Goal: To understand how light acts like a particle

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Goal To understand how light acts like a particle

• Objectives
• Quantization
• Blackbody radiation
• The photon
• Photoelectric Effect

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Exam 3

• Ave 89.5
• High 99
• Low 78 (note that with 100 on HW, lab, and
  recitation that this would be a B)

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MC 1-4

• 1) A light ray is incident at angle ?1 gt 0 into
  a medium of lower index. Its angle of refraction
• (A) is greater than ?1. (B) is less than ?1. (C)
  equals ?1. (D) equals 0 exactly.
• 2) Signals may be transmitted inside optical
  fibers because of the phenomenon of
• (A) refraction. (B) total internal
  reflection. (C) dispersion.
• (D) polarization.
• 3) As viewed from earth a spacecraft traveling
  close to the speed of light reaches a distant
  star in 100 years. How much time passes by for
  the crew of that ship?
• A) no time B) the crew travels back in time C)
  100 years exactly
• D) more than 100 years E) less than 100 years
• 4) As light refracts into a medium of higher
  index, its
• (A) frequency increases. (C) wavelength
  decreases.
• (B) frequency decreases. (D) speed increases.

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MC 5-7

• 5) An object is located exactly two focal lengths
  from a convex lens. Its image is
• (A) real (B) upright. (C) enlarged. (D)
  virtual.
• 6) Nearsightedness is otherwise known as
• (A) glaucoma (B) astigmatism (C)
  presbyopia (D) myopia
• 7) A farsighted person has a far point
• (A) at 8. (B) at 25 cm. (C) less than 25
  cm. (D) between 25 cm and 8.

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MC 8-10

• 8) Which of the following correctly describes an
  image of magnification of -1.5 formed by a single
  concave or convex lens?
• (A) virtual, upright (B) virtual, inverted
  
• (C) real, inverted (D) real, upright
• 9) As viewed from earth a spacecraft traveling
  close to the speed of light has an observed
  length of 100 m. How long is the ship for the
  crew of that ship?
• a) less 100 m b) 100 m exactly c) more than
  100 m d) no length e) infinite length
• 10) The Brewster angle for light reflecting off
  water from air is 53º. This means that all light
  rays that reflect at 53º will be
• (A) totally reflected. (B) totally
  refracted.
• (C) totally linearly polarized. (D) totally
  unpolarized.
• (E) totally absorbed

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Unpolarized light of intensity I0 is incident
upon Polarizer 1 (at transmission angle 0). The
light emerges from Polarizer 1 with intensity I1
and then enters Polarizer 2 (at transmission
angle 30). The light emerges from Polarizer 2
with intensity I2 and then enters Polarizer 3 (at
transmission angle 90).

• (A) intensity I1. Give answer in terms of I0
• I1 0.5 I0
• (B) intensity I2. Give answer in terms of I0
• I2 I1 cos2(30 - 0) 0.375 I0
• (C) intensity I3. Give answer in terms of I0
• I3 I2 cos2(90 - 30) 0.094 I0
• (D) In 1 sentence describe a way to decrease the
  value of I3 by only making 1 change to the 2nd or
  3rd polarizer. Try to be specific although you
  do not have to calculate how much the change is.
• A few ways
• 1) lower the angle of polarizer 2.
• 2) Raise the angle of polarizer 2 to exactly 90
  degrees (well, actually just need to go to over
  60 degrees).
• 3) raise the angle of polarizer 3.

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A convex lens of focal length 20 cm is located 50
cm to the left of a concave lens of focal length
10 cm. An upright object sits 100 cm to the left
of the convex lens.

• (A) Calculate the location of the final image
  created by the two lenses. Express your answer as
  a distance from the concave lens.
• q1 -f1p1 / (f1 p1) -20 100 / (20-100)
• q1 25 cm
• p2 s q1 50 cm 25 cm 25 cm
• q2 -f2p2 / (f2 p2)
• -(-10) 25 / (-10 25) -7.1 cm
• So, q2 is 7.1 cm to the left of the concave lens

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A convex lens of focal length 20 cm is located 50
cm to the left of a concave lens of focal length
10 cm. An upright object sits 100 cm to the left
of the convex lens.

• (B) Calculate the overall magnification of the
  final image.
• M M1 M2 q1 q2 / (p1 p2)
• M 25 (-7.1) / (100 25)
• M -0.71
• (C) Is your final image REAL or VIRTUAL?
• Since the 2nd lens is concave this forms a
  Virtual image.
• D) Final image should have been drawn smaller and
  upside down.

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A ship of rest length of 1200 m launches from
Earth headed for the closest sun like star t
Ceti. t Ceti is located 11.8 light years from the
earth. If the ship travels at 98 of the speed
of light then

• A) How long will it take the ship to get to t
  Ceti from the point of view of the observers on
  Earth?
• T d / v 11.8 lyr / (0.97 lyr / year)
• T 12.04 years
• B) How much time will pass by for the crew of the
  spacecraft?
• Tcraft Tearth / ?
• ? (1 v2/c2)-1 5.025
• T 12.04 yrs / 5.025 2.4 years
• C) How long does the spacecraft appear to be for
  observers on the earth?
• Lob Lrest / ? 1200 m / 5.025 239 m
• D) How long does the spacecraft appear to be for
  the crew of the ship?
• Since for the crew the ship is at rest then L
  1200 m.

