Double Slit is VERY IMPORTANT because it is evidence of waves. Only waves interfere like this.

Superposition of Sinusoidal Waves

• Assume two waves are traveling in the same direction, with the same frequency, wavelength and amplitude

• The waves differ in phase

• y1 = A sin (kx - ωt)• y2 = A sin (kx - ωt + φ)• y = y1+y2

= 2A cos (φ/2) sin (kx - ωt + φ/2)

Resultant Amplitude Depends on phase: Spatial Interference Term

1-D Wave Interferencey = y1+y2 = 2A cos (φ/2) sin (kx - ωt + φ/2)

• The resultant wave function, y, is also sinusoidal• The resultant wave has the same frequency and

wavelength as the original waves• The amplitude of the resultant wave is 2A cos (φ/2) • The phase of the resultant wave is φ/2

Constructive Destructive Interference

0φ = φφ π=

1-D Wave Interferencey = y1+y2 = 2A cos (φ/2) sin (kx - ωt + φ/2)

Constructive Destructive Interference

0φ = φφ π=

( )

Resultant Amplitude: 2 cos2

Constructive Interference (Even ): 2 , 0,1, 2,3...Destructive Interference Odd : (2 1) , 0,1,2,3...

A

n nn n

φ

π φ ππ φ π

⎛ ⎞⎜ ⎟⎝ ⎠

Δ = =

Δ = + =

1-D Sound Wave Interference

2-D Wave Interference?

P

02Phase Difference at P:

Amplitude at P: 2Acos( )2

rπφ φλ

φ

Δ = Δ +

Δ

2-D Phase Difference at P: is different from the phase difference between the two source waves!It depends on the path difference traveled by the two waves!

φφ

Δ

Constructive or Destructive?(Indentical in phase sources)

P

Resultant Amplitude: 2 cos2

Constructive Interference: , 2 , 0,1, 2,3...

Destructive Interference: (2 1) , (2 1) , 0,1,2,3...2

A

r n n n

r n n n

φ

λ φ πλ φ π

Δ⎛ ⎞⎜ ⎟⎝ ⎠

Δ = Δ = =

Δ = + Δ = + =

2 (1 ) 2

!Constructive

πφ λ πλ

Δ = =

02Phase Difference at P: rπφ φλ

Δ = Δ +

Constructive or Destructive?(Source out of Phase by 180 degrees)

2 (1 ) 3

!Destructive

πφ λ π πλ

Δ = + =

P

Resultant Amplitude: 2 cos2

Constructive Interference: , 2 , 0,1, 2,3...

Destructive Interference: (2 1) , (2 1) , 0,1,2,3...2

A

r n n n

r n n n

φ

λ φ πλ φ π

Δ⎛ ⎞⎜ ⎟⎝ ⎠

Δ = Δ = =

Δ = + Δ = + =

02Phase Difference at P: rπφ φλ

Δ = Δ +

Constructive or Destructive?(Indentical in phase sources)

P

02Phase Difference at P: rπφ φλ

Δ = Δ +

Constructive or Destructive?(Sources out of phase)

P

02Phase Difference at P: rπφ φλ

Δ = Δ +

Exam Extra Credit

02Phase Difference at P: rπφ φλ

Δ = Δ +

Exam Extra Credit

I ∝ 2A

0

1 2

2Phase Difference at P:

2 cos( / 2)sin( / 2)

Amplitude at P: 2 cos( )2

r

y y y A kx t

A

πφ φλ

φ ω φφ

Δ = Δ +

= + = Δ − + ΔΔ

2max cos ( / 2)I I φ= Δ

In Phase or Out of Phase?Constructive? Destructive?

BA

Loud Max Quiet Min

Quiet MinLoud Max

Light Waves: Coherent Sources

Intensity

02Phase Difference at P: , 0rπφ φλ

Δ = Δ Δ =

Interference of 2 Light Sources

Conditions for InterferenceUse a Double Slit!

• To observe interference in light waves, the following two conditions must be met:1) The sources must be

coherent• They must maintain a

constant phase with respect to each other

2) The sources should be monochromatic

• Monochromatic means they have a single wavelength

Interference Patterns

• Constructive interference occurs at point P

• The two waves travel the same distance– Therefore, they arrive in

phase

• As a result, constructive interference occurs at this point and a bright fringe is observed

