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1 Chapter 31 PROPERTIES OF LIGHT 31.1 The speed of light c = 299,792,458 m/s ~ 3.0x10 8 m/s Measurements: astronomical, rotating toothed wheel, rotating mirror, etc. Example 1. Reproduce Fizeau’s determination of the speed of light using a rotating toothed wheel (f = 17.262 rev/s) that has N = 840 teeth at a distance L = 10 km from the mirror 3 1 8 2 / , 1/ 2 2 10 10 840 17.262 2.9 10 / c L t t Nf c LNf m s ms = ∆= = = =

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Page 1: Chapter 31 PROPERTIES OF LIGHT - Department of Physics › phy101 › PHY205 › Spring2018 › Chapter31.pdf · 2009-04-07 · 1 Chapter 31 PROPERTIES OF LIGHT 31.1 The speed of

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Chapter 31PROPERTIES OF LIGHT

31.1 The speed of light

c = 299,792,458 m/s ~ 3.0x108 m/s

Measurements: astronomical, rotating toothed wheel, rotating mirror, etc.

Example 1. Reproduce Fizeau’s determination of the speed of light using a rotating toothed wheel (f = 17.262 rev/s) that has N = 840 teeth at a distance L = 10 km from the mirror

3 1 82 / , 1/ 2 2 10 10 840 17.262 2.9 10 /c L t t Nf c LNf m s m s−= ∆ ∆ = ⇒ = = ⋅ ⋅ ⋅ ⋅ = ⋅

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31.2 The propagation of lightIn general, the propagation of light is governed by the wave equation. The empirical description can be done using Huygens’s principle (a.k.a. Huygens’s construction),

Each point on a primary wavefront serves asa source spherical secondary wavelets that

advance with a speed and frequency equal tothat in the primary wave. The wavefront is an

envelope of these wavelets,if one ignores the back wavelets

Planewave

Sphericalwave

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The path taken by light traveling from onepoint to another is such that the time of

travel is minimal.

(very useful when describing reflection and refraction)

and by Fermat’s principle,

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31.3 Reflection and refractionThe speed of light v in any transparent medium (air, water, glass, etc.) is always less than its speed c in vacuum. The ratio

n = c/v > 1

is called the index of refraction. For water, n ~ 1.33; for glass n ~ 1.5 – 1.7, for air n ~ 1.0003When a beam strikes a boundary between two separate media, part of the energy is reflected, and part continues into the second medium, but at a different angle. This change in angle is called refraction.

The law of reflection: 1 1 'θ θ=The law of refraction (Snell’s law):

( )1 11 1 2 2 1 1 2 2sin sin sin sinn n vor vθ θ θ θ− −= =

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The reflection and refraction laws are simple consequences of the energy and momentum conservation laws for photons (particles of light). Since the energy (frequency) remains the same, the wavelength changes when light passes from one medium into another:

1 1 1 2 2 2

1

1 2

1 11 2 2 22

, / , / ,/ /

f v f vv v

f fn n

λλ λ λ

λλ

==

= == ⇒

Specular vs. diffuse reflection

P’ is called the image point

Reflection from a smooth surface (mirror) is called the specular reflection.

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Reflection from a rough surface is called diffuse reflection (reflection in random directions)

Intensity of reflected and transmitted light

is a complicated function of angles and indexes of refraction. In the simplest case of normal incidence, the reflected intensity is

( )( )

1 2

1 2

1 2

1 2

2

2

,

1

n nreflected incident n n

n ntransmitted incident n n

I I

I I

−+

−+

=

= −

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Example 2. Light enters water from air with an angle of incidence 34o. What is the angle of refraction?

The refraction indexes for air and water are na = 1and nw = 1.33. According to Snell’s law,

( )1 2 2 1

2

sin sin , sin sin /1.33

sin 34 /1.33 0.420 24.8 .a w

o o

n nθ θ θ θ

θ

= =

= = ⇒ =

Total internal reflection

As the angle of incidence is increased, the angle of refraction

( )2 1 2 1sin / sinn nθ θ=increases as well. If/when it reaches 900, it cannot increase anymore and all the light is reflected(total internal reflection).

