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Chapter 37 : Interference of

Eye See YOU!! Human Retina Sharp Spot: Fovea Blind Spot: Optic Nerve Human Vision An optical Tuning Fork Optical Antennae: Rods & Cones Rods: Cones: Where does light actually come from? Light comes from the acceleration of charges. Light is emitted when an in an atom jumps between energy levels either by excitation or collisions. Atomic Emission of Light

Each chemical element produces its own unique set of spectral lines when it burns Hydrogen Spectra Incandescent Light Bulb Full Spectrum of Light All excited! Light

E = Emax cos (kx – ωt) B = Bmax cos (kx – ωt) E 1 c  B με 00 in a vacuum: 186,000 miles per second v c/ n 300,000 kilometers per second 3 x 10^8 m/s The Electromagnetic Spectrum Visible Light

• Different correspond to different • The range is from red (λ ~ 7 x 10-7 m) to violet (λ ~4 x 10-7 m)

Limits of Vision

Electron Waves

11 e  2.4xm 10 Double Slit is VERY IMPORTANT because it is evidence of waves. Only waves interfere like this. Intereference of 2-D Coherent Waves

Quiet Min Loud Max Quiet Min Loud Max

2 Difference at P:   r ,   0  0 Intereference of 2-D Coherent Light Waves depends on SLIT WIDTH: the smaller the width, relative to , the more bending and diffraction. Single Slit Interference Is called Diffraction (Chapter 38) Single Slit

If light were made of hard bullets, there would be no interference pattern. In reality, light does show an interference pattern.

Light acts like a going through the slits but arrive at the detector like a particle. Double Slit for shows ! Key to Theory! Interference pattern builds one electron at a time.

Electrons act like waves going through the slits but arrive at the detector like a particle. with Crystals (Chapter 38) Electron Microscope (Chapter 38)

Electron microscope picture of a fly. The resolving power of an optical lens depends on the wavelength of the light used. An electron-microscope exploits the wave-like properties of particles to reveal details that would be impossible to see with visible light.

Young’s 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 Intereference of 2-D Coherent Light Waves

Derive Fringe Equations

• For bright fringes

λL y m( m  0,,  1  2 ) bright d

• For dark fringes

λL 1 ydark  m ( m  0,,  1  2 ) d 2 Double Slit Interference Reality Combination of Single and Double

Prelab

A Young’s interference experiment is performed with monochromatic light. The separation between the slits is 0.500 mm, and the interference pattern on a screen 3.30 m away shows the first side maximum 3.40 mm from the center of the pattern. What is the wavelength? Problem Red light (=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 y m( m  0,,  1  2 ) bright d Measuring the wavelength of light

A double-slit interference pattern is observed on a screen 1.0 m behind two slits spaced 0.30 mm apart. 9 bright fringes span a distance of 1.7cm. What is the wavelength of light? Double 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 y m( m  0,,  1  2 ) bright d Double Slit Interference Reality Combination of Single and Double

Double Slit Interference Reality Combination of Single and Double Intensity of Light Waves

E = Emax cos (kx – ωt) B = Bmax cos (kx – ωt) E ωE max    c Bmax k B E B E22 c B I S max max  max  max 2 av 2μ 2 μ c 2 μ o o o I  Emax 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2 cos  sin ωt  22   

– EP has the same as the light at the slits

– The at P is given by 2Eo cos (φ / 2) • Intensity is proportional to the square of the amplitude: 2 IIcos2 ( / 2) I  A max 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 θ   πd  IIImaxcos  max cos y  λ   λL  Double Slit Intenisty

22πd sin θ   πd  IIImaxcos  max cos y  λ   λL  You Try Intensity

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 22πd sin θ   πd  b. 700 IIImaxcos  max cos y  c. 500 λ   λL  d. 300 e. 600 Double Slit Interference Dependence on Slit Separation

λL y m( m  0,,  1  2 ) bright d http://web.phys.ksu.edu/vqmorig/programs/java/makewave/Slit/vq_mws.htm

Thin Film Interference Diffraction & Interference

RGB Color Theory Additive Complementary Colors 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 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 • For a vacuum, n = 1 speed of light in a vacuum c λ – We assume n = 1 for n    air also speed of light in a medium v λ n • 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 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. From Low to High, a phase change of pi! From High to Low, a phase change? NO!

Interference in Thin Films

• The wavelength of 1 in the film is /n • For constructive interference

2t = (m + ½) /n (m = 0, 1, 2 …)

This takes into account both the difference in for the two rays and the 180° phase change • For destructive interference

2t = m/n (m = 0, 1, 2 …) Problem: Thin Films

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. What color do you see? Thin Film Interference

The light reflected from a (n = 1.40) appears red ( = 640 nm). What is the minimum thickness (in nm)? a. 124 b.104 c. 114 d.134 e. 234 Double-Slit with Thin Film

In a Young’s double-slit experiment using light of wavelength λ, a thin piece of Plexiglas having index of refraction n covers one of the slits. If the center point on the screen is a dark spot instead of a bright spot, what is the minimum thickness of the Plexiglas? Double-Slit with Thin Film If the center point on the screen is to be a dark spot rather than bright, passage through the plastic must delay the light by one-half wavelength. Calling the thickness of the plastic t,there is destructive interference if the phase difference is: 

  film   air

22  tt / n

 t  21n  

• 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 c3.0 x 108 m / s 0 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 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