Transcript 04_Elements of wave optics

The Nature of Light
Chapter Five
Elements of wave optics.
Plan
1.
2.
3.
4.
5.
6.
7.
Nature of light
Thermal radiation
Black body radiation
Stefan-Boltzman law
Wien’s law
Absorbtion
Polarization
Speed of Light
• In 1850 Fizeau and Foucalt also experimented with light by
bouncing it off a rotating mirror and measuring time
• The light returned to its source at a slightly different position
because the mirror has moved during the time light was
traveling
• The deflection angle depends on the speed of light and the
dimensions of the apparatus.
Light: spectrum and color
• Newton found that the white light from the Sun is composed
of light of different color, or spectrum (1670).
Light has wavelike property
• Young’s Double-Slit Experiment indicated light behaved as
a wave (1801)
• The alternating black and bright bands appearing on the
screen is analogous to the water waves that pass through a
barrier with two openings
Light is Electromagnetic Radiation
•
•
The nature of light is electromagnetic radiation
In the 1860s, James Clerk Maxwell succeeded in describing all
the basic properties of electricity and magnetism in four
equations: the Maxwell equations of electromagnetism.
Light: Wavelength and Frequency
• Example
– FM radio, e.g., 103.5 MHz (WTOP station) => λ = 2.90 m
– Visible light, e.g., red 700 nm => ν = 4.29 X 1014 Hz
• The speed of light in the vacuum
– C = 299,792.458 km/s, or
– C = 3.00 X 105 km/s = 3.00·108 m/s
Electromagnetic Spectrum
• Visible light falls in the 400 to
700 nm range
• In the order of decreasing
wavelength
– Radio waves: 1 m
– Microwave: 1 mm
– Infrared radiation: 1 μm
– Visible light: 500 nm
– Ultraviolet radiation: 100 nm
– X-rays: 1 nm
– Gamma rays: 10-3 nm
Radiation depending on Temperature
• A general rule:
The higher an object’s temperature, the more intensely
the object emits electromagnetic radiation and the
shorter the wavelength at which emits most strongly
The example of heated iron bar.
As the temperature increases
– The bar glows more
brightly
– The color of the bar also
changes
Thermal radiation is electromagnetic radiation
emitted from the surface of an object which is due to
the object's temperature. An example of thermal
radiation is the infrared radiation emitted by hot
objects. A person near a fire will feel the radiated
heat of the fire, even if the surrounding air is very
cold. You cannot see the thermal radiation but you
can see other energies.
10
How does the sun warm you on a hot day?
Earth is warmed by heat energy from the Sun.
How does this heat energy travel from the Sun to the Earth?
?
infrared
radiation
There are no particles
between the Sun and the
Earth so the heat cannot
travel by conduction or by
convection.
The heat travels to Earth by
infrared waves. They are
similar to light waves and
are able to travel through
empty space.
11
INFRARED WAVES
Heat can move by travelling as infrared waves.
These are electromagnetic waves, like light waves,
but with a longer wavelength.
This means that infrared waves act like light waves:
 They can travel through a vacuum.
 They travel at the same speed as light – 300,000,000 m/s.
 They can be reflected and absorbed.
Infrared waves heat objects that absorb them and so
can be called thermal radiation.
12
Radiation in Equilibrium with Matter
Typically, radiation emitted by a hot body, or from a laser
is not in equilibrium: energy is flowing outwards and must be
replenished from some source. The first step towards
understanding of radiation being in equilibrium with matter was
made by Kirchhoff, who considered a cavity filled with radiation,
the walls can be regarded as a heat bath for radiation.
The walls emit and absorb e.-m. waves. In equilibrium, the
walls and radiation must have the same temperature T. The energy
of radiation is spread over a range of frequencies, and we define
uS (,T) d as the energy density (per unit volume) of the
radiation with frequencies between  and +d. uS(,T) is the
spectral energy density. The internal energy of the photon gas:

