The Nature of Light

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The Nature of Light. Elements of wave optics. Chapter Five. Plan. Nature of light Thermal radiation Black body radiation Stefan-Boltzman law Wien’s law Absorbtion Polarization. Speed of Light.

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The Nature of Light

Elements of wave optics.

Chapter Five

Plan
• Nature of light
• 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

• 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
• Microwave: 1 mm
• Visible light: 500 nm
• X-rays: 1 nm
• Gamma rays: 10-3 nm
• 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.

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

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.

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 absorbthem and so can be called thermal radiation.

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:

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 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

A human body at room temperature emits most strongly at infrared light

The Sun as a Blackbody

Three Temperature Scales

Temperature in unit of Kelvin is often used in physics

TK = TC +273

TF = 1.8 (TC+32)

• 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:
• s=Stefan-Boltzmann constant =5.67 x10-8 W/m2K4
• 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 l gives us
• 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

Light rays

Ocean

Beach

Diffraction of Light

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

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.

Light source

S1

S2

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.

The Superposition Principle
• The resultant displacement of two simul-taneous 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.

s1

s1

s1

Constructive

s2

s2

s2

Destructive

Constructive

Young’s Interference Pattern

Bright fringe

Dark fringe

Bright fringe

p1

l

l

l

p2

p3

p4

Path difference

Bright fringes: Dp = nl, n = 0, 1, 2, . . .

Dp = 0, l , 2l, 3l, …

Conditions for Bright Fringes

Bright fringes occur when the difference in path Dpis an integral multiple of one wave length l.

p1

p2

l

n = odd

p3

l

n = 1,3,5 …

p3

Dark fringes:

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.

x

Path difference determines light and dark pattern.

d sin q

s1

d

q

s2

p1

y

Dp = p1 – p2

Dp = d sin q

p2

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

Dark fringes: d sin q = nl/2 , n = 1, 3, 5, . . .

Analytical Methods for Fringes

From geometry, we recall that:

x

d sin q

s1

d

q

s2

p1

y

p2

Bright fringes:

Dark fringes:

Analytical Methods (Cont.)

So that . . .

x

s1

d sin q

q

s2

y

n = 1, 3, 5

Dark fringes:

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

l = 600 nm; y = ?

d sin q = 5(l/2)

The third dark fringe occurs when n = 5

x

s1

x = 2 m; d = 0.08 mm

d sin q

q

l = 600 nm; y = ?

s2

y

n = 1, 3, 5

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?

y = 3.75 cm

d sin q

d

q

d sin q = nl

n = 1, 2, 3, …

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.

3l

The grating equation:

2l

l

1st order

d = slit width (spacing)

l= wavelength of light

6l

4l

q= angular deviation

2l

2nd order

n = order of fringe

The Grating Equation

n = 2

300 lines/mm

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

l = 600 nm

n = 2

300 lines/mm

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

Angular deviation of second order fringe is:

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.

Relative intensity

Pattern Exaggerated

Interference From Single Slit

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

The interference results from the fact that not all paths of light travel the same distance some arrive out of phase.

a/2

a

1

2

a/2

3

4

5

Single Slit Interference Pattern

Each point inside slit acts as a source.

For rays 1 and 3 and for 2 and 4:

First dark fringe:

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

a/2

a

1

2

a/2

3

4

5

Single Slit Interference Pattern

First dark fringe:

Other dark fringes occur for integral multiples of this fraction l/a.

l = ?

x = 1.5 m

q

y

a = 0.35 mm

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 = 600 nm

D

Circular diffraction

Diffraction for a Circular Opening

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.

Clear image of each object

Separate images barely seen

d2

d1

Resolution of Images

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

Resolution limit

d2

Separate images

Resolution Limit

Resolution Limit

Images are just resolved when central maximum of one pattern coincides with first dark fringe of the other pattern.

D

q

Limiting angle

Resolving Power of Instruments

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

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

Limiting angle of resolution:

p

D

q

so

q

Limiting Angle of Resolution:

Limiting angle qo

Resolution and Distance

p

q

so

q

Eye

Tail lights

D

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 =3.11 km

Young’s Experiment:

x

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

d sin q

s1

d

q

s2

p1

y

p2

Bright fringes:

Dark fringes:

Summary

The grating equation:

d = slit width (spacing)

q= angular deviation

l= wavelength of light

n = order of fringe

Summary (Cont.)

Relative Intensity

Pattern Exaggerated

Summary (Cont.)

Interference from a single slit of width a:

p

D

q

so

q

Limiting Angle of Resolution:

Limiting angle qo

Summary (cont.)

The resolving power of instruments.