physics beyond 2000

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 × 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/

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

Electromagnetic 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.

Radio Waves• 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.

direction of propagation ofradio wave

oscillating a.c.

oscillating electric fieldtransmitter

Radio Waves• Receiver:

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

direction of propagation ofradio wave

oscillating a.c.

oscillating electric field

transmitter receiver

Spectrum of Radio WavesRadio waves Wavelength

Long waves 1 km – 10 km

Medium waves 100 m – 1 km

Short waves 10 m – 100 m

Very high frequency (VHF)

1 m – 10 m

Ultra high frequency

(UHF)

0.1 m – 1 m

Microwaves

• Microwave is polarized along the length of the dipole.

direction of propagation ofmicrowave

oscillating a.c.



Microwaves

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


oscillating a.c.



metal rod

no response

Microwaves

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


oscillating a.c.


transmitter receivermetal rod

Interference of Microwaves• At P, the wave from the transmitter meets

the reflected wave. Interference occurs.

P

transmitter

imageof transmitter

metal plate

Interference of Microwaves• We may consider it as an interference from two

coherent sources, the transmitter and its image.

P

transmitter

imageof transmitter

metal plate

Interference of Microwaves• The two sources are in anti-phase because

there is a phase change of on reflection.

P

transmitter

imageof transmitter

metal plate

Interference of Microwaves• If the path difference at P = n., there is

destructive interference.

P

transmitter

imageof transmitter

metal plate

Interference of Microwaves

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

P

transmitter

imageof transmitter

metal plate

)2

1( n

Microwave Cooking

• 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

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

Microwave in satellite communications

• Reading the following

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

http://www.s-t.au.ac.th/~supoet/satel.htm#Slide%201

Infrared Radiation

• Self-reading.

Ultraviolet Radiation

• Self-reading.

Visible light

• Self-reading.

Colored Video Pictures

• Self-reading.

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.

incident light

atomoscillating E-field

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.

atom

scattered light

axis along which the atom oscillating

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.

atom

scattered light

axis along which the atom oscillating

strongeststrongest

Why is the sky blue at noon and red at sunrise and sunset?

• Why is the sky blue in daytime?

• http://physics.about.com/science/physics/library/weekly/aa051600a.htm

• Why is the sky red in sunset/sunrise?

• http://physics.about.com/science/physics/library/weekly/aa052300a.htm

http://physics.about.com/science/physics/library/weekly/aa051600a.htm





• 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 fromthe sun

Blue lightis most scattered

Red light is least scattered


• At sunset, we see the least scattered light.

• Red light is least scattered. white light fromthe sun

Red lightis least scattered

Blue light is most scattered

Polarization of light• Light is transverse wave so it exhibits polari

zation.• Unpolarized light: the electric field is not co

nfined to oscillate in a plane.• Plane-polarized light: the electric field at ev

ery point oscillates in the same fixed plane.• Plane of polarization: the plane in which the

electric field of a plane polarized light oscillates.

Polarization of Light

• Plane polarized light:

• Unpolarised light:

electric vector

electric vector

Polarization by Absorption • An array of parallel conducting wires.

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

Conducting wires are verticalPlane-polarized microwave

No microwave

E-field is vertial.


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

Conducting wires are verticalPlane-polarized microwave

E-field is horizontal Plane-polarized microwave


• It can be a polarizer of microwaves

Conducting wires are verticalUnpolarized microwave

Plane-polarized microwave

Polarization by Absorption • 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.

E-field is vertical

Chains of molecules are vertical

No light

Plane-polarized light

Polarization by Absorption

• 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.

E-field is horizontal

Chains of molecules are verticalPlane-polarized light


Polarization by Absorption • 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 verticalUnpolarized light


Polarization by Reflection

plane-polarizedincident light

plane-polarizedrefracted light

No reflected light

air

glass

Assume that the direction of the reflected light and that of the refracted light are perpendicular.


plane-polarizedincident light

plane-polarizedrefracted light

No reflected light

air

glass

•The electric field sets theelectrons in the glass tooscillate at right angles tothe refracted ray.•The intensity perpendicularto the axis of oscillation isstrongest The refracted ray is bright.•The intensity parallel to the axis of oscillation is zero no reflected ray.


Unpolarizedincident light

Unpolarizedrefracted light

Polarized reflected light

air

glass

The plane of polarizationis parallel to the surface of medium.

Assume that the direction of the reflected light and that of the refracted light are perpendicular.

The Brewster’s Angle

p = Brewster’s angler = Angle of refraction.

p p

r

incidentray

reflected ray is completely polarized

refractedray

air

medium

The Brewster’s Angle

• n = tan p where n is the refractive index of the medium.

p p

r

incidentray

reflected ray

refractedray

air

medium

Prove it!