Interference Patterns• The upper wave has

to travel farther than the lower wave to reach point Q

• The upper wave travels one wavelength farther– Therefore, the waves

arrive in phase• A second bright fringe

occurs at this position

Interference Patterns• The upper wave travels

one-half of a wavelength farther than the lower wave to reach point R

• The trough of the bottom wave overlaps the crest of the upper wave

• This is destructive interference– A dark fringe occurs

Interference of 2 Light Sources

Double Slit InterferenceHyperphysics Website

Light Intensity

• The interference pattern consists of equally spaced fringes of equal intensity

• This result is valid only if L >> d and for small values of θ

2 2max max

sin cos cosI I Iπd θ πd yλ λL

⎛ ⎞ ⎛ ⎞= ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Reality

Double-Slit

Young’s Double-Slit Experiment: Geometry

• The path difference, δ, is found from the tan triangle

• δ = r2 – r1 = d sin θ– This assumes the paths

are parallel, L>>d– Not exactly true, but a

very good approximation if L is much greater than d

2 2 sin π πφ δ d θλ λ

= =

rΔ

Interference AssumptionsIf L>>d, then the rays are approximated parrallel, and the pink triangle is a right triangle and the angles are equal as shown and the path difference is:

, sin>> Δ =If L d then r d θ

Constructive: , 0,1, 2,3,...r m mλΔ = =

1Destructive: ( ) , 0,1, 2,3,...2

r m mλΔ = + =

Interference Conditions

Bright fringe: sin , 0,1, 2,3,...r d m mθ λΔ = = =

1Dark fringe : sin ( ) 0,1, 2,3,...2

r d m mθ λΔ = = + =

rΔ

Interference Fringes

Derive Fringe Equations

• For bright fringes

• For dark fringes

bright ( 0 1 2 ), ,λLy m md

= = ± ± K

dark1 ( 0 1 2 )2

, ,λLy m md⎛ ⎞= + = ± ±⎜ ⎟⎝ ⎠

K

Red light (λ=664nm) is used in Young’s double slit as shown. Find the distance y on the screen between the central bright fringe and the third order bright fringe.Find the width of the central bright maxima.

Problem

bright ( 0 1 2 ), ,λLy m md

= = ± ± K

You Try 2 Slit

The image shows the light intensity on a screen behind a double slit. The slit spacing is 0.20 mm and the wavelength of light is 600 nm. What is the distance from the slits to the screen?

bright ( 0 1 2 ), ,λLy m md

= = ± ± K

Double Slit InterferenceDependence on Slit Separation

bright ( 0 1 2 ), ,λLy m md

= = ± ± K

http://web.phys.ksu.edu/vqmorig/programs/java/makewave/Slit/vq_mws.htm

Light Waves

E = Emax cos (kx – ωt) B = Bmax cos (kx – ωt)

max

max

E ω E cB k B

= = =

2 2max max max max

av 2 2 2I

o o o

E B E c BSμ μ c μ

= = = = I ∝ 2maxE

Double Slit InterferenceHyperphysics Website

Intensity DistributionResultant Field

• The magnitude of the resultant electric field comes from the superposition principle– EP = E1+ E2 = Eo[sin ωt + sin (ωt + φ)]

• This can also be expressed as

– EP has the same frequency as the light at the slits– The amplitude at P is given by 2Eo cos (φ / 2)

• Intensity is proportional to the square of the amplitude:

2 cos sin2 2P oφ φE E ωt⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

I ∝ 2A 2max cos ( / 2)I I φ= Δ

Light Intensity

• The interference pattern consists of equally spaced fringes of equal intensity

• This result is valid only if L >> d and for small values of θ

2 2max max

sin cos cosI I Iπd θ πd yλ λL

⎛ ⎞ ⎛ ⎞= ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

HO 2 Slit

You TryIn a double-slit experiment, the distance between the slits is 0.2 mm, and the distance to the screen is 150 cm. What wavelength (in nm) is needed to have the intensity at a point 1 mm from the central maximum on the screen be 80% of the maximum intensity?

a.900b.700c.500d.300e.600

Thin Film Interference

RGB Color Theory

Additive Complementary ColorsYellow, Cyan, Magenta

The color you have to add to get white light.

Red + Green = YellowBlue + Green = CyanRed + Blue = Magenta

Red + Blue + Green = White

White light – yellow light = ??

White light – red light = ??

The Index of Refraction• Refraction: Light Bends in

Transmission• The speed of light in any

material is less than its speed in vacuum

• The index of refraction, n, of a medium can be defined as

speed of light in a vacuum cnspeed of light in a medium v

≡ =

in vacuumin a mediumn

λ λnλ λ

⎛ ⎞= ⎜ ⎟

⎝ ⎠

• For a vacuum, n = 1– We assume n = 1 for

air also• For other media, n > 1• n is a dimensionless

number greater than unity, not necessarily an integer

Some Indices of Refraction

Interference in Thin Films• Assume the light rays are

traveling in air nearly normal to the two surfaces of the film

• Ray 1 undergoes a phase change of 180° with respect to the incident ray

• Ray 2, which is reflected from the lower surface, undergoes no phase change with respect to the incident wave

Thin FilmsWhen reflecting off a medium of greater refractive index, a

light wave undergoes a phase shift of ½ a wavelength.Wave 1 undergoes a phase shift of 180 degrees.

