The Nature of LightChapter Five

Elements of wave optics.

Plan

1. Nature of light2. Thermal radiation3. Black body radiation4. Stefan-Boltzman law5. Wien’s law6. Absorbtion7. Polarization

• 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.

Speed of Light

Light: spectrum and color• Newton found that the white light from the Sun is composed

of light of different color, or spectrum (1670).

• 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 has wavelike property

• 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 is Electromagnetic Radiation

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

• 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

Electromagnetic Spectrum

• 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

Radiation depending on Temperature

The example of heated iron bar. As the temperature increases– The bar glows more

brightly– The color of the bar also

changes

10

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.

11

Earth is warmed by heat energy from the Sun.

How does this heat energy travel from the Sun to the Earth?

There are no particles between the Sun and the Earth so the heat cannot travel by conduction or by convection.

?infraredradiation

The heat travels to Earth by infrared waves. They are similar to light waves and are able to travel through empty space.

How does the sun warm you on a hot day?

12

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.

INFRARED WAVES

13

Radiation in Equilibrium with MatterTypically, 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:

dTuTu S

0

,

14

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.

• 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 Radiation

Blackbody curve

• 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

Blackbody Radiation

The Sun as a BlackbodyA human body at room temperature emits most strongly at infrared light

Three Temperature ScalesTemperature 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:

=Stefan-Boltzmann constant =5.67 x10-8 W/m2K4 24 mW TEb

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 = T4

• F = energy flux, in joules per square meter of surface per second

• = 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

• Integrating this function over all gives us

constant sBoltzmann'J/K 1038051

Kμm 104391

mμmW 1074232

where

μmmW 1

23

42

24821

2

25

1

-o

o

b

x.k

x.khcC

x.hcC

TCexpCTE

4 bE T

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

• 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

Dual properties of Light: Planck’s Law

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

IoIocos2θ

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 LightDiffraction is the ability of light waves to bend around obstacles placed in their path.

Ocean Beach

Water waves easily bend around obstacles, but light waves also bend, as evidenced by the lack of a sharp shadow on the wall.

Fuzzy Shadow

Light rays

Water WavesA 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 WavesAn interference pattern is set up by water waves leaving two slits at the same instant.

Young’s ExperimentIn 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 simul-

taneous waves (blue and green) is the algebraic sum of the two displacements.

The superposition of two coherent light waves results in light and dark fringes on a screen.

• The composite wave is shown in yellow.

Constructive InterferenceConstructive Interference Destructive InterferenceDestructive Interference

Young’s Interference Patterns1

s2

s1

s2

s1

s2

Constructive

Constructive

Bright fringe

Bright fringe

Dark fringe

Destructive

Conditions for Bright FringesBright fringes occur when the difference in path Dp is an integral multiple of one wave length l..

pp11

pp22

pp33

pp44

Path difference

p = 0, , 2, 3, …

Bright fringes: p = n, n = 0, 1, 2, . . .

Conditions for Dark FringesDark fringes occur when the difference in path Dp is an odd multiple of one-half of a wave length l/2.

pp11

pp22 2

pp33

pp33

2p n

n n = odd= oddn n = = 1,3,5 …1,3,5 …

Dark fringes: 1, 3, 5, 7, . . .2

p n n

Analytical Methods for Fringes

x

y

d sin s1

s2

d

p1

p2

Bright fringes: d sin = n, n = 0, 1, 2, 3, . . .

Dark fringes: d sin = n, n = 1, 3, 5, . . .

p = p1 – p2

p = d sin

Path difference determines Path difference determines light and dark pattern.light and dark pattern.

Analytical Methods (Cont.)x

y

d sin s1

s2

d p1

p2

From geometry, we recall From geometry, we recall that:that:

Bright fringes:

, 0, 1, 2, ...dy n nx

Dark fringes:

, 1, 3, 5...2

dy n nx

sin tan yx

So that . . .So that . . .

sin dydx

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 mml = 600 nm; y = ?

