1 / 161

# Physics Beyond 2000 - PowerPoint PPT Presentation

Physics Beyond 2000. Chapter 11 Electromagnetic Waves. -. What are electromagnetic waves?. EM waves are energy emitted resulting from acceleration of electric charges. EM Waves. They can travel through vacuum. In vacuum, their speed = 3 × 10 8 ms -1 c = f.λ

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Physics Beyond 2000' - september-middleton

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Physics Beyond 2000

Chapter 11

Electromagnetic Waves

What are electromagnetic waves?

• EM waves are energy emitted resulting from acceleration of electric charges.

• They can travel through vacuum.

• In vacuum, their speed = 3 × 108 ms-1

• c = f.λ

• An EM wave consists of electric and magnetic fields, oscillating in phase and at right angles to each other.

http://www.geo.mtu.edu/rs/back/spectrum/

• The range of the wavelength of EM waves is enormous.

• 10-14 m – 1 km

• The electromagnetic spectrum is named according to the range of the wavelength and the method of production.

direction of propagation of

oscillating a.c.

• Production:

• Apply an a.c. voltage of high frequency to a pair of metal rods (dipole).

• If the rods are vertical, the radio wave is also said to be vertically polarized.

transmitter

• The receiver dipole is parallel to the direction of polarization. In this case, it is vertical.

transmitter

oscillating electric field

direction of propagation of

oscillating a.c.

• Microwave is polarized along the length of the dipole.

transmitter

oscillating electric field

direction of propagation of

microwave

oscillating a.c.

• Vertical metal rod can absorb the energy of the microwave.

metal rod

transmitter

oscillating electric field

direction of propagation of

microwave

no response

oscillating a.c.

• Horizontal metal rod cannot absorb the energy of the microwave.

metal rod

transmitter

oscillating electric field

direction of propagation of

microwave

oscillating a.c.

of transmitter

metal plate

P

transmitter

Interference of Microwaves

• At P, the wave from the transmitter meets the reflected wave. Interference occurs.

of transmitter

metal plate

P

transmitter

Interference of Microwaves

• We may consider it as an interference from two coherent sources, the transmitter and its image.

of transmitter

metal plate

P

transmitter

Interference of Microwaves

• The two sources are in anti-phase because there is a phase change of  on reflection.

of transmitter

metal plate

P

transmitter

Interference of Microwaves

• If the path difference at P = n., there is destructive interference.

of transmitter

metal plate

P

transmitter

Interference of Microwaves

• If the path difference at P = , there is constructive interference.

• One possible frequency of microwave is 2.45 GHz which is equal to the natural frequency of water molecules.

• Microwave can set water molecules into oscillation. The water molecules absorb the energy from microwave.

http://www.gallawa.com/microtech/howcook.html

Microwave in satellite communications

http://www.s-t.au.ac.th/~supoet/satel.htm#1

atom

oscillating E-field

Scattering of Light

• Light energy is absorbed by an atom or molecule.

• The atom (molecule) re-emits the light energy in all direction.

• The intensity of light in initial direction is reduced.

atom

Scattering of Light

• Light energy is absorbed by an atom or molecule.

• The atom (molecule) re-emits the light energy in all directions.

• The intensity of light in initial direction is reduced.

axis along which the atom oscillating

atom

Scattering of Light

• Note that there is not any scattered light along the direction of oscillation of the atom.

• The scattered light is maximum at right angle to the axis.

axis along which the atom oscillating

strongest

strongest

• Why is the sky blue in daytime?

• Why is the sky red in sunset/sunrise?

• At noon, we see the most scattered light.

• Note that the natural frequency of air molecules is in the ultraviolet region. Blue light is easily scattered by air molecules.

white light from

the sun

Blue light

is most scattered

Red light is least scattered

• At sunset, we see the least scattered light.

• Red light is least scattered.

white light from

the sun

Red light

is least scattered

Blue light is most scattered

• Light is transverse wave so it exhibits polarization.

