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Chapter 22 Wave

Chapter Goal: To understand and apply the wave model of .

© 2013 Pearson Education, Inc. Slide 22-2 © 2013 Pearson Education, Inc. Models of Light

. The wave model: Under many circumstances, light exhibits the same behavior as sound or water waves. The study of light as a wave is called wave optics. . The ray model: The properties of prisms, mirrors, and lenses are best understood in terms of light rays. The ray model is the basis of ray optics. Chapter 23 . The photon model: In the quantum world, light behaves like neither a wave nor a particle. Instead, light consists of photons that have both wave-like and particle-like properties. This is the quantum theory of light. Physics 43

© 2013 Pearson Education, Inc. Slide 22-29 depends on SLIT WIDTH: the smaller the width, relative to , the more bending and diffraction. Ray Optics: Ignores Diffraction and Interference of waves! Diffraction depends on SLIT WIDTH: the smaller the width, relative to wavelength, the more bending and diffraction. Ray Optics assumes that λ<1mm: This approximation is good for the study of interference. Geometric RAY Optics (Ch 23)

nn1sin 1 2 sin 2 ir James Clerk Maxwell 1860s

Light is wave. Speed of Light in a vacuum:

1 8 186,000 miles per second c3.0 x 10 m / s 300,000 kilometers per second 0 o 3 x 10^8 m/s 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 maxmax  max  max 2 av 2μ 2 μ c 2 μ o o o I  E m ax The Electromagnetic Spectrum Visible Light

• Different correspond to different colors • 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 Interference of Light Wave Optics Double Slit is VERY IMPORTANT because it is evidence of waves. Only waves interfere like this. If light were made of hard bullets, there would be no interference pattern. In reality, light does show an interference pattern. Photons

Light acts like a wave going through the slits but arrive at the detector like a particle. Particle Wave Duality Photons & Electrons! Interference pattern builds one electron at a time.

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

11 e  2.4xm 10 Double Slit for Electrons shows ! Key to Quantum Theory! Electron Diffraction with Crystals Electron Microscope

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. Intereference of 2-D Coherent Light Waves RECALL: Intereference of 2-D Coherent Sound Waves

Quiet Min Loud Max Quiet Min Loud Max

2 Phase Difference at P:   r ,   0  0 Intereference of 2-D Coherent Light Waves 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 2 Phase Difference at P:   r 

Constructive :   2m  ,  r  m  , m  0,1,2,3... 1 Destructive :   (2m  1) ,  r  ( m  ), m  0,1,2,3... 2 Fig. 37-3, p. 1086 2 Phase Difference at P:   r 

Constructive :   2m  ,  r  m  , m  0,1,2,3... 1 Destructive :   (2m  1) ,  r  ( m  ), m  0,1,2,3... 2 Fig. 37-5, p. 1087 Fringe Equations

“m” is the fringe order.

• Maxima: bright fringes d sin θ mλ λL y m( m  0,,  1  2 ) bright d • Minima: dark fringes

1 d sin θ m λ 2 λL 1 ydark  m ( m  0,,  1  2 ) d 2 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.

λL y m( m  0,,  1  2 ) bright d Measuring the wavelength of light

λL y m( m  0,,  1  2 ) bright d

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?

 L  y  Fringe Spacing. d Double Slit λL y m( m  0,,  1  2 ) bright d 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? Double Slit Interference Dependence on Slit Separation

λL y m( m  0,,  1  2 ) bright d QuickCheck 22.5 λL y m( m  0,,  1  2 ) bright d A laboratory experiment produces a double-slit interference pattern on a screen. If green light is used, with everything else the same, the bright fringes will be

A. Closer together B. In the same positions. C. Farther apart. D. There will be no fringes because the conditions for interference won’t be satisfied.

Slide 22-44 QuickCheck 22.5 λL y m( m  0,,  1  2 ) bright d A laboratory experiment produces a double-slit interference pattern on a screen. If green light is used, with everything else the same, the bright fringes will be

A. Closer together. B. In the same positions. C. Farther apart. D. There will be no fringes because the conditions for interference won’t be satisfied.

 L y  and green light has a shorter wavelength. d

Slide 22-45 QuickCheck 22.6 λL y m( m  0,,  1  2 ) bright d A laboratory experiment produces a double-slit interference pattern on a screen. If the slits are moved closer together, the bright fringes will be

A. Closer together. B. In the same positions. C. Farther apart. D. There will be no fringes because the conditions for interference won’t be satisfied.

Slide 22-46 QuickCheck 22.6 λL y m( m  0,,  1  2 ) bright d A laboratory experiment produces a double-slit interference pattern on a screen. If the slits are moved closer together, the bright fringes will be

A. Closer together. B. In the same positions. C. Farther apart. D. There will be no fringes because the conditions for interference won’t be satisfied.

L y  and d is smaller. d

Slide 22-47 QuickCheck 22.3 λL y m( m  0,,  1  2 ) bright d A laboratory experiment produces a double-slit interference pattern on a screen. What is the path difference the light travels from the two slits to the point on the screen marked with a dot?

A. 1.0 . B. 1.5 . C. 2.0 . D. 2.5 . E. 3.0 .

Slide 22-35 QuickCheck 22.3 λL y m( m  0,,  1  2 ) bright d A laboratory experiment produces a double-slit interference pattern on a screen. What is the path difference the light travels from the two slits to the point on the screen marked with a dot?

