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Double Slit is VERY IMPORTANT because it is evidence of . Only waves interfere like this. Superposition of Sinusoidal Waves • Assume two waves are traveling in the same direction, with the same , and • The waves differ in

• 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 Interference

y = y1+y2 = 2A cos (φ/2) sin (kx - ωt + φ/2) • The resultant , 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 φ = 0 φ = π φ 1-D

y = y1+y2 = 2A cos (φ/2) sin (kx - ωt + φ/2)

⎛⎞φ Resultant Amplitude: 2A cos⎜⎟ ⎝⎠2 Constructive Interference (Even πφ ): Δ= 2 πnn , = 0,1,2,3... Destructive Interference () Odd πφ : Δ= (2 πnn + 1) , = 0,1,2,3...

Constructive Destructive Interference φ = 0 φ = π φ 1-D Wave Interference 2-D Wave Interference? 2-D Phase Difference at P: Δφ is different from the phase difference φ between the two source waves! It depends on the path difference traveled by the two waves! P 2π Phase Difference at P: Δφ =Δ+r φ λ 0 Δφ Amplitude at P: 2Acos( ) 2 Constructive or Destructive? (Indentical in phase sources) P 2π Phase Difference at P: Δφ =Δ+r φ λ 0

2π Δ=φ (1λπ ) = 2 λ Constructive!

⎛⎞Δφ Resultant Amplitude: 2A cos⎜⎟ ⎝⎠2 Constructive Interference: Δ=rnλφ , Δ = 2 π n , n = 0,1,2,3... λ Destructive Interference: Δ=rn (2 + 1) , Δφπ = (2 n + 1) , n = 0,1,2,3... 2 Constructive or Destructive? (Source out of Phase by 180 degrees) P 2π Phase Difference at P: Δφ =Δ+r φ λ 0

2π Δ=φ (1λπ ) + = 3π λ Destructive!

⎛⎞Δφ Resultant Amplitude: 2A cos⎜⎟ ⎝⎠2 Constructive Interference: Δ=rnλφ , Δ = 2 π n , n = 0,1,2,3... λ Destructive Interference: Δ=rn (2 + 1) , Δφπ = (2 n + 1) , n = 0,1,2,3... 2 Constructive or Destructive? (Indentical in phase sources) 2π Phase Difference at P: Δφ =Δ+r φ λ 0

P Constructive or Destructive? (Sources out of phase) 2π Phase Difference at P: Δφ =Δ+r φ λ 0

P Exam Extra Credit

2π Phase Difference at P: Δφ =Δ+r φ λ 0 Exam Extra Credit 2π Phase Difference at P: Δ=φφ Δ+r λ 0

yy=+=12 y2 A cos( Δφωφ / 2)sin( kxt −+Δ/ 2) Δφ Amplitude at P: 2A cos( ) 2

2 2 I ∝ A II=Δmax cos (φ / 2) In Phase or Out of Phase? Constructive? Destructive? A B Quiet Min Loud Max Quiet Min Loud Max

Light Waves: Coherent Sources

Intensity

2π Phase Difference at P: Δφφ=ΔΔ=r , 0 λ 0 Interference of 2 Sources Conditions for Interference Use 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 Interference Hyperphysics Website Light

• 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

22⎛⎞πd sin θπ ⎛⎞d II=≈max cos ⎜⎟Imax cos ⎜⎟y ⎝⎠λλ ⎝⎠L 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 22ππ φδ==d sin θ much greater than d λλ Interference Assumptions If 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:

Δr If L>> d,s then Δ r = d inθ Interference Conditions Constructive: Δrm==λ , m 0,1,2,3,... 1 Destructive: Δ=rm ( + )λ , m = 0,1,2,3,... 2 Interference Fringes

Bright fringe: Δrd===sinθ mλ , m0,1,2,3,... 1 Dark fringe : Δ=rd sinθλ = ( m + ) m = 0,1,2,3,... 2

Δr Derive Fringe Equations

• For bright fringes

λL ymm==±±(012),, K bright d

• For dark fringes

λL ⎛⎞1 ymmdark =+⎜⎟(012) =±±,, K d ⎝⎠2 Problem 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. λL ymm==±±(012),, K bright d 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? λL ymm==±±(012),, K bright d Double Slit Interference Dependence on Slit Separation

λL ymm==±±(012),, K bright d 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) E ω E max = ==c BkBmax EB E22 cB I ==S max max = max = max 2 av 222μ μ c μ ooo I ∝ Emax Double Slit Interference Hyperphysics Website Intensity Distribution Resultant Field • The magnitude of the resultant electric field comes from the

– EP = E1+ E2 = Eo[sin ωt + sin (ωt + φ)] • This can also be expressed as ⎛⎞φφ ⎛ ⎞ EEPo=+2cossin⎜⎟ ⎜ωt ⎟ ⎝⎠22 ⎝ ⎠

– 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 2 I ∝ A II=Δmax cos (φ / 2) 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

22⎛⎞πd sin θπ ⎛⎞d II=≈max cos ⎜⎟Imax cos ⎜⎟y ⎝⎠λλ ⎝⎠L HO 2 Slit You Try

In 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.900 b.700 c.500 d.300 e.600 Thin Film Interference RGB Theory Additive Complementary Yellow, Cyan, Magenta The color you have to add to get white light.

Red + Green = Yellow Blue + Green = Cyan Red + Blue = Magenta

Red + Blue + Green = White White light – red light = ?? White light – yellow light = ?? The Index of • Refraction: Light Bends in Transmission • The in any material is less than its speed in vacuum •The index of refraction, n, of a medium can be defined as • For a vacuum, n = 1 speed of light in a vacuum c – We assume n = 1 for n ≡=air also speed of light in a medium v • For other media, n > 1 λλ⎛⎞in vacuum • n is a dimensionless n = ⎜⎟number greater than λλn ⎝⎠in a medium 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 • 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 Films When reflecting off a medium of greater , 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 Rings Interference 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

• A ray of light is split into two rays by the Mo – The mirror is at 45o to the incident beam – The mirror is called a • It transmits half the light and reflects the rest

• After reflecting from M1 and M2, the rays eventually recombine at Mo and 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=ETLG5SLFMZo http://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 Maxwell 1860s

Light is wave. The medium is the Ether. 1 cxm==3.0 108 / s με0 o Measure the Speed of the Ether Wind

The was imagined by physicists since as the invisible "vapor" or "gas aether" filling the universe and hence as the carrier of heat and light. Rotate arms to produce interference fringes and find different speeds of light. http://www.youtube.com/watch?v=XavC4w_Y9b8&feature=related http://www.youtube.com/watch?v=4KFMeKJySwA&feature=related Michelson-Morely Experiment 1887 The speed of light is independent of the and is always c. The speed of the Ether wind is zero. OR…. Lorentz Contraction The apparatus shrinks by a factor :

1/− vc22 On the Electrodynamics of Moving Bodies 1905 Clocks slow down and rulers shrink in order to keep the speed of light the same for all observers! Time is Relative! Space is Relative! Only the SPEED OF LIGHT is Absolute! 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 shows Wave Interference If 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 the 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. -