Example 1

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

Polarization by Scattering

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

The atom absorbsthe wave energy.

incident ray

atom

The atom re-radiates the wave energy.


vertically polarizedlight

water mixed with milk



vertically polarizedlightno scattered light

no scattered light


horizontally polarizedlight





no scattered light

no scattered light


unpolarizedlight


unpolarizedlight





Polaroid Sunglasses

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

Study p.233 of the textbook.

Interference of Light

• Light is a kind of wave.

• Interference is a wave property.

Interference of LightConditions 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.

Interference of Light

• Young’s double-slit experiment

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

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

•The incident ray is split into two coherent sourcesS1 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.



Young’s double-slit experiment

a

P

S1

S2

central line

Suppose that there is a maximum at point P.A constructive interference occurs at P.

screenD


a

parallel rays meet at point P

As point P is far away from the double slit,the light rays of the same fringe are parallel.

The path difference = a.sin

S1

S2central line


a

parallel rays meet at point P withmaximum intensity.


S1

S2central line

= a.sin = m. where m = 0, 1, 2,… m is the order of the fringes.

For points with constructive interference,


a

parallel rays meet at a point withminimumintensity.


S1

S2central line

= a.sin = (m + ). where m = 0, 1, 2,…

For points with destructive interference,

2

1

separation between the fringes

a

P

S1

S2

central line

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 slitand the screen is D.

ym

M

D


a

P

S1

S2

central line

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

ym

M

D

ym = D.tan D.sin = Da

m.


ym = Da

m.

(1)

By similar consideration, for the m+1 bright fringe

ym+1= Da

m.

)1( (2)

The separation between the two fringes is

s = ym+1 – ym = a

D


• 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.

s = ym+1 – ym = a

D

a

1

Variation of intensity• If the slits are sufficiently narrow, light spreads

out evenly from each slit and the bright fringes are equally bright.

Variation of intensity

• 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.

Variation of intensity

• 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

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

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

Variation of intensity• The intensity of the fringes is enclosed in an

envelope as shown.

Example 2

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

White light fringes

s =a

D s

Separation of violet fringes is shortest.Separation of red fringes is longest.





Submerging in a Liquid• If the Young’s double-slit experiment is

done in a liquid, what would happen to the separation of fringes?

Liquid with refractive index n

ym

Submerging in a Liquid

• The wavelength changes!

Let be the wavelength in vacuum/airand n the wavelength in liquid.Let n be the refractive index of the liquid.

nn

The fringe separation in liquid

sn =

n

s

Submerging in a Liquid

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

nn

The fringe separation in liquid

sn =

n

s

Optical path

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

medium ofrefractive index

nIncidentlight

vacuum

thickness = t

thickness = optical path

Light requires the sametime to travel throughthe two paths.

Optical path

• Show that the optical path = n.t


nIncidentlight

vacuum

thickness = t



Optical path

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


nIncidentlight

vacuum

thickness = t



Example 3

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

Shifting a System of Fringes

Note that ray A has passes through a medium of refractiveindex n and thickness t.We need to find the path difference in terms of the optical path.

A

B


Without the medium, the central maximum is atthe central line. (Path difference = 0)Now the central maximum shifts to another position.Find the central maximum.

The central maximum shifts through a distance

a

tDny

)1(


If the central line now has the mth bright fringe,the central maximum has shifted through m fringes.

tn

m)1(

Multiple-slit

• More slits than two.

3 slits

4 slits

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

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

Multiple-slit

• 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

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

Multiple-slit• N = number of slits.• Compare N = 2 with N = 3

N = 2 N = 3Maximum occurs at positions with a.sin = m.

Maximum occurs at positions with a.sin = m.

The intensity at the maximum is 4.Io


Between two maxima, it is a minimum.

Between two maxima, it is a peak.

The width of bright fringes is large.

The width of bright fringes is less

Multiple-slit• N = number of slits.• Compare N = 2 with large N.

N = 2 Large N Maximum occurs at positions with a.sin = m.

Maximum occurs at positions with a.sin = m.


The intensity at the maximum is N2.Io

Between two maxima, it is a minimum.

Between two maxima, there are (N-2) peaks. The intensity of the peak is almost zero

The width of bright fringes is very narrow

Multiple-slit• N = 3• Maximum occurs at a.sin = m. (same as N = 2).

Central maximum 1st order maximum

m= 0 = 0All three rays are in phase.

m= 1 a.sin = All three rays are in phase.

a

a

a

a

phase difference= 0

phase difference= 0

Multiple-slit

• 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.