A. 1.0 . B. 1.5 . C. 2.0 . D. 2.5 . E. 3.0 .

Slide 22-36 Start Using Exam 2 Equation Sheet! Fringe Equations

“m” is the fringe order.

• Maxima: bright fringes d sin θ mλ λL y m( m  0,,  1  2 ) bright d • Minima: dark fringes

1 d sin θ m λ 2 λL 1 ydark  m ( m  0,,  1  2 ) d 2 Single Slit Interference Is called Diffraction Light interferes with itself Single Slit Diffraction  Dark fringe: sin mm   1,  2,  3,... a Single Slit Problem  Dark fringe: sin mm   1,  2,  3,... a A narrow slit is illuminated with sodium yellow light of wavelength 589 nm. If the central maximum extends to ±30, how wide is the slit? Width of the Central Maximum If the incident light has =690nm (red) and a = 4 x 10-6m, find the width of the central bright fringe when the screen is 0.4m away.  Dark fringe: sin mm 0,1,2,3,... a

yLmm tan

wy 2  Dark fringe: sin m a Compare Fringe Equations for Single and Double Slits maxima:d sinθ mλ ( m  0 ,  1 ,  2 ) 1 minima: d sinθ m  λ ( m  0 ,  1 ,  2 ) 2

minima: a sindark  m , m   1,  2,  3,... Light Intensity: Ignoring Single Slit Diffraction • The interference pattern consists of equally spaced fringes of equal intensity – it is an idealization • This result is valid only if L >> d and for small values of θ and is an approximation of the inside fringe pattern. Reality: Combined Effects Interference maximum coincides with the first diffraction minimum. Light Intensity: Ignoring Diffraction IIcos2 ( / 2) • The interference max pattern consists of 2 equally spaced Phase Difference at P:   r  fringes of equal intensity

d sin θ mλ λL y m( m  0,,  1  2 ) bright d Recall: Sound Power The power transmitted through a closed surface by a wave is proportional to the amplitude of the wave. Intensity~  Amplitude2 RECALL: 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 - wt)

• y2 = A sin (kx - wt + )

• y = y1+y2 = 2A cos (/2) sin (kx - wt + /2)

Resultant Amplitude Depends on phase: Spatial Interference Term Intensity Distribution Resultant 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 φφ    EEPo2 cos  sin 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 2 φ • The maximum intensity at P IEEP P4 o cos  2 is 4 times one source!!!!!!!!! Amplitude is twice as big But intensity is proportional to the amplitude SQUARED so the Intensity is four times as big as the source. This is energy being conserved! The light energy is redistributed on the screen. φφ    EEPo2 cos  sin ωt  22   

2 2 2 φ IEEP P4 o cos  2 Light Intensity: Ignoring Diffraction IIcos2 ( / 2) • The interference max pattern consists of 2 equally spaced Phase Difference at P:   r  fringes of equal intensity

d sin θ mλ λL y m( m  0,,  1  2 ) bright d 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? WARNING! We are going to ignore the intensity drop due to distance in inverse square law. We assume that the amplitude remains constant over the short distances considered. We will only considering the intensity change due to interference! This is not a bad approximation using as sources.

P W I  4mr 22 Combined Effects Interference maximum coincides with the first diffraction minimum. Two-Slit Diffraction Patterns, Maxima and Minima • To find which interference maximum coincides with the first diffraction minimum d sin θ mλ d    m a sin θ λ a – The conditions for the first interference maximum • d sin θ = mλ – The conditions for the first diffraction minimum • a sin θ = λ Intensity of Two-Slit Diffraction The Complete Equation 2 2 πd sin θ sinπa sin θ/ λ II max cos  λ πa sin θ/ λ

Section 38.2 Hyperphysics Hyperphysics Hyperphysics Hyperphysics

Multiple Slits: Diffraction Gratings

For N slits, the intensity of the primary maxima is N2 times greater than that due to a single slit.

Section 37.3 The Diffraction Grating Multiple Slits: Diffraction Gratings

. The figure shows a diffraction grating in which N slits are equally spaced a distance d apart. . This is a top view of the grating, as we look down on the experiment, and the slits extend above and below the page. . Only 10 slits are shown here, but a practical grating will have hundreds or even thousands of slits. Slide 22-53 Reflection Gratings

. In practice, most diffraction gratings are manufactured as reflection gratings. . The interference pattern is exactly the same as the interference pattern of light transmitted through N parallel slits.

Slide 22-62 The Diffraction Grating

Bright fringes will

occur at angles m, such that

d sinm = m where m = 0, 1, 2, 3, …

The y-positions of these fringes are:

Slide 22-54 Circular Aperature

 Dark fringe: sin mm 0,1,2,3,... D

Circular Diffraction Patterns At the beginning of the 19th century, the idea that light does not simply propagate along straight lines gained traction. published his double-slit experiment in 1807. The original Arago spot experiment was carried out a decade later and was the deciding experiment on the question of whether light is a particle or a wave. It is

Penny: Central Bright Spot: Poisson Spot thus an example of an . Fig 38-3, p.1207 The images show simulated Arago spots in the shadow of a disc of varying diameter (4 mm