Phase Changes Due To Reflection

• An electromagnetic wave undergoes a phase change of 180°upon reflection from a medium of higher index of refraction than the one in which it was traveling– Analogous to a pulse

on a string reflected from a rigid support

Phase Changes Due To Reflection

• There is no phase change when the wave is reflected from a boundary leading to a medium of lower index of refraction– Analogous to a

pulse on a string reflecting from a free support

Problem Solving with Thin Films• Phase differences have two causes

– differences in the distances traveled– phase changes occurring on

reflection• Both causes must be considered

when determining constructive or destructive interference

• The interference is constructive if the path difference is an integral multiple of λ and destructive if the path difference is an odd half multiple of λ

A thin film of gasoline floats on a puddle of water. Sunlight falls almost

perpendicularly on the film and reflects into your eyes a yellow hue. Interference

in the the thin gasoline film has eliminated blue (469nm in vacuum) from the

reflected light. The refractive indices of the blue light in gasoline and water are

1.40 and 1.33 respectively.Determine the minimum nonzero

thickness of the film.

Newton’s RingsInterference Pattern

Newton’s Rings

• Another method for viewing interference is to place a plano-convex lens on top of a flat glass surface

• The air film between the glass surfaces varies in thickness from zero at the point of contact to some thickness t

• A pattern of light and dark rings is observed– These rings are called Newton’s rings– The particle model of light could not explain the origin

of the rings• Newton’s rings can be used to test optical lenses

Michelson Interferometer• A ray of light is split into two rays by

the mirror Mo– The mirror is at 45o to the incident

beam– The mirror is called a beam splitter

• It transmits half the light and reflects the rest

• After reflecting from M1 and M2, the rays eventually recombine at Moand form an interference pattern

• The fringe pattern shifts by one-half fringe each time M1 is moved a distance λ/4

• The wavelength of the light is then measured by counting the number of fringe shifts for a given displacement of M1

Michelson Interferometer

The fringe pattern shifts by one-half fringe each time M1 is moved a distance λ/4

http://www.youtube.com/watch?v=ETLG5SLFMZohttp://www.youtube.com/watch?v=Z8K3gcHQiqk&feature=related

Michelson Interferometer –Applications

• The Michelson interferometer was used to disprove the idea that the Earth moves through an ether

• Modern applications include– Fourier Transform Infrared Spectroscopy

(FTIR)– Laser Interferometer Gravitational-Wave

Observatory (LIGO)

James Clerk Maxwell1860s

Light is wave. The medium is the Ether.

8

0

1 3.0 10 /o

c x m sμ ε

= =

Measure the Speed of the Ether Wind

The Luminiferous Aether was imagined by physicists since Isaac Newton as the invisible "vapor" or "gas aether" filling the universe and hence as the carrier of heat and light.

http://www.youtube.com/watch?v=XavC4w_Y9b8&feature=relatedhttp://www.youtube.com/watch?v=4KFMeKJySwA&feature=related

Rotate arms to produce interference fringes and find different speeds of light.

Michelson-MorelyExperiment

1887The speed of light is independent of the motion and

is always c. The speed of the Ether wind is zero. OR….

Lorentz ContractionThe apparatus shrinks by a factor :

2 21 /− v c

On the Electrodynamics of Moving Bodies1905

Clocks slow downand rulers shrink

in order to keep thespeed of light the

same for all observers!

Time is Relative!Space is Relative!Only the SPEED

OF LIGHT isAbsolute!

As you approach c, lengths contract.

LIGO in Richland, Washington

http://www.youtube.com/watch?v=RzZgFKoIfQI&feature=related

LISA

http://www.youtube.com/watch?v=DrWwWcA_Hgw&feature=related

Double Slit for Electronsshows Wave Interference

If electron were hard bullets, there would be no interference pattern.

In reality, electrons do show an interference pattern, like light waves.

Electrons act like waves going through the slits but arrive at the detector like a particle.

Interference pattern builds one electron at a time.

Electrons act like waves going through the slits but arrive at the detector like a particle.

Particle Wave Duality

Particle Picture: Trying to detect which slit the electron went through

destroys the wave behavior

Feynman version of theUncertainty Principle

It is impossible to design an apparatus to determine which hole the electron passes through, that will not at the

same time disturb the electrons enough to destroy the interference pattern.

-Richard Feynman