u T    uS  , T  d
0
13
In equilibrium, uS (,T) is the same everywhere in the
cavity, and is a function of frequency and temperature only. If the
cavity volume increases at T=const, the internal energy U = u (T)
V also increases. The essential difference between the photon gas
and the ideal gas of molecules: for an ideal gas, an isothermal
expansion would conserve the gas energy, whereas for the photon
gas, it is the energy density which is unchanged, the number of
photons is not conserved, but proportional to volume in an
isothermal change.
A real surface absorbs only a fraction of the radiation
falling on it. The absorptivity  is a function of  and T; a surface
for which ( ) =1 for all frequencies is called a black body.
14
Blackbody Radiation
• A blackbody is a hypothetical
object that is a perfect absorber of
electromagnetic radiation at all
wavelengths
– The radiation of a blackbody is
entirely the result of its
temperature
– A blackbody does not reflect any
light at all
• Blackbody curve: the intensities
of radiation emitted at various
wavelengths by a blackbody at a
given temperature
– The higher the temperature, the
shorter the peak wavelength
– The higher the temperature, the
higher the intensity
Blackbody curve
Blackbody Radiation
• Hot and dense objects act like a blackbody
• Stars, which are opaque gas ball, closely approximate the behavior
of blackbodies
• The Sun’s radiation is remarkably close to that from a blackbody at
a temperature of 5800 K
The Sun as a Blackbody
A human body at room temperature
emits most strongly at infrared light
Three Temperature Scales
Temperature in unit of Kelvin is
often used in physics
TK = TC +273
TF = 1.8 (TC+32)
Blackbody Radiation
• Blackbody – a perfect emitter & absorber of
radiation; it absorbs all incident radiation, and no
surface can emit more for a given temperature
and wavelength
• Emits radiation uniformly in all directions – no
directional distribution – it’s diffuse
• Example of a blackbody: large cavity with a small
hole Joseph Stefan (1879)– total radiation
emission per unit time & area over all
wavelengths and in all directions:

Eb  sT 4 W m2
 s=Stefan-Boltzmann
W/m2K4
constant

=5.67
x10-8
Blackbody radiation:
Stefan-Boltzmann Law
• The Stefan-Boltzmann law states that a blackbody radiates
electromagnetic waves with a total energy flux F directly
proportional to the fourth power of the Kelvin
temperature T of the object:
F = sT4
• F = energy flux, in joules per square meter of surface per
second
• s = Stefan-Boltzmann constant = 5.67 X 10-8 W m-2 K-4
• T = object’s temperature, in kelvins
Planck’s Distribution Law
• Sometimes we care about the radiation in a certain
wavelength interval
• For a surface in a vacuum or gas
C1
Ebl T   5
W m 2  μm
l expC2 lT   1
where