Multiple-slit (N = 3)m = 0m = 1 m = 1

0sin a

sina

sin

a2sin

a2

sin

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

Multiple-slit (N = 3)

a2sin

There is a peak at position with

a

a

θθ

Δ1

Δ2

Δ1 = 2

phase difference = π

Δ2 =λ phase difference = 0

Add three rotating vectors for the resultant wave.


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

m = 0m = 1 m = 1

0sin a

sina3

2sin

a2sin

a2

sin

a3sin


a3sin

There is a minimum at position with

a

a

θθ

Δ1

Δ2

3


Δ1 = phase difference

= 3

2

Δ2 = phase difference =

3

43

2


a3

2sin

There is a minimum at position with

a

a

θθ

Δ1

Δ2


3

2Δ1 = phase difference

= 3

4

Δ2 = phase difference =

3

83

4

Diffraction grating

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

Diffraction grating• The mth order maximum is given by

a.sin = m.

m = 0

m = 1

m = 1

m = 2

m = 2

m = 3

m = 3

Diffraction grating• The maximum order is given by

m = 0

m = 1

m = 1

m = 2

m = 2

m = 3

m = 3

a

m max

Intensity of diffraction grating

coarse grating and fine grating

• 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.

Colour Spectrum from a Diffraction Grating

m = 0

m = 1

m = 1

m = 2

m = 2

m = 3

m = 3

m = 1

m = 2

m = 3

white light

m = 1

m = 2

m = 3

a..sinm = m. The spatial angle m depends on

Example 4

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

Example 5

• Overlapping of colour spectrum

Blooming of lenses

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

without coating

incident rayreflected ray

with coating

incident ray

no reflected ray

glass glassfilm

Blooming of lenses

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

with coating

incident rayReflected ray A

Blooming of lenses

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

with coating


Reflected ray B

Blooming of lenses

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

with coating


Reflected ray B

Blooming of lenses• Suppose that the incident ray is normal to

the lens.• The reflected light rays are also along the

normal.

incident ray reflected rays

film

glass glass

film

Blooming of lenses• Let n’ be the refractive index and t be the

thickness of the thin film.

• Let be the wavelength of the incident light.


film

glass glass

film

Blooming of lenses• To have destructive interference for the

reflected rays,


film

glass glass

film

'4min nt

Blooming of lenses• Energy is conserved. As there is not any

reflected light rays, the energy goes to the transmitted light ray.

incident ray No reflected rays

film

glass glass

film

transmitted ray

Blooming of lenses• The reflected ray from the bottom of the

glass is so dim that it can be ignored.

incident ray

This reflected rayis ignored.

film

glass glass

film

Blooming of lenses• Limitation:

• For normal incident ray only.

• For wave of one particular wavelength only.

incident ray

This reflected rayis ignored.

film

glass glass

film

Examples

• Example 6

Find the minimum thickness of the thin film.

• Example 7

Find the wavelength.

Air Wedge:Experimental setup

Air Wedge

reflected ray A

normal incidentray

air wedge

A normal incident ray is reflected at the boundarybetween the slide and the air wedge.

slide

glass block

Air Wedge

reflected ray A

reflected ray Bnormal incidentray

air wedget

The normal incident ray goes into the wedge, passingthrough a distance t and reflected at the boundary betweenthe air wedge and the glass block.

slide

glass block

Air Wedge

reflected ray A


air wedget

The ray into the glass block is ignored.

slide

glass block

Air Wedge

normal incidentray

air wedge

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

slide

glass block

Air Wedge

• 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.

Air Wedge

reflected ray A


air wedget

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

slide

glass block

2

1

constructiveinterference

Air Wedge

reflected ray A


air wedget

slide

glass block

constructiveinterference

•Ray B has a phase change on reflection.•The path difference must be m.

Air Wedge

reflected ray A


air wedget

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

slide

glass block

destructiveinterference

Note that ray B has a phase change on reflection.

Air Wedge

reflected ray A


air wedget

slide

glass block



•Ray B has a phase change on reflection.•The path difference must be (m + ).

2

1

Air Wedge

reflected ray A


air wedget

slide

glass block



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

Air WedgeTo find the separation s between two successive bright fringes

air wedgetN

slide

glass block

Nth fringe (N+1)th fringe

stN+1

DL

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

Air WedgeTo find the separation s between two successive bright fringes

air wedgetN

slide

glass block

Nth fringe (N+1)th fringe

stN+1

DL

Angle of inclination of the slide can be found from

L

Dsin

Air Wedge

Separation s between two successive bright fringes

tN+1 – tN = 2

tN

tN+1

s

ss .2

2tan

Air Wedge

Separation s between two successive bright fringes

tN

tN+1

s

s.2tan

L

Dsin

and

For small angle , sin tan

D

Ls

.2

.

