[ SHIVOK   SP212 ]
A.
Maxwell’s Rainbow
March   17,   2016
1.
As the figure shows, we now know a wide spectrum (or range) of electromagnetic waves: Maxwell’s rainbow.
In the wavelength scale in the figure, (and similarly the corresponding frequency scale), each scale marker represents a change in wavelength (and correspondingly in frequency) by a factor of 10.
2.
________________________________________________________________________________
_________________________________________________________________________________________
________________________________________________________________________________________.
3.
Visible Spectrum:
CH:   16   Review   items:
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[ SHIVOK   SP212 ] March   17,   2016
B.
The Traveling Electromagnetic Wave, Qualitatively
1.
Some electromagnetic waves, including x rays, gamma rays, and visible light, are ________________________________________ from sources that are of atomic or nuclear size.
Figure 33 ‐ 3 shows the generation of such waves.
At its heart is an LC oscillator, which establishes an angular frequency
____________________________________________.
Charges and currents in this circuit vary sinusoidally at this frequency.
2.
Figure 33 ‐ 4 shows how the electric field and the magnetic field change with time as one wavelength of the wave sweeps past the distant point P in the last figure; in each part of Fig.
33 ‐ 4, the wave is traveling directly out of the page.
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[ SHIVOK   SP212 ] March   17,   2016
3.
At a distant point, such as P, the curvature of the waves is small enough to neglect it.
At such points, the wave is said to be a
_________________________________________.
4.
Here are some key features regardless of how the waves are generated: a) The electric and magnetic fields and are always perpendicular to the direction in which the wave is traveling.
The wave is a transverse wave.
b) The electric field is always ______________________to the magnetic field.
c) The cross product ______________________always gives the direction in which the wave travels.
d) The fields always vary ________________________________________.
The fields vary with the same frequency and are ______________________ with each other.
5.
We can write the electric and magnetic fields as sinusoidal functions of position x (along the path of the wave) and time t:
6.
Here E m and B m are the amplitudes of the fields and,  and k are the angular frequency and angular wave number of the wave, respectively.
7.
________________________________________________________________________________________
________________________________________________________________________________________________.
8.
The speed of the wave (in vacuum) is given by c.
Its   value   is   about   3.0
x10
8
m/s
9.
The ratio of amplitudes of the Electric and Magnetic fields are also related to the speed of light as follows:
(Eq   33 ‐ 4)
10.
The magnitudes of the fields at every instant and at any point are related by:
(Eq   33 ‐ 5)
Let’s   now   prove   equations   33 ‐ 4   and   33 ‐ 5   with   Calculus.
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[ SHIVOK   SP212 ]
C.
The Traveling Electromagnetic Wave, Quantitatively
1.
Let’s represent an Electromagnetic wave as in Fig 33 ‐ 5
March   17,   2016
2.
The dashed rectangle of dimensions dx and h in Fig.
33 ‐ 6 is fixed at point P on the x axis and in the xy plane.
3.
As the electromagnetic wave moves rightward past the rectangle, the magnetic flux B through the rectangle changes and—according to Faraday’s law of induction—induced electric fields appear throughout the region of the rectangle.
We take E and E + dE to be the induced fields along the two long sides of the rectangle.
These induced electric fields are, in fact, the electrical component of the electromagnetic wave.
a) Starting with Maxwell’s Equation that relates the induced electric field to the changing magnetic flux:
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b) c)
[ SHIVOK   SP212 ]
Apply it to our drawings of 33 ‐ 5 and 33 ‐ 6; thus:
March   17,   2016
The flux through the rectangle is:
d) e)
Finding 
Therfore substitute and thus
f) g)
SO (Eq 33 ‐ 11)
Thus from Eqs 33 ‐ 1 and 33 ‐ 2
h) So rewriting Eq 33 ‐ 11 we get: i) Finally you can see that
where   =   c
(Eq 33 ‐ 13)
If   we   divide   Eq   33 ‐ 1   by   33 ‐ 2   and   then   substitute   in   Eq   33 ‐ 13   we   get   Eq   33 ‐ 5.
You   can   prove   on   your   own.
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4.
[ SHIVOK   SP212 ]
Now let’s prove the Equation 33 ‐ 3 using Calculus.
a) Examine the Magnetic field
March   17,   2016
b)
Fig.
33 ‐ 7 The sinusoidal variation of the electric field through this rectangle, located (but not shown) at point P in Fig.
33 ‐ 5b, E induces magnetic fields along the rectangle.
The instant shown is that of Fig.
33 ‐
6: is decreasing in magnitude, and the magnitude of the induced magnetic field is greater on the right side of the rectangle than on the left.
c) Starting with Maxwell’s Equation that relates the induced magnetic flux to the changing electrical field:
d) Apply it to our drawing of 33 ‐ 7 thus:
e) The electric field through the rectangle is:
f) This means g) So h) Which leads to
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i)
[ SHIVOK   SP212 ]
Just as we did before substituting we get: j) k)
Now
Which finally means that
March   17,   2016
1.
The rate of energy transport per unit area is called the Poynting Vector.
2.
Instantaneous Energy flow rate.
3.
Average Energy Transported over time or Intensity a)
(1)
Remember   that   the   average   for   Sin 2 f
for   any   f
is   ½.
(2) The energy density u (= ) within an electric field, can be written as:
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b)
[ SHIVOK   SP212 ]
Variation of Intensity with Distance
March   17,   2016
(1) The intensity I (power per unit area) measured at the sphere must be
Show   all   work
4.
Sample Problem: A 10‐kW radio station radiates spherical electromagnetic waves. The maximum value (amplitude) of the wave’s oscillating electric field at a distance of 5.0
km from the station is closest to:
A. 3.6 V/m
B. 0.16 V/m
C. 0.56 V/m
D. 1.6 V/m
E. 16 V/m
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[ SHIVOK   SP212 ] March   17,   2016
E.
Radiation Pressure
1.
Electromagnetic waves have linear momentum and thus can exert a pressure on an object when shining on it.
2.
During the interval  t, the object gains an energy  U from the radiation .
If the object is free to move and that the radiation is entirely absorbed (taken up) by the object, then the momentum change  p is given by
3.
If the radiation is entirely reflected back along its original path, the magnitude of the momentum change of the object is twice that given above, or
4.
Since and it follows that
5.
Finally, the radiation pressure in the two cases are
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[ SHIVOK   SP212 ] March   17,   2016
6.
Show   all   work/Explain:
Sample Problems: a) The next two problems deal with an electromagnetic wave in a material where the electric field has a y ‐ component only and, in SI units, is given by E y
= (40.0
V/m) sin[(1.4x10
7 m ‐ 1 ) x – (4.2x10
15 rad/s)t)] where x is in meters and t is in seconds.
(1) The wavelength and direction of travel of the wave are closest to
A. 449 nm in the positive x direction.
B. 449 nm in the negative x direction.
C. 333 nm in the positive x direction.
D. 333 nm in the positive x direction.
E. 282 nm in the positive x direction.
(2) If the electromagnetic wave is fully reflected by a surface, the radiation pressure is closest to
A. 2.03 x 10
-8
N/m
2
.
B. 3.45x 10
-7
N/m
2
.
C. 1.42 x 10
-8
N/m
2
.
D. 7.08 x 10
-9
N/m
2
.
E. 5.63 x 10 -9 N/m
2
.
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10

