Major Concepts of Physics PHY 102
Lecture #9
Everything about
electromagnetic (EM) waves
Monday, February 22nd
Spring 2016
Prof. Liviu Movileanu
hppt://movileanulab.syr.edu/MajorConceptsPhysics2016.html
lmovilea@syr.edu
Room 211, Physics Bldg., 443-8078
Major Concepts of Physics PHY102 – Lecture #9
2016Syracuse University
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Lecture objectives-Electromagnetic waves
1. Review of single-slit diffraction: quantitative
understanding
 A couple of examples pertaining to Homework #3
2. What are the electromagnetic (EM) waves?
 The electromagnetic spectrum
 Example of EM waves and their applications
3. Lecture demonstrations/Movie (EM waves)
4. Speed of EM waves in vacuum and matter
5. Announcements
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Review – Diffraction
1) Recall:
An important result of our study of diffraction:
Let any wave go through a narrow slit.
Then, the light spreads out after leaving the slit.
Its spread can be characterized by the result
Sin θ = λ/a.
Here, a is the width of the slit-opening, and λ is the wavelength of the
wave.
Also, θ is the angle for the first minimum for wave-intensity.
Hence, objects of angular size smaller than θ cannot be resolved.
That is why θ gives the diffraction limit.
. Concepts of Physics PHY102 – Lecture #9
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2016Syracuse University
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Review - Diffraction
2) Questions to test your understanding:
a) A biologist wishes to see in detail a small object in a cell whose size
is close to the diffraction limit for light waves.
Should red light or blue light be used?
Hint: λ(blue) = 400 nm (roughly).
λ(red) = 700 nm (roughly).
b)
A biologist wishes to
see a structure in a cell whose size is smaller than the diffraction limit for
light waves.
She uses an electron microscope
(Electrons are waves, as well as particles!)
Is the wavelength of the electrons smaller or larger than that of light?
Remark: We will understand later,
how electrons can behave as waves. This is a subject known as quantum mechanics.
Major Concepts of Physics PHY102 – Lecture #9
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Minima of diffraction: single-slit experiment
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Single-slit diffraction Problem 47
Section 25.7 Diffraction by a single slit
47. The central bright fringe in a single-slit diffraction
pattern from light of wavelength 476 nm is 2.0 cm wide on
a screen that is 1.05 m from the slit. (a) How wide is the
slit? (b) How wide are the first two bright fringes on
either side of the central bright fringe? (Define the width
of a bright fringe as the linear distance from minimum to
minimum.)
.
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Single-slit diffraction Solutions
Problem 47
46.(a) Strategy The distance from the center of the diffraction pattern to the first minimum is half the width of the
central bright fringe, or 1.0 cm. Since 1.0 cm << 1.05 m, assume that sin   tan   x D , where x  1.0 cm
and D  1.05 m. Use Eq. (25-12).
Solution Find the width of the slit.
(476 109 m)(1.05 m)
a sin   m  (1) and sin   tan   x D , so a 

 0.050 mm .
x
0.010 m
D
(b) Strategy and Solution For small angles, the minima are approximately evenly spaced. This spacing is half
the width of the central bright fringe, or 1.0 cm. If we define the width of a bright fringe as the linear distance
from minimum to minimum, then the width of the first two bright fringes on either side of the central bright
fringe is 1.0 cm.
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Single-slit diffraction Problem 52
52. Light of wavelength 490 nm is incident on a narrow slit.
The diffraction pattern is viewed on a screen 3.20 m
from the slit. The distance on the screen between the
central maximum and the third minimum is 2.5 cm. What
is the width of the slit?
.
51.Strategy Since the distance between the third minimum and the center of the screen is much smaller than the slitscreen distance, the small angle approximation can be used. Refer to Figure 25.33, but let x be the third minimum
to center distance. Use Eq. (25-12).
Solution From the figure, tan   x D . The single-slit diffraction minima are given by a sin   m. Since
x << D,  is small, so tan   sin  and ax D  m. Calculate a from this and the given information.
m D 3(490 109 m)(3.20 m)
a

 0.19 mm
x
0.025 m
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Electromagnetic Waves
We know that light is a wave.
But:
What is it that is “waving”?
Mystery until 1861.
Then: A breakthrough.
James Clerk Maxwell found the answer.
He put together all the basic laws
known about a “different” subject:
Electricity and Magnetism
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The results of Maxwell
Whenever a charge accelerates, it sends waves
outward in all directions.
The waving quantity:
An electric field and a magnetic field.
He found:
The wave-speed v in empty space has one value only.
Its value is c=300,000 km/s= 3108 m/s
We call this value c.
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The Electromagnetic Spectrum
The entire set of frequencies for EM waves, is called the
electromagnetic spectrum.
For example, EM waves with frequency f just below red are called
infrared waves.
EM waves with f just above violet are called ultraviolet.
Major Concepts of Physics PHY102 – Lecture #9
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Surprise of Maxwell
The surprise of Maxwell:
The value of c is the same as the known speed of light!
The only reasonable conclusion:
Light must be an EM wave.
He answered the question “What is waving in a light wave?”
Before Maxwell:
Light was one subject.
Electricity and magnetism was another subject.
After Maxwell: Both subjects are the same phenomena.
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Recall
Three laws you already studied last semester:
1) Coulomb’s Law:
It gives the electric field produced by a static charge.
2) Oersted and Ampere’s experiments:
Gives the magnetic field produced by a moving charge.
3) Faraday’s Law:
A changing magnetic field produces an electric field.
Maxwell combined these three results, and wrote a consistent
formulation for electricity and magnetism.
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Maxwell work was theoretical
But: His ideas were later confirmed by the experiments of Hertz
(1880)
Hertz showed:
Accelerating charges produce EM waves.
The Frequencies of Light Waves
Suppose a charge performs simple harmonic motion.
Let f be its frequency.
Then: The EM wave it produces must have the same frequency f.
Find its value for light.
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Recall
v = fλ
This is valid for any wave.
But Maxwell showed: v = c for all EM waves in empty space.
Conclude: c = fλ for any EM wave in empty space.
Now, we already know the values for λ for light waves.
From the double-slit Young experiment or from the diffraction
experiments:
λ for light is between 400 nm (violet or blue) and 700 nm (red).
Use this to solve for the frequency f, using
c = fλ.
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Results
f = 430 x 1012 Hertz for red light.
f = 750 x 1012 Hertz for violet light.
1 Hertz = 1 Hz = 1 s-1 = 1 full cycle per second.
Important note: This is a very narrow range.
It is less than a factor of two.
The human eye is “color-blind” to EM waves not in this range.
But:
Hertz and others showed:
Other f than this small range also produce EM waves.
The eye, happens to not
have sensitivity to anything other than the range from 400 to 750
trillion Hertz.
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Visible electromagnetic waves
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Electromagnetic waves - Summary
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Infrared electromagnetic waves
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Ultraviolet electromagnetic waves
a. The large star coral (Montastraea cavernosa) is dull brown when
illuminated by white light.
b. When illuminated by an UV source, the coral absorbs UV and emits
visible light that appears bright yellow.
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Announcements
1. Reading: Chapter 22nd,
Sections
22.1 and 22.3 pp. 822-823 and 826-831
Conceptual example 22.2
2. Homework The homework #3 is due on this week’s
workshop (Feb 22-26).
3. This week’s lab: Production of light by solids and
emission spectroscopy
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