1 cθ θ= 02 290 , sin 1θ θ= =

2 2

11sin 90sin n o

nn

c nθ ==The critical angle at which is

The total internal reflection can occur only if the incident light is in the medium with higher index of refraction!

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Example 3. What are the critical angles for light coming into air a) from water? b) from glass with n = 1.6? c) from glass into water?

2

1sin n

c nθ =

2

1

11.33sin 0.752, 48.8n o

c cnθ θ= = = =

2

1

11.6sin 0.625, 38.5n o

c cnθ θ= = = =

2

1

1.331.6sin 0.831, 56.3n o

c cnθ θ= = = =

The critical angle is

For water/air, the angle is

For glass/air, the angle is

For glass/water, the angle is

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Example 4. From the bottom of a pool you see the objects above the water level only in the circle of radius 3 m. How deep are you?

You see the objects only at angles less than the water/air critical angle 48.6o. According to the Figure,

0tan / tan 48.8 1.14,/ tan 3 /1.14 2.63

c

c

R yy R m m

θθ

= = == = =

Since the critical angle for total internal reflection for glass/air with n=1.6 glass is 38.50, and the incident angle insidethese two prisms is 450, all light will get out of the prisms as shown in the Figure.

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An interesting application is fiberoptics: light does not com out of bended fibers

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Dispersion

The dependence of the index of refraction on wavelength is called dispersion. Usually, n decreases slightly as the wavelength increases. Since the angle of refraction is inversely proportional to n, the refraction angle goes up with increasing wavelength.

Examples are the rainbow or light out of the prism (Newton). In both the prism and the rainbow (water drops) the light is refracted twice – when entering and leaving glass (water drops).

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31.4 PolarizationE

B

E

In electromagnetic waves, the electric field vector

(magnetic field vector ) oscillates perpendicularly to

oscillates along one direction), the direction of propagation. Thus, the wave can have linear polarization (

circular polarization (clock- and counterclockwise), and elliptic polarization. Circular and elliptic polarization are essentially combinations of two linearly polarized waves.

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20 cosI θ θ

Polarization by absorption. Some crystals can transmit light only when the electric field is directed along the transmission axis which is determined by the structure and orientation of molecules. The intensity of the linearly polarized transmitted beam is where is the angle between incident electric field and transmission axis

Example 5. Unpolarized light of intensity 5.4 W/m2 is incident on two polarizing sheets whose transmission axes make an angle of 470 . What is the intensity of transmitted light coming out of the second sheet?

The incident light is unpolarized. Therefore, the intensity of light transmitted by the first polarizing sheet is one half of the incident intensity,

11 02I I=

The second polarizing sheet further reduces the intensity as

2 2 2 01 12 1 02 2cos cos 5.4cos 47 1.26I I I W Wθ θ= = = =

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Polarization by reflection. The reflected light is always partially polarized. When the reflected and refracted rays are perpendicular to each other, thereflected light is completely linearly polarized; the corresponding incident angle is called the polarizing or Brewster’s angle.

( )

01 2

2

2

1

2

01 2 2

sin sin , 90 ,

sin sin 90 c

tan /

os ,p p

p p p

p

n n

n

n

n n

n

θ θ θ θ

θ θ θ

θ

= = −

=

= − =

At polarizing angle, the reflected light is polarized perpendicularly to the incidence plane.

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Polarization by scattering. Scattering is absorption and re-radiation of light by molecules (atoms) of the medium. The scattering molecules can sometimes work as dipole antennas with the maximum radiation in the direction perpendicular to the dipole moment and electric field in the plane of the dipole moment (“antenna”).