• Unpolarized light: the electric field is not confined to oscillate in a plane.

• Plane-polarized light: the electric field at every point oscillates in the same fixed plane.

• Plane of polarization: the plane in which the electric field of a plane polarized light oscillates.

• Plane polarized light:

• Unpolarised light:

electric vector

electric vector

• An array of parallel conducting wires.

• It can absorb electric field of microwave oscillating in a plane parallel to its conducting wires.

Plane-polarized

microwave

Conducting wires are vertical

No microwave

E-field is vertial.

• An array of parallel conducting wires.

• It cannot absorb electric field of microwave oscillating in a plane perdpndicular to its conducting wires.

Plane-polarized

microwave

Conducting wires are vertical

E-field is horizontal

Plane-polarized

microwave

• An array of parallel conducting wires.

• It can be a polarizer of microwaves

Unpolarized

microwave

Conducting wires are vertical

Plane-polarized

microwave

• Polaroid is a plastic sheet consisting of long chains of molecules parallel to one another.

• It can absorb electric field of light oscillating in a plane parallel to its chains of molecules.

Chains of molecules are vertical

Plane-polarized light

No light

E-field is vertical

• Polaroid is a plastic sheet consisting of long chains of molecules parallel to one another.

• It cannot absorb electric field of light oscillating in a plane perpendicular to its chains of molecules.

Chains of molecules are vertical

Plane-polarized light

Plane-polarized light

E-field is horizontal

• Polaroid is a plastic sheet consisting of long chains of molecules parallel to one another.

• It can be a polarizer of light.

Chains of molecules are vertical

Unpolarized light

Plane-polarized light

incident light

No reflected light

air

glass

plane-polarized

refracted light

Polarization by Reflection

Assume that the direction of the reflected light and that of the refracted

light are perpendicular.

incident light

No reflected light

air

glass

plane-polarized

refracted light

Polarization by Reflection

• The electric field sets the

• electrons in the glass to

• oscillate at right angles to

• the refracted ray.

• The intensity perpendicular

• to the axis of oscillation is

• strongest  The refracted ray is bright.

• The intensity parallel to

• the axis of oscillation is zero  no reflected ray.

Assume that the direction of the reflected light and that of the refracted

light are perpendicular.

The plane of polarization

is parallel to the surface

of medium.

Unpolarized

incident light

Polarized reflected light

air

glass

Unpolarized

refracted light

ray is completely

polarized

incident

ray

p

p

air

medium

r

refracted

ray

The Brewster’s Angle

• p = Brewster’s angle

• r = Angle of refraction.

ray

reflected

ray

p

p

r

refracted

ray

The Brewster’s Angle

• n = tan pwhere n is the refractive index of the medium.

air

medium

Prove it!

• The Brewster’s angle for glass is about 56.3o.

• When light energy is absorbed by an atom, the atom re-radiates the light.

incident ray

atom

The atom absorbs

the wave energy.

the wave energy.

vertically polarized

light

no scattered light

vertically polarized

light

water mixed with milk

vertically polarized

light

vertically polarized

light

no scattered light

horizontally polarized

light

no scattered light

horizontally polarized

light

water mixed with milk

horizontally polarized

light

no scattered light

horizontally polarized

light

horizontally polarized

light

vertically polarized

light

unpolarized

light

water mixed with milk

unpolarized

light

vertically polarized

light

horizontally polarized

light

• Why are the polaroid sunglasses designed to absorb horizontally polarized light?

Study p.233 of the textbook.

• Light is a kind of wave.

• Interference is a wave property.

Conditions for an observable interference pattern of light:

• Coherent sources : two sources emit light of the same frequency and maintain a constant phase difference.

• The light waves are of same frequency and almost equal amplitude.

• The separation of the two sources is of the same order as the wavelength.

• The path difference must be not too large.

• Young’s double-slit experiment

http://surendranath.tripod.com/DblSlt/DblSltApp.html

http://members.tripod.com/~vsg/interfer.htm

• The incident ray is split into two coherent sources

• S1 and S2 by the double-slit.