[ SHIVOK   SP212 ] March   17,   2016
F.
Polarization
1.
Polarized  ______________________________________________________________________.
a) Diagram ‐ Figure 33 ‐ 9a shows an electromagnetic wave with its electric field oscillating parallel to the vertical Y ‐ axis.
2.
Polarized randomly or unpolarized
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3.
[ SHIVOK   SP212 ] March   17,   2016
(1) Intensity of Unpolarized Light
(a) If the intensity of original unpolarized light is I o
, then the intensity of the emerging light through the polarizer, I, is half of that.
Polarizing sheets a) We can transform unpolarized visible light into polarized light by sending it through a polarizing sheet, as shown below.
(1) ______________________________________________________________________
_______________________________________________________________________________
_______________________________________________________________________________
______________________________________________________________________________.
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12

4.
[ SHIVOK   SP212 ] March   17,   2016
Intensity of Polarized Light a) Suppose now that the light reaching a polarizing sheet is already polarized.
b) Figure 33 ‐ 12 shows a polarizing sheet in the plane of the page and the electric field of such a polarized light wave traveling toward the sheet (and thus prior to an absorption).
c) We can resolve E into two components relative to the polarizing direction of the sheet: parallel component E y is transmitted by the sheet and perpendicular component E z is absorbed.
Since q is the angle between and the polarizing direction of the sheet, the transmitted parallel component is d) Since , then
(1) ______________________________________________________________________
_______________________________________________________________________________
______________________________________________________________________________.
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13

[ SHIVOK   SP212 ] March   17,   2016   e) As shown in figure 33 ‐ 13, often unpolarized light will be sent through at least two sheets.
The first sheet is often called a polarizer, and the additional sheets are called analyzers.
f) If the two sheets are parallel all the light passed by the first is also passed by the second.
If the sheets are perpendicular (the sheets are said to be crossed), no light is passed by the second sheet (figure 33 ‐ 14).
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14

[ SHIVOK   SP212 ] March   17,   2016   g) So obviously there is the case where they are not parallel or perpendicular.
The following is a sample problem on how this case is handled.
In   the   figure   on   the   left,   a   beam   of   light,   with   Intensity
43W/m 2   and   polarization   parallel   to   the   y   axis,   is   sent   into   a   system   of   two   polarizing   sheets   with   polarizing   directions   at   angles   of   q
1
=70 ±   and   q
2
=90 ±   to   the   Y ‐ axis.
What   is   the   Intensity   of   the   light   transmitted   by   the   two ‐ sheet   system?
(1) Solution:
(a) The angle between the direction of polarization of the light incident on the first polarizing sheet and the polarizing direction of that sheet is 
1
= 70°.
If I
0 is the intensity of the incident light, then the intensity of the light transmitted through the first sheet is:
(b) The direction of polarization of the transmitted light makes an angle of 70° with the vertical and an angle of 
2
=
20° with the horizontal.

2 is the angle it makes with the polarizing direction of the second polarizing sheet.
Consequently, the transmitted intensity is:
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[ SHIVOK   SP212 ] March   17,   2016
5.
When unpolarized light is passed through two polarizing filters in succession, its intensity is decreased by 80%.
The angle, θ , between the transmission axes of the filters is:
A. 78.5°.
B. 63.4°.
C. 26.6°.
D. 36.9°.
E. 50.8°.
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Show   all   work
6.
A laser produces unpolarized light with an intensity of 5.0
W/cm 2 .
The light passes through three sheets of Polaroid film as shown.
The transmission axis of the second Polaroid makes a 30  angle with that of the first, and the axis of the third makes a 60Þ angle with that of the second (and 90  angle with that of the first).
The intensity of the light that emerges from the third
Polaroid is closest to:
A. 0
B. 1.4 W/cm 2
C. 1.1 W/cm 2
D. 0.16 W/cm
2
E. 0.47 W/cm
2
7.
Light can be polarized by means other than polarizing sheet…such as by scattering or reflection.
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[ SHIVOK   SP212 ] March   17,   2016
G.
Reflection and Refraction a) Law of Reflection
b) Law of
Refraction (Snell’s Law)
c) Table of Indexes of Refraction
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17