δ090δ =

0180δ =

Polarization by birefringence. In birefringentmaterials, the speed of light is anisotropic and depends on polarization and direction of propagation. The incident light splits into the ordinary and extraordinary rays which are polarized in mutually perpendicular directions and travel with different speeds and in different directions. Due to the difference in speed and paths, the rays emerge with a phase difference

In a quarter-wave plate,

In a half-wave plate,

.

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31.5 Derivation of the laws of reflection and refraction from Fermat principle

Reflection.

The optical path from A to B is exactly the same as from A’ to B. Thus the incident and reflection angle are equal.

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Refraction. The time for light to reach B from A is

( )( )( )

2 21 1 2 2 1

222

/ / /

/

t L v L v a x v

b d x v

= + = +

+ + −

Minimization of this time with respect to x yields

( ) ( )( )

( ) ( )( )

22 2 2

22 2 2

1 2

1 2

2 2 21 11 22 2

1 2

sin sin1 1 2 2

0 / / / ,

/ / ,

sin sin

x x da x b d x

x x da x b d x

v v

dt dx v v

v v

n nθ θ θ θ

+ + −

+ + −

= = +

= −

= ⇒ =

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31.6 Wave-particle dualityVisible light is electromagnetic radiation (electromagnetic waves) with wavelengths

400 700 nmλ ÷∼The energy of particles of light – photons – is

34 15

6

,6.626 10 4.136 10 ,1.24 10

/

0h J s eE hf h

V shc

c

eV m

λ− −

= ⋅ ⋅ = ⋅

=

=

=

where h is called Plank’s constant.

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31.7 Light spectra

White light is a mixture of light of all colors (wavelengths) of approximately equal intensity. Decomposition of light into individual colors is done by means of refraction which is based on the fact that the angle of refraction (e.g. in glass) depends slightly on the wavelength.

The spectrum of light (wavelength composition) can be continuous (as in sunlight) or discrete (line spectrum) as in fluorescent lamp.

31.8 Sources of light

The most common sources of visible light are transitions of electrons from their excited (higher) energy state to the lowest (ground) energy states in atom with (often spontaneous) emission of photons responsible for energy difference:

Line spectra

/higher groundE E E hf hc λ∆ = − = =Such transitions with well-defined energies are responsible for discrete (line) spectrum.

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The energies that correspond to the shortest and longest wavelengths of the visible light are

6 9

6 9

/ 1.240 10 / 400 10 3.1 ,/ 1.240 10 / 700 10 1.77

E hc eV m m eVE hc eV m m eV

λ

λ

− −

− −

∆ = = ⋅ ⋅ ⋅ =

∆ = = ⋅ ⋅ ⋅ =Continuous spectra appear because of spread of electrons over energies as a result of thermal motion, collisions, interaction, etc.

In general, emission and absorption of light correspond to transitions of electrons between higher and lower energy states.

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(a)elastic (Rayleigh) scattering hf’ = hf(b) inelastic Stokes Raman scattering hf’ < hf

(c) anti-Stokes Raman scattering hf’ > hf(d) resonance absorption hf’ = 0(e) spontaneous emission hf = 0

(f) photoelectric effect: electron is emitted; no emitted photon(g) stimulated emission (an additional hf photon emitted)

(h) Compton scattering (in contrast to photoelectric effect, there is an emitted photon)

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Example 6. The excited states of potassium are 1.62 eV, 2.61 eV, and 3.07 eVabove the ground state. What are the possible wavelengths and frequencies of spontaneously emitted photons?