• S1 and S2 are in phase.

• The screen is far away from the slit. D>> a.

• The angles are very small.

Suppose that there is a maximum at point P.

A constructive interference occurs at P.

P

S1

a

central line

S2

D

screen

As point P is far away from the double slit,

the light rays of the same fringe are parallel.

parallel rays meet at point P

S1

a

central line

S2

The path difference  = a.sin

• = a.sin = m. where m = 0, 1, 2,…

m is the order of the fringes.

For points with constructive interference,

parallel rays meet at point P with

maximum intensity.

S1

a

central line

S2

The path difference  = a.sin

For points with destructive interference,

• = a.sin = (m + ).  where m = 0, 1, 2,…

parallel rays meet at a point with

minimumintensity.

S1

a

central line

S2

The path difference  = a.sin

Suppose the the order of the fringe at P is m.

The distance from P to the central line is ym.

The distance between the double slit

and the screen is D.

P

ym

S1

a

M

central line

S2

D

ym = D.tan  D.sin  =

separation between the fringes

The line from mid-point M to P makes the same angle  with the central line.

P

ym

S1

a

M

central line

S2

D

s = ym+1 – ym =

separation between the fringes

ym =

(1)

By similar consideration, for the m+1 bright fringe

ym+1=

(2)

The separation between the two fringes is

s = ym+1 – ym =

separation between the fringes

• The fringes are evenly separated.

• For well separated fringes.

• s  D  Place the screen far away from the slit.

• s   Different separation for waves of different wavelength.

• s   The slits should be close.

• If the slits are sufficiently narrow, light spreads out evenly from each slit and the bright fringes are equally bright.

• If the intensity on the screen using one slit is Io,

the intensity is 4.Io at the position of bright fringes and

the intensity is 0 at the position of dark fringes.

• Energy is re-distributed on the screen.

• In practice, light waves do not diffract evenly out from each slit. There is an angle of spread.

http://numerix.us.es/numex/numex2.html

http://bc1.lbl.gov/CBP_pages/educational/java/duality/duality2.html

• The intensity of the fringes is enclosed in an envelope as shown.

• Separation of fringes in Young’s double-slit experiment.

White light fringes

 s  

Separation of violet fringes is shortest.

Separation of red fringes is longest.

http://surendranath.tripod.com/DblSlt/DblSltApp.html

http://members.tripod.com/~vsg/interfer.htm

• If the Young’s double-slit experiment is done in a liquid, what would happen to the separation of fringes?

ym

Liquid with refractive index n

• The wavelength changes!

Let  be the wavelength in vacuum/air

and nthe wavelength in liquid.

Let n be the refractive index of the liquid.

The fringe separation in liquid

sn =

• The fringe separation is reduced by a factor of n.

The fringe separation in liquid

sn =

• Optical path of light in a medium is the equivalent distance travelled by light in vacuum.

thickness = t

medium of

refractive index

n

Light requires the same

time to travel through

the two paths.

Incident

light

thickness = optical path

vacuum

• Show that the optical path = n.t

thickness = t

medium of

refractive index

n

Light requires the same

time to travel through

the two paths.

Incident

light

thickness = optical path

vacuum

• The number of waves in the medium = the number of waves in the optical path

thickness = t

medium of

refractive index

n

Light requires the same

time to travel through

the two paths.

Incident

light

thickness = optical path

vacuum

• Note that light rays pass through different media. We need to consider their path difference in terms of the optical paths.

B

Shifting a System of Fringes

Note that ray A has passes through a medium of refractive

index n and thickness t.

We need to find the path difference in terms of the optical

path.

Without the medium, the central maximum is at

the central line. (Path difference = 0)

Now the central maximum shifts to another position.

Find the central maximum.

The central maximum shifts through a distance

If the central line now has the mth bright fringe,

the central maximum has shifted through m fringes.