Air Wedge

• For flat surfaces of glass block and slide– the fringes are parallel and evenly spaced.

Air Wedge

• For flat surface of glass block and slide with surface curved upwards– the fringes are parallel and become more

closely packed at higher orders.

Air Wedge• For flat surface of glass block and slide with

surface curved downwards– the fringes are parallel and become more

widely separated at higher orders.

Air Wedge• We can use this method to check a flat glass

surface.

The surface is flat. The surface is not flat.

Measuring the Diameter D of a Wire

air wedgeD

s

LD

.2

. Measure the quantities on the right

hand side and calculate D.

L

Example 8

• There are 20 dark fringes m = 19.

• L 19.s s = D = 2

.19

m = 0m = 19

DL

19.s

19

L

Soap film

• Why soap film is coloured?

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


Soap film

ray A

ray B

incident ray

The two transmitted rays A and B may have interferencedepending on the thickness t of the film and the wavelength.

t

soapwater

Soap film

ray A

ray B

incident ray

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

t

Soap film

ray A

ray B

incident ray

To observe bright fringes, the path difference = m. 2.t = m.

tconstructiveinterference

Soap film

ray A

ray B

incident ray

To observe dark fringes, the path difference = (m+ ). 2.t = (m+ ).

tconstructiveinterference

2

1

2

1

Soap filmThe two reflected rays A and B may have interferencedepending on the thickness t of the film and the wavelength.

ray A

ray B

incident rayt

soapwater

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

ray A

ray B

incident rayt

soapwater

Soap film

Find out how the interference depends on t and

ray A

ray B

incident rayt

soapwater

interference

Soap film

• 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 film • 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.

Soap film

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

incident ray

C

C

C

C

D

D

D

D

C: constructiveinterferenceD: destructive interference

reflected rays transmitted rays

Example 9

• The reflected rays have constructive interference.

• Note that there is a phase change on reflection.

Newton’s rings:Experimental setup

Newton’s rings• What do we see through the travelling micr

oscope with white incident light?

Newton’s rings

• If we use red incident light,



Newton’s rings

reflected ray A

air

lens

glass block

reflected ray Bincident ray

tt = thickness of air gap

interference

The two reflected rays have interference depending onthe thickness t of the air gap and the wavelength .

Newton’s rings

reflected ray A

air

lens

glass block



interference

For bright fringes, 4

).12(

mt

Newton’s rings

reflected ray A

air

lens

glass block



interference

For dark fringes, 2

.mt

Newton’s rings

reflected ray A

air

lens

glass block



interference

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

Newton’s rings

• The spacing of the rings are not even.

• Near the center, the rings are widely separated.

• Near the edge, the rings are close together.

Newton’s rings

• Find the radius of the mth dark ring.

Rm

Newton’s rings

air

lens

glass block

t t = thickness of air gap

R = radius of curvature of the lens

Rm

C

O

A B

tRtRRRm 2)( 22

t

Newton’s rings

air

lens

glass block



Rm

C

O

A B

t

tRRm 2

2

mt for the mth dark fringe

Newton’s rings

air

lens

glass block



Rm

C

O

A B

t

mRRm

Newton’s rings

• Separation between two successive ringss = Rm+1 – Rm = Rmm ).1(

The separation approaches zero for high orders.

Example 10

• 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.

Thin filmsLight is incident obliquely onto a thin film of refractive index n and thickness t.

thin filmtn

incidentray

Thin films

The reflected rays have interference depending on the angleof view and wavelength .

thin filmtn

incidentray

Thin film

incident white lightcolouredspectrum



Diffraction of Light

single slitlaser tube

screen

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

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


• Intensity variation

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

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.sinm = m.

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

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


• Variation of intensity

• http://www.fed.cuhk.edu.hk/sci_lab/download/project/javapm/java/slitdiffr/Default.htm

http://www.fed.cuhk.edu.hk/sci_lab/download/project/javapm/java/slitdiffr/Default.htm



Theory of diffractionFormation 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

d

A1

A2

1

1

Prove that d

1sin

Theory of diffractionFormation 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

d

A1

A2

Prove that d

2sin 2

A3

A4

2

Theory of diffractionFormation of 2nd order dark fringe.

d

A1

A2

d

mm

sin

A3

A4

2

In general for the mth dark fringe,

Diffraction of Light• 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

d

22.1sin 1

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