Transitions to the ground state yield the following energy differences, frequencies and the wavelengths:

15 1,0

15 2,0

1,0 1,0

141.621,0 1,0 1,04.136 10

2,0 2,0

142.612,0 2,0 2,04.136 10

3,0 3,0

3.073,0 3,0 4.13

1.62 ;

/ 3.917 10 ; 766 ;

2.61 ;

/ 6.310 10 ; 475 ;

3.07 ;

/

cf

cf

E hf eV

f E h Hz Hz nm

E hf eV

f E h Hz Hz nm

E hf eV

f E h

λ

λ

∆ = =

= ∆ = = ⋅ = =

∆ = =

= ∆ = = ⋅ = =

∆ = =

= ∆ = 15 3,0

143,06 10

7.422 10 ; 403 ;cfHz Hz nmλ−⋅

= ⋅ = =

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Transitions between the first excited state and higher states yield:

15 2,1

15 3,1

2,1 2,1

140.992,1 2,1 2,14.136 10

3,1 3,1

141.453,1 3,1 3,14.136 10

(2.61 1.62) 0.99 ;

/ 2.393 10 ; 1253 ;

(3.07 1.62) 1.45 ;

/ 3.506 10 ; 856

cf

cf

E hf eV eV

f E h Hz Hz nm

E hf eV eV

f E h Hz Hz nm

λ

λ

∆ = = − =

= ∆ = = ⋅ = =

∆ = = − =

= ∆ = = ⋅ = =

And, finally, the transition between the second and the third level yields

15 3,2

3,2 3,2

140.463,2 3,2 3,24.136 10

(3.07 2.61) 0.46 ;

/ 1.112 10 ; 2697cf

E hf eV eV

f E h Hz Hz nmλ−⋅

∆ = = − =

= ∆ = = ⋅ = =

Some of these wavelengths are too large or too short to be visible (infrared and ultraviolet light).

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LASERS(Light Amplification by Stimulated Emission of Radiation)

Successful operation of a laser requires a build-up of an excessive (non-equilibrium) population of electrons inthe metastable excited state E2.

This build-up is ensured by absorption of the (broad spectrum of) light emitted by the gaseous flashtube.

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One end of the (ruby) crystal is silvered (a perfect 100% mirror!), while the second one – the output coupler – is 85% reflective. As a result, there is a build-up of spontaneously emitted photons

A laser beam is coherent, narrow, and intense.

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Review of Chapter 31Visible light, λ ~ 400 – 700 nm.

Photon energy & Plank’s constant 34 15

6

/ , 6.626 10 4.136 10 ,1.240 10

E hf hc h J s eV shc eV m

λ − −

= = = ⋅ ⋅ = ⋅ ⋅

= ⋅ ⋅Atoms in dilute gases emit discrete set of wavelength (line spectra) that correspond transitions of electrons between atomic orbits,

/E hf hc λ∆ = =Atoms in dense liquids, solids, and gases have continuous energy bands and emit a continuous spectrum

Spontaneous emission of light with electron transition to a lower level occurs in 10-8 s

Stimulated emission requires presence of photon with proper energy / frequency

Speed of light c = 299,792,458 m/s ~ 3.0x108 m/s

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Lasers produce intense, coherent, and narrow beams of photons as a result of stimulated emission from a state with population inversion

In transparent medium, v = c/n, n = c/v > 1 (n is the index of refraction)

Huygens’s principle: Each point on a primary wavefront serves as a source spherical secondary wavelets that advance with a speed and frequency equal to that in the primary wave. The wavefront is an envelope of these wavelets, if one ignores the back wavelets.

Fermat’s principle: The path taken by light traveling from one point to another is such that the time of travel is minimal.

Laws of reflection and refraction (Snell’s law):

1 1 1 1 2 2', sin sinn nθ θ θ θ= =Reflected intensity at normal incidence:

( ) ( )1 2 1 2

1 2 1 2

2 2, 1n n n n

reflected incident transmitted incidentn n n nI I I I− −+ +

= = −

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Critical angle for total internal reflection

2 2

1 1sin sin 90n no

c n nθ = =Dispersion: dependence of the speed of light and the refraction index in the medium on the wavelength as a result of which white light splits into monochromatic components

θ

20 cosI I θ=

When two polarizers have transmission axes at angle , the intensity transmitted by the second polarizer drops as (Malus’s law)

Light can be polarized by absorption, reflection, scattering, and birefringence.