Multiple-slit

• More slits than two.

4 slits

http://bednorzmuller87.phys.cmu.edu/demonstrations/optics/interference/demo323.html

• N = number of slits.

• Compare N = 2 with N = 3

http://wug.physics.uiuc.edu/courses/phys114/spring01/Discussions/html/wk3/multiple/html/3-extra2.htm

• N = number of slits.

• Compare N = 2 with N = 3

• N = number of slits.

• Compare N = 2 with large N.

• N = 3

• Maximum occurs at a.sin = m. (same as N = 2).

Central maximum

1st order maximum

phase difference

= 0

a

a

phase difference

= 0

a

a

m= 1  a.sin = 

All three rays are in phase.

m= 0   = 0

All three rays are in phase.

• Textbook, p.238. Fig. 30.

• Position of maximum when N = 2 is still a maximum.

• There are peak(s) between two successive maxima.

• The number of peaks = N – 2.

• The intensity of peaks drops with N.

m = 0

m = 1

Multiple-slit (N = 3)

Why is there a peak between two maxima when N = 3?

• phase difference

= π

Δ1 =

Δ2 =λ

• phase difference

= 0

Multiple-slit (N = 3)

a

θ

θ

Δ1

a

Δ2

Add three rotating vectors for the resultant wave.

m = 1

m = 0

m = 1

Why are there 2 minima between two maxima when N = 3?

• phase difference

=

Multiple-slit (N = 3)

There is a minimum at position with

a

θ

θ

Δ2 =

• phase difference

=

Δ1

a

Δ2

Add three rotating vectors for the resultant wave.

Δ1 =

• phase difference

=

Δ2 =

• phase difference

=

Multiple-slit (N = 3)

a

θ

θ

Δ1

a

Δ2

Add three rotating vectors for the resultant wave.

• A diffraction grating is a piece of glass with many equidistant parallel lines.

• The mth order maximum is given by

a.sin = m.

m = 3

m = 2

m = 1

m = 0

m = 1

m = 2

m = 3

• The maximum order is given by

m = 3

m = 2

m = 1

m = 0

m = 1

m = 2

m = 3

• The separation between lines on a coarse grating is longer than that of a fine grating.

• Example of a coarse grating: 300 lines/cm.

• Example of a fine grating: 3000 lines/cm.

Find the maximum order of the above two gratings.

m = 3

a..sinm = m. 

The spatial angle mdepends on 

m = 2

m = 3

m = 2

m = 1

m = 1

white light

m = 0

m = 1

m = 1

m = 2

m = 3

m = 3

m = 2

• Monochromatic light = light with only one colour (frequency)

• Overlapping of colour spectrum

incident ray

no reflected ray

without coating

with coating

Blooming of lenses

• Coat a thin film on a lens to reduce the reflection of light.

reflected ray

film

glass

glass

• Ray A is reflected at the boundary between air and the film.

Reflected ray A

incident ray

with coating

• Ray B is reflected at the boundary between the thin film and the glass.

Reflected ray A

incident ray

Reflected ray B

with coating

• It is designed to have destructive interference for the reflected light rays.  No reflected light ray.

Reflected ray A

incident ray

Reflected ray B

with coating

• Suppose that the incident ray is normal to the lens.

• The reflected light rays are also along the normal.

reflected rays

incident ray

film

film

glass

glass

• Let n’ be the refractive index and t be the thickness of the thin film.

• Let  be the wavelength of the incident light.

incident ray

reflected rays

film

film

glass

glass

• To have destructive interference for the reflected rays,

incident ray

reflected rays

film

film

glass

glass

• Energy is conserved. As there is not any reflected light rays, the energy goes to the transmitted light ray.

No reflected rays

incident ray

film

film

glass

glass

transmitted ray

• The reflected ray from the bottom of the glass is so dim that it can be ignored.

This reflected ray

is ignored.

incident ray

film

film

glass

glass

• Limitation:

• For normal incident ray only.