C1  2hco2  3.742 x108 W  μm4 m 2
C2  hco k  1.439 x10 4 μm  K
k  1.3805 x10- 23 J/K  Boltzmann' s constant
• Integrating this function over all l gives
uss T 4
E 
b
Blackbody Radiation: Wien’s Law
•Wien’s law states that the dominant wavelength at
which a blackbody emits electromagnetic radiation is
inversely proportional to the Kelvin temperature of the
object
For example
– The Sun, λmax = 500 nm  T = 5800 K
– Human body at 37 degrees Celcius, or 310 Kelvin  λmax =
9.35 μm = 9350 nm
Dual properties of Light:
(1) waves and (2) particles
• Light is an electromagnetic radiation wave, e.g, Young’s
double slit experiment
• Light is also a particle-like packet of energy - photon
– Light particle is called photon
– The energy of phone is related to the wavelength of light
• Light has a dual personality; it behaves as a stream of
particle like photons, but each photon has wavelike
properties
Dual properties of Light: Planck’s Law
• Planck’s law relates the energy of a photon to its
wavelength or frequency
– E = energy of a photon
– h = Planck’s constant
= 6.625 x 10–34 J s
– c = speed of light
– λ= wavelength of light
• Energy of photon is inversely proportional to the
wavelength of light
• Example: 633-nm red-light photon
– E = 3.14 x 10–19 J
– or E = 1.96 eV
– eV: electron volt, a small energy unit = 1.602 x 10–19 J
Kirchhoff’s Laws on Spectrum
• Three different spectrum: continuous spectrum, emission-line
spectrum, and absorption line spectrum
Kirchhoff’s Laws on Spectrum
• Law 1- Continuous spectrum: a hot opaque body, such as a
perfect blackbody, produce a continuous spectrum – a complete
rainbow of colors without any spectral line
• Law 2 – emission line spectrum: a hot, transparent gas
produces an emission line spectrum – a series of bright spectral
lines against a dark background
• Law 3 – absorption line spectrum: a relatively cool, transparent
gas in front of a source of a continuous spectrum produces an
absorption line spectrum – a series of dark spectral lines
amongst the colors of the continuous spectrum. Further, the
dark lines of a particular gas occur at exactly the same
wavelength as the bright lines of that same gas.
Polarizers and analysers
• A polarizer (like polaroid) can be used to
polarize light
Polarizers and analysers
• A polarizer can also be used to determine
if light is polarized. It is then called an
analyser.
Malus’ Law
• The intensity of polarised light that passes
through a polarizer is proportional to the
square of the cosine of the angle between
the electric field of the polarized light and
the angle of the polarizer!
Malus’ law
I = Iocos2θ
Io
Iocos2θ
Optical activity
• Some substances can change the plane of
polarized light. We say they are optically
active
Optical activity
Sugar solution is optically active. The
amount of rotation of the plane of
polarization depends on the concentration
of the solution.
Interference and Diffraction
Diffraction of Light
Diffraction is the ability of light waves to
bend around obstacles placed in their path.
Ocean
Beach
Light rays
Fuzzy Shadow
Water waves easily bend around obstacles, but light
waves also bend, as evidenced by the lack of a sharp
shadow on the wall.
Water Waves
A wave generator sends periodic water
waves into a barrier with a small gap,
as shown below.
A new set of waves is observed
emerging from the gap to the wall.
Interference of Water Waves
An interference pattern is set up by water
waves leaving two slits at the same instant.
Young’s Experiment
In Young’s experiment, light from a monochromatic source
falls on two slits, setting up an interference pattern
analogous to that with water waves.
Light
source
S1
S2
The Superposition Principle
• The resultant displacement of two simultaneous waves (blue and green) is the
algebraic sum of the two displacements.
• The composite wave is shown in yellow.
Constructive Interference Destructive Interference
The superposition of two coherent light waves
results in light and dark fringes on a screen.
Young’s Interference Pattern
s1
Constructive
Bright fringe
s2
s1
s2
s1
Dark fringe
Destructive
s2
Constructive
Bright fringe
Conditions for Bright Fringes
Bright fringes occur when the difference in path Dp is
an integral multiple of one wave length l.
p1
l
l
l
p2
p3
p4
Path difference
Dp = 0, l , 2l, 3l, …
Bright fringes:
nl, n = 0, 1, 2, . . .
Dp =
Conditions for Dark Fringes
Dark fringes occur when the difference in path Dp is an
odd multiple of one-half of a wave length l/2.
l
2
p1
p2
l
l
p3
p3
Dark fringes:
Dp  n
Dp  n
l
2
l
2
n = odd
n=
1,3,5 …
n  1, 3, 5, 7, . . .
Analytical Methods for
Fringes
x
d sin q
s1
d
s2
Path difference determines
light and dark pattern.
q
p1
y
p2
Bright fringes: d sin q = nl, n = 0, 1, 2, 3, . . .
Dark fringes:
d sin q = nl/2 , n = 1, 3, 5, . . .
Dp = p1 – p2
Dp = d sin q
Analytical Methods (Cont.)
From geometry, we recall
that:
x
d sin q
s1
d
s2
q
p1
y
p2
y
sin q  tan q 
x
So that . . .
dy
d sin q 
x
Bright fringes:
dy
 nl , n  0, 1, 2, ...
x
Dark fringes:
dy
l
 n , n  1, 3, 5...
x
2
Example 1: Two slits are 0.08 mm apart, and the
screen is 2 m away. How far is the third dark
fringe located from the central maximum if light of
wavelength 600 nm is used?
x = 2 m; d = 0.08 mm
x
s1
l = 600 nm; y = ?
d sin q = 5(l/2)
The third dark fringe occurs when n = 5
Dark fringes:
dy
l
 n , n  1, 3, 5...
x
2
q
d sin q
s2
y
n = 1, 3, 5
dy 5l

x
2
Example 1 (Cont.): Two slits are 0.08 mm apart, and the
screen is 2 m away. How far is the third dark fringe
located from the central maximum if l = 600 nm?
x = 2 m; d = 0.08 mm
x
s1
l = 600 nm; y = ?
dy 5l

x
2
q
d sin q
s2
y
n = 1, 3, 5
5l x 5(600 x 10-9 m)(2 m)
y

2d
2(0.08 x 10-3m)
y = 3.75 cm
The Diffraction Grating
A diffraction grating consists of thousands of parallel slits
etched on glass so that brighter and sharper patterns can
be observed than with Young’s experiment. Equation is
similar.
d sin q
d
q
d sin q  nl
n = 1, 2, 3, …
The Grating Equation
3l
The grating equation:
d sin q  nl n  1, 2, 3, ...
2l
l
1st order
d = slit width (spacing)
l = wavelength of light
q = angular deviation
6l
4l
2l nd
2
n = order of fringe
order
Example 2: Light (600 nm) strikes a grating ruled with
300 lines/mm. What is the angular deviation of the 2nd
order bright fringe?
To find slit separation, we
take reciprocal of 300
lines/mm:
Lines/mm  mm/line
n=2
300 lines/mm
d  3 x 10 m
-6
Example (Cont.) 2: A grating is ruled with 300
lines/mm. What is the angular deviation of the 2nd
order bright fringe?
l = 600 nm
d  3 x 10-6 m
d sinq  nl
n2
n=2
300 lines/mm
2l 2(600 x 10-9 m)
sin q 