The third dark fringe occurs when n = 5

x

y

d sin s1

s2

n = 1, 3, 5

Dark fringes:

, 1, 3, 5...2

dy n nx

d sin q = 5(l/2)

52

dyx

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

l = 600 nm; y = ?

x

y

d sin s1

s2

n = 1, 3, 5

52

dyx

-9

-3

5 5(600 x 10 m)(2 m)2 2(0.08 x 10 m)

xyd

y = 3.75 cm

The Diffraction GratingA 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

d

d sin n

n = 1, 2, 3, …

The Grating EquationThe grating equation:

sin 1, 2, 3, ...d n n

d = slit width (spacing)

= wavelength of light

= angular deviation

n = order of fringe

1st order

2nd 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?

300 lines/mm

n = 2

To find slit separation, we take reciprocal of 300 lines/mm:

Lines/mm mm/line

-6 3 x 10 md

Example (Cont.) 2: A grating is ruled with 300 lines/mm. What is the angular deviation of the 2nd

order bright fringe?

-9

-6

2 2(600 x 10 m)sin ;3.33 x 10d

sin 0.360

2 = 21.10Angular deviation of second order Angular deviation of second order fringe is:fringe is:

300 lines/mm

n = 2

sin 2d n n

-6 3 x 10 md = 600 nm

A compact disk acts as a diffraction grating. The colors A compact disk acts as a diffraction grating. The colors and intensity of the reflected light depend on the and intensity of the reflected light depend on the orientation of the disc relative to the eye.orientation of the disc relative to the eye.

Interference From Single Slit

Pattern ExaggeratedPattern Exaggerated

When monochromatic light strikes a single slit, diffraction from the edges produces an interference pattern as illustrated.

Relative intensity

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/2

aa/2

sin2a

1

2

4

3

5

Each point inside slit acts as a source.

For rays 1 and 3 and for 2 and 4:

sin2ap

First dark fringe:

sin2 2a

For every ray there is another ray that differs by this path and therefore interferes destructively.

Single Slit Interference Pattern

a/2

aa/2

sin2a

1

2

4

3

5

sin2 2a

First dark fringe:First dark fringe:

sina

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?

x = 1.5 m

y

a = 0.35 mm

= ?

sina

ysin tan ; ; x

y yax a x

(0.002 m)(0.00045 m)1.50 m

= 600 nm

Diffraction for a Circular Opening

Circular diffractionCircular diffraction

D

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 ImagesConsider light through a pinhole. As two objects get closer the interference fringes overlap, making it difficult to distinguish separate images.

d2

Separate images barely seen

d1

Clear image of each object

Resolution Limit

d2

Resolution limitImages are Images are just resolvedjust resolved when when central central maximummaximum of one pattern coincides of one pattern coincides with with first dark fringefirst dark fringe of the other of the other pattern.pattern.

Resolution LimitResolution LimitSeparate imagesSeparate images

Resolving Power of Instruments

The resolving power of an instrument is a measure of its ability to produce well-defined separate images.

0 1.22D

Limiting angle of resolution:

For small angles, For small angles, sin sin ,,and the limiting angle of resolution for and the limiting angle of resolution for a circular opening is:a circular opening is:

Limiting angleLimiting angle

D

Resolution and Distance

Limiting Angle of Resolution: 0

0 1.22 sD p

ssoo

pp

D

Limiting angle Limiting angle oo

Example 4: The tail lights ( = 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?

ssoo

pp

EyeEye

D

Tail lightsTail lights

00 1.22 s

D p 0

1.22s Dp

-9

(1.2 m)(0.002 m)1.22(632 x 10 m)

p p = 3.11 km

Summary

Bright fringes:

, 0, 1, 2, ...dy n nx

Dark fringes:

, 1, 3, 5...2

dy n nx

Young’s Experiment:Young’s Experiment:

Monochromatic light falls on Monochromatic light falls on two slits, producing two slits, producing interference fringes on a interference fringes on a screen.screen.

x

y

d sin s1

s2

d p1

p2

sin dydx

Summary (Cont.)

The grating equation:

sin 1, 2, 3, ...d n n

d = slit width (spacing)

= wavelength of light

= angular deviation

n = order of fringe

Summary (Cont.)

Pattern ExaggeratedPattern Exaggerated

Relative Intensity

Dark Fringes: sin 1, 2, 3, . . .n na

Interference from a single slit of width a:

Summary (cont.)

Limiting Angle of Resolution: 0

0 1.22 sD p

ssoo

pp

D

Limiting angle Limiting angle oo

The resolving power of instruments.