• For wave of one particular wavelength only.

This reflected ray

is ignored.

incident ray

film

film

glass

glass

• Example 6

Find the minimum thickness of the thin film.

• Example 7

Find the wavelength.

Experimental setup

A normal incident ray is reflected at the boundary

between the slide and the air wedge.

reflected ray A

normal incident

ray

slide

air wedge

glass block

normal incident

ray

reflected ray B

air wedge

t

Air Wedge

The normal incident ray goes into the wedge, passing

through a distance t and reflected at the boundary between

the air wedge and the glass block.

slide

glass block

normal incident

ray

reflected ray B

air wedge

t

Air Wedge

The ray into the glass block is ignored.

slide

glass block

The ray reflected at the top of the slide is ignored.

normal incident

ray

slide

air wedge

glass block

• Depending on the the distance t and the wavelength  of the incident wave, the two reflected rays may have interference.

• The pattern is a series of bright and dark fringes when we view through the travelling microscope.

normal incident

ray

reflected ray B

air wedge

t

Air Wedge

To produce bright fringes, 2.t = (m - )., m = 1, 2, 3,…

constructive

interference

slide

glass block

normal incident

ray

reflected ray B

air wedge

t

Air Wedge

• Ray B has a phase change  on reflection.

• The path difference must be m.

constructive

interference

slide

glass block

normal incident

ray

reflected ray B

air wedge

t

Air Wedge

To produce dark fringes, 2.t = m ., m = 0, 1, 2,…

destructive

interference

slide

Note that ray B has a phase change on reflection.

glass block

normal incident

ray

reflected ray B

air wedge

t

Air Wedge

• Ray B has a phase change  on reflection.

• The path difference must be (m + ).

destructive

interference

slide

Note that ray B has a phase change on reflection.

glass block

normal incident

ray

reflected ray B

air wedge

t

Air Wedge

• At the vertex, t = 0  m = 0  dark fringe.

destructive

interference

slide

Note that ray B has a phase change on reflection.

glass block

To find the separation s between two successive

bright fringes

Let D be the height of the high end of the slide.

Let L be the length of the slide.

(N+1)th fringe

Nth fringe

slide

L

D

air wedge

tN+1

tN

s

glass block

To find the separation s between two successive

bright fringes

Angle of inclination of the slide can be found from

(N+1)th fringe

Nth fringe

slide

L

D

air wedge

tN+1

tN

s

glass block

tN+1

tN

s

Air Wedge

Separation s between two successive bright fringes

tN+1 – tN =

tN+1

tN

s

Air Wedge

Separation s between two successive bright fringes

For small angle , sin   tan 

and

• For flat surfaces of glass block and slide

• the fringes are parallel and evenly spaced.

• For flat surface of glass block and slide with surface curved upwards

• the fringes are parallel and become more closely packed at higher orders.

• For flat surface of glass block and slide with surface curved downwards

• the fringes are parallel and become more widely separated at higher orders.

• We can use this method to check a flat glass surface.

The surface is not flat.

The surface is flat.

air wedge

Measuring the Diameter D of a Wire

Measure the quantities on the right

hand side and calculate D.

L

m = 0

L

D

19.s

Example 8

• There are 20 dark fringes  m = 19.

• L  19.s  s =  D =

• Why soap film is coloured?

http://www.cs.utah.edu/~zhukov/applets/film/applet.html

The two transmitted rays A and B may have interference

depending on the thickness t of the film and the wavelength

.

ray A

t

incident ray

ray B

soap

water

Note that ray B has two reflections. No phase change is due

to reflection.

ray A

t

incident ray

ray B

To observe bright fringes, the path difference = m.

 2.t = m. 

ray A

t

constructive

interference

incident ray

ray B

To observe dark fringes, the path difference = (m+ ).