;
-6
d
3.33 x 10
Angular deviation of second order
fringe is:
sinq  0.360
q2 = 21.10
A compact disk acts as a diffraction grating. The colors
and intensity of the reflected light depend on the
orientation of the disc relative to the eye.
Interference From Single Slit
When monochromatic light strikes a single slit, diffraction from
the edges produces an interference pattern as illustrated.
Relative intensity
Pattern Exaggerated
The interference results from the fact that not all paths of light
travel the same distance some arrive out of phase.
Single Slit Interference Pattern
a
sin q
2
For rays 1 and 3 and for 2
and 4:
a/2
a
1
a/2
Each point inside slit acts
as a source.
2
3
4
5
a
Dp  sin q
2
First dark fringe:
a
l
sin q 
2
2
For every ray there is another ray that differs by this path and
therefore interferes destructively.
Single Slit Interference
Pattern
a
sin q
2
a
l
sin q 
2
2
First dark fringe:
a/2
a
1
a/2
2
sin q 
l
a
3
4
5
Other dark fringes occur for
integral multiples of this
fraction l/a.
Example 3: Monochromatic light shines on a
single slit of width 0.45 mm. On a screen 1.5 m
away, the first dark fringe is displaced 2 mm
from the central maximum. What is the
wavelength of the
light?
l=?
sin q 
l
x = 1.5 m
q
a
y
sin q  tan q  ;
x
a = 0.35 mm
y l
 ;
x a
(0.002 m)(0.00045 m)
l
1.50 m
ya
l
x
l = 600 nm
y
Diffraction for a Circular Opening
D
Circular diffraction
The diffraction of light passing through a circular opening
produces circular interference fringes that often blur images.
For optical instruments, the problem increases with larger
diameters D.
Resolution of Images
Consider light through a pinhole. As two objects get closer the
interference fringes overlap, making it difficult to distinguish
separate images.
Clear image of each
object
d1
Separate images barely
seen
d2
Resolution Limit
Images are just resolved when central
maximum of one pattern coincides
with first dark fringe of the other
pattern.
Resolution limit
d2
Separate images
Resolution Limit
Resolving Power of
Instruments
The resolving power of an instrument
is a measure of its ability to produce
well-defined separate images.
D
q
Limiting angle
For small angles, sin q  q, and the limiting angle of resolution for a
circular opening is:
Limiting angle of resolution:
q 0  1.22
l
D
Resolution and Distance
p
q
so
D
q
Limiting angle qo
Limiting Angle of
Resolution:
l
s0
q 0  1.22 
D p
Example 4: The tail lights (l = 632 nm) of an
auto are 1.2 m apart and the pupil of the eye is
around 2 mm in diameter. How far away can
the tail lights be resolved as separate images?
p
q
so
q
D
Eye
Tail lights
l
s0
q 0  1.22 
D p
(1.2 m)(0.002 m)
p
1.22(632 x 10-9 m)
s0 D
p
1.22l
p = 3.11 km
Summary
Young’s Experiment:
x
d sin q
s1
Monochromatic light falls on
two slits, producing
interference fringes on a
screen.
d
s2
q
p1
y
p2
dy
d sin q 
x
Bright fringes:
dy
 nl , n  0, 1, 2, ...
x
Dark fringes:
dy
l
 n , n  1, 3, 5...
x
2
Summary (Cont.)
The grating equation:
d sin q  nl n  1, 2, 3, ...
d = slit width (spacing)
q = angular deviation
l = wavelength of light
n = order of fringe
Summary (Cont.)
Interference from a single slit of width a:
Relative Intensity
Pattern Exaggerated
Dark Fringes: sin q  n
l
a
n  1, 2, 3, . . .
Summary (cont.)
The resolving power of instruments.
p
q
so
D
q
Limiting angle qo
Limiting Angle of
Resolution:
l
s0
q 0  1.22 
D p