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A ray of light is incident at angle 100 to the
sloped face of glass prism (n 2.50) as shown
above. The ray refracts at angle ?2 into the
prism and then refracts at angle ?3 on the other
slop of the prism to ?4. Assume that the prism is
surrounded by air. Calculate

• (A) ?2.
• n1 sin(?1) n2 sin(?2)
• ?2 3.98 degrees
• (B) ?3.
• ?3 80 - ?2 76.02 degrees
• (C) What is the critical angle for light going
  into the prism (if any)?
• None there is only a critical angle if you go
  to lower index
• (D) What is the critical angle for the light
  exiting the prism (if any)?
• Sin(?c) n2 / n1 1 / 2.5
• ?c 23.6 degrees
• (E) ?4. Explain your result for ?4.
• Since ?3 gt ?c there is NO refraction. Therefore
  there is no ?4!

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Quantization

• Okay this is an ugly looking word.
• All it means is that energy is restricted to
  specific values.
• Think of a staircase.
• Each stair has a difference potential energy.
• However, you are only allowed to have the
  potential energies that are on the stairs.
• Any potential energy between two stairs is not
  allowed.

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Blackbody Spectrum continuous

• All objects which absorb most of the light which
  passes through them emit energy as a blackbody.
  The shape of the blackbody spectrum is always
  the same.
• The strength of the spectrum (i.e. how much light
  it emits) and the peak wavelength of the spectrum
  depend on the temperature of the object.

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http//theory.uwinnipeg.ca/users/gabor/foundations
/quantum/slide6.html
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The photon

• A single light particle is called a photon.
• Photons have energy which depend on wavelength or
  frequency.
• E hf hc / ?
• h is a constant called Plancks constant
• h 6.626 10-34 J s
• Often times though photons are measured in units
  of eV (electron-Volts).
• Note that eVs are energies.

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What did Einstein win a noble prize for?
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What did Einstein win a noble prize for?

• The Photoelectric Effect!

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Photoelectric Effect

• When light is shined onto a metal electrons can
  be stripped from that metal because of the impact
  of the photons.
• Some specific amount of the photon gets absorbed
  and the rest may be used on the electron.
• The absorbed amount is called the Work Function
  (F).
• The maximum energy the electron can leave the
  metal with is called Kmax
• So, Kmax hf - F

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Example

• An alloy has a work function of 1.2 eV (1 eV
  1.602 10-19 J).
• A photon of light with a frequency of 2 1015 Hz
  strikes the alloy.
• What is the maximum kinetic energy the emitted
  electron can have?
• Give answer in terms of eV

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Example

• An alloy has a work function of 1.2 eV (1 eV
  1.602 10-19 J).
• A photon of light with a frequency of 2 1015 Hz
  strikes the alloy.
• What is the maximum kinetic energy the emitted
  electron can have?
• Kmax hf F
• 6.626 10-34 Js 21015 1/s ( 1 eV / 1.602
  10-19 J) 1.2 eV
• 6.626 2 / 1.602 eV 1.2 eV 7.07 eV

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Example 2

• A laser is shined onto a metal strip.
• The work function of the strip is 2.3 eV.
• If the highest energy electron emitted has an
  energy of 3.5 eV then what is the frequency of
  the laser beam?

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Example 2

• A laser is shined onto a metal strip.
• The work function of the strip is 2.3 eV.
• If the highest energy electron emitted has an
  energy of 3.5 eV then what is the frequency of
  the laser beam?
• Kmax hf F
• 3.5 eV hf 2.3 eV, so hf 5.8 eV
• f 5.8 eV (1.602 10-19 J / 1eV) / 6.626
  10-34 Js
• f 1.4 1015 Hz

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But what happens if

• What happens if the energy of your light photon
  is less than the Work Function?
• Well if that happens you dont emit electrons.
• Therefore there is a minimum photon energy to
  knock loose photons.
• This energy of course corresponds to some
  frequency.
• This frequency is called the threshold frequency
  (fo).

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Threshold Frequency

• Kmax hf F
• At the threshold frequency Kmax is 0, that is you
  barely get an electron.
• So, 0 hfo F
• Or, fo F / h
• Example
• If the work function of a metal is 4.2 eV then
  what is the Threshold Frequency?

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Threshold Frequency

• Kmax hf F
• At the threshold frequency Kmax is 0, that is you
  barely get an electron.
• So, 0 hfo F
• Or, fo F / h
• Example
• If the work function of a metal is 4.2 eV then
  what is the Threshold Frequency?
• fo F / h
• fo 4.2 eV ((1.602 10-19 J / 1eV) / 6.626
  10-34 Js
• fo 1.02 1015 Hz

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A few other topics

• Compton Scattering
• When a photon hits an electron the electron can
  absorb the photon.
• Since it gains energy the photon will move off
  (just like in P218 when you had collisions).
• Energy and momentum are conserved so a photon is
  emitted.
• However, it is emitted at some angle determined
  by the recoil of the electron.
• Also, since the electron steals some of the
  energy from the photon, the new photon will have
  less energy than the old photon.

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Compton Shift

• This causes a Compton Shift
• ?new ?old (h/mec)(1-cos(?))
• me is the mass of the electron
• ? is the angle between the direction of the old
  photon and the new one.

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Conclusion

• We learned about the particle nature of light.
• We saw how light is emitted form most objects.
• We learned about the Photoelectric effect and how
  to calculate work function and threshold
  frequency.