 2.t = (m+ ). 

ray A

t

constructive

interference

incident ray

ray B

t

ray A

ray B

soap

water

Soap film

The two reflected rays A and B may have interference

depending on the thickness t of the film and the wavelength

.

t

ray A

ray B

soap

water

Soap film

Note that this time there is a phase change due to reflection.

t

ray A

ray B

soap

water

Soap film

Find out how the interference depends on t and 

interference

• As the interference depends on , there will be a colour band for white incident light.

• In each colour band, violet is at the top and red is at the bottom.

• Soap water tends to move downwards due to gravity.

• The soap film has a thin vertex and a thick base. The fringes are not evenly spaced.

• The fringes are dense near the bottom and less dense near the vertex.

C

D

C

D

C

D

C

D

Soap film

• The pattern of the reflected rays and that of the transmitted rays are complementary.

C: constructive

interference

D: destructive

interference

transmitted rays

reflected rays

• The reflected rays have constructive interference.

• Note that there is a phase change on reflection.

Experimental setup

• What do we see through the travelling microscope with white incident light?

• If we use red incident light,

http://www.cs.utah.edu/~zhukov/applets/film/applet.html

The two reflected rays have interference depending on

the thickness t of the air gap and the wavelength .

reflected ray A

interference

incident ray

reflected ray B

lens

air

t = thickness of air gap

t

glass block

For bright fringes,

reflected ray A

interference

incident ray

reflected ray B

lens

air

t = thickness of air gap

t

glass block

For dark fringes,

reflected ray A

interference

incident ray

reflected ray B

lens

air

t = thickness of air gap

t

glass block

At the center of the lens, there is a dark spot.

reflected ray A

interference

incident ray

reflected ray B

lens

air

t = thickness of air gap

t

glass block

• The spacing of the rings are not even.

• Near the center, the rings are widely separated.

• Near the edge, the rings are close together.

• Find the radius of the mth dark ring.

Rm

C

R = radius of curvature of the lens

lens

B

A

Rm

air

t

t

t = thickness of air gap

O

glass block

C

for the mth dark fringe

R = radius of curvature of the lens

lens

B

A

Rm

air

t

t

t = thickness of air gap

O

glass block

C

R = radius of curvature of the lens

lens

B

A

Rm

air

t

t

t = thickness of air gap

O

glass block

s = Rm+1 – Rm =

Newton’s rings

• Separation between two successive rings

The separation approaches zero for high orders.

• To find the radius of curvature of a lens by Newton’s rings.

• If there is distortion of the Newton’s rings, the lens is not a good one.

ray

n

t

thin film

Thin films

Light is incident obliquely onto a thin film of refractive

index n and thickness t.

ray

n

t

thin film

Thin films

The reflected rays have interference depending on the angle

of view  and wavelength .

coloured

spectrum

incident white light

http://www.cs.utah.edu/~zhukov/applets/film/applet.html

single slit

Diffraction of Light

screen

http://surendranath.tripod.com/SnglSlt/SnglSltApp.html

• Intensity variation

http://arborsci.com/Oscillations_Waves/Diffraction.htm

• Consider a light source which is far away from the single slit and the light is normal to the slit.

• In general, the position of the mth dark fringe due to a slit of width d is given by

d.sinm = m.

http://www-optics.unine.ch/research/microoptics/RigDiffraction/aper/aper.html

• Variation of intensity

A1

1

1

d

A2

Theory of diffraction

Formation of 1st order dark fringe.

Divide the slit into two equal sections A1 and A2.

The light from section A1 cancels the light from section A2

Prove that

Formation of 2nd order dark fringe.

Divide the slit into 4 equal sections A1 , A2, A3 and A4.

The light from section A1 cancels the light from section A2.

The light from section A3 cancels the light from section A4

Prove that

A1

2

A2

d

A3

A4

Formation of 2nd order dark fringe.

In general for the mth dark fringe,

A1

2

A2

d

A3

A4

• Consider a light source which is far away from the single slit and the light is normal to the slit.

• For a circular hole with diameter d, the center is a bright spot and the 1st dark ring is given by