Fraunhofer Diffraction – lens used when screen is far away to focus the
parallel rays.
Fresnel Diffraction – no lens is used when the screen is close.
DEMOS: Laser through single slit or through a pin hole in a piece of paper. Form Single Slit with
2 fingers!
Why does this happen?
Huygen theorized that as a wave front passes through a medium, each point hit by
the wave produces a circular wave front that interferes with the other circular
fronts produced in its vicinity. When the resulting wave front hits a barrier with an
opening or slit in it, only the points inident on the slit can produce circular wave
fronts. It is these points that act as individual point sources of waves. Hence, light
from one portion of the slits can interfere with light from another portion. The
light interferes with…itself. Whoa. Video demonstrating Huygens Principle
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Derivation of Single Slit equation that relates slit length d with λ
If we break the slit into 2 equal parts and analyze three coherent rays:
o
Rays 1 and 2: 180 phase difference
o
Rays 1 and 3: 0 phase difference
i. If we choose a θ that relates d with λ, we get:
sinθ =
2
d
2
Thus the condition for destructive interference is:
dsinθ = 1λ
ii. If we break the slit into 4 equal parts,
sinθ =
2
d
4
or:
dsinθ = 2λ (destructive)
iii. If we break the slit into 6 equal parts,
sinθ =
2
d
6
hence dsinθ = 3λ
Formula for finding angle to 1st dark fringe using single slit:
dsinθ = mλ
m = +/-1, +/-2, +/- 3…
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What does the intensity graph look like for the single slit.
Remember: the single slit pattern has a larger central maximum (about
twice the size of the next weaker nth order maximum).
#25) Monochromatic light with a wavelength of 750 nm passes through a slit 1 µm
wide. The distance from the screen to the slit is 20 cm.
The first dark band appears at point p.
(A) Calculate angle θ (48.6°)
(B) What is the width of the central maximum? (θ gets you to the
minima, so use tanθ = x/L) (0.23 m)
Now go home and do Ch 26 HW – #38, 40
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26.7 Polarization –
Polarization is the process of turning unpolarized light into
polarized light.
Polarization can ONLY BE DONE with transverse waves.
Therefore, only electromagnetic radiation and light can be
polarized.
SOUND CANNOT BE POLARIZED, EVER, EVER, EVER!!!!
What do electromagnetic waves really look like?
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35.0
Unpolarized light – Light with EM fields oscillating in more
than one plane of vibration.
Define: Polarized light – Light with with EM fields oscillating
in a single plane of vibration
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=692.0
(show two waves, move around polarizers. Read directions to adjust parameters)
What are polarizing filters made of?
In 1938, Edwin H. Land discovered a material, which he called Polaroid, that could polarize
light through selective absorption. He made a solid sheets of long-chained hydrocarbons, such
as polyvinyl alcohol (PVA used in making slime), that are stretched during the manufacture
process to align the molecules in one direction. The sheets are then dipped in a solution
containing iodine; the iodine atoms impregnate the PVA molecules and provide electrons that
can move along the length of the chain. Thus when light with an electric field vector E parallel
to the hydrocarbon chains is incident on the sheet, the field causes electrons to move, thus
absorbing the energy of the incident wave. Light with E perpendicular to the chains cannot be
absorbed and passes through the sheet. Land is the founder of the Polaroid Corporation,
which I remember as a kid because he invented the instamatic camera, which could take a
picture, spit out the photographic sheet, and then we would all watch for sixty seconds as the
picture developed in front of our eyes.
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What happens when unpolarized light enters a vertical polarizer?
Repeat for a horizontal polarizer.
What happens when a horizontal and vertical polarizer are 90o apart?
Practical applications
1. In polarized sunglasses- These types of sunglasses reduce glare. Glare is reflected light off of a surface that
interferes with you actually seeing the surface. It is important to note that glare is polarized as it is reflected
off a surface from above some extreme angle that is surface dependent. If glare is reflected off of a horizontal
surface, it is horizontally polarized, and off of a vertical surface, it is vertically polarized. For drivers since most
glare is road glare, how would the polarizers in the glasses need to be oriented ?
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3d-movie technology- Anyone who has seen an Imax 3d movie or has been to Universal’s Islands of Adventure
to take a ride on “the Amazing Adventures of Spiderman” knows that modern 3d technology seems so lifelike
compared to the old red/blue 3d technology from the early 1900s.
The recording system is expensive because it employs 2 cameras, each filming the same image through
separate polarizers The projector has a rotating polarizer that is frequency matched with one moving film that
has both sets of camera images on it alternating left and right. The rotating polarizing filter is there to allows
for every other picture to be plane polarized 90º to each other and offset a few centimeters. The viewer
wears 3d glasses also fitted with polarizers. How would the polarizers in the glasses need to be oriented now?
What is the problem with plane-polarized 3d technology? You cannot turn your head or lie down, as the 3d
effect would be lost. Thus Circular Polarization is employed in current movies and 3d LCD TVs
Chapter 23, 24 and 26: Done.