d)
[ SHIVOK   SP212 ]
Effects of different index mediums
March   17,   2016
(1) If n
2
is equal to n
1
, then q
2
= q
1
and the beam continues un‐ deflected.
(2) If n
2
> n
1
, then q
2
< q
1
and the beam is bent away from the un‐ deflected direction toward the normal.
(3) If n
2
< n
1
, then q
2
> q
1
and the beam is bent away from the un‐ deflected direction and away from the normal.
(4) It is a memory aid to think toward the medium with the
HIGHER index. See in (b) it bends down and in (c) it bends up.
(5) Refraction CANNOT bend a beam so much that the refracted ray is on the same side of the normal as the incident ray! e) Example Problem:
Light   in   a   vacuum   is   incident   on   the   surface   of   an   unknown   medium.
They   Physics   Lab   student   decides   she   can   figure   out   the   medium   if   she   knows   the   index   of   the   unknown   material.
She   measures   the   angle   of   the   light   in   the   unknown   material   and   gets   21.28
± .
In   the   vacuum   the   beam   of   light   makes   and   angle   of   32.00
±   with   the   normal   to   the   surface.
In   your   opinion   what   is   the   material   made   of?
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18

[ SHIVOK   SP212 ]
(1) Solution:
The law of refraction states
March   17,   2016
(2) We take medium 1 to be the vacuum, with n
1
= 1 and 
1
=
32.0°. Medium 2 is the unknown, with 
2
= 21.28°.
(3) We solve for n
2
:
(4) So looking ________________ the material is _______________________ .
H.
Chromatic Dispersion
1.
The index of refraction n encountered by light in any medium except vacuum depends on the wavelength of the light.
2.
The dependence of n on wavelength implies that when a light beam consists of rays of different wavelengths the rays will be refracted at different angles by a surface; that is, the light will be spread out by the refraction.
3.
The spreading of the light is called Chromatic Dispersion.
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19

4.
[ SHIVOK   SP212 ]
Chromatic Dispersion of White light
March   17,   2016
5.
Rainbows
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20

I.
Total Internal Reflection
[ SHIVOK   SP212 ] March   17,   2016
1.
For angles of incidence large than  c
, such as for rays f and g above, there is no refracted ray and all the light is reflected; this effect is called Total Internal
Reflection.
2.
 c is called the _______________________:
3.
Which means
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21

[ SHIVOK   SP212 ] March   17,   2016
1.
2.
3.
Page
22

[ SHIVOK   SP212 ] March   17,   2016
K.
Sample problems for Reflection/Refraction
1.
A beam of light traveling in air strikes the surface of a solution of corn syrup in water at an angle of 30  to the vertical.
If the beam is refracted at an angle of 19  to the vertical, then the speed of the light in the corn syrup solution is closest to:
A.
1.5
x   10 8   m/s
B.
1.7
x   10 8   m/s
C.
2.0
x   10 8   m/s
D.
2.3
x   10 8   m/s
E.
2.6
x   10 8   m/s
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2.
A flat piece of glass (with index of refraction 1.50) has a layer of ethanol (with index of refraction 1.36) floating on top of it.
Light traveling in the glass strikes the glass ‐ ethanol surface.
The critical angle for total internal reflection in the glass is closest to:
A. 65

B. 59

C. 47

D. 55

E. 39

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[ SHIVOK   SP212 ] March   17,   2016
3.
A layer of water ( n = 1.33) exists on a slab of glass ( n = 1.46).
A laser beam in the glass is incident on the glass ‐ water interface.
Relative to the perpendicular to the interface, the smallest angle for total internal reflection is closest to:
A. 57.3
° .
B. 65.6
° .
C. 40.1
°
.
D. 42.3
° .
E. 74.8
° .
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4.
The angle of incidence (relative to the normal to the surface) for which the light reflected from the water ‐ diamond surface is completely polarized is closest to:
A. 57.6º.
B. 59.5º.
C. 61.2º.
D. 66.3º.
E. 64.1º.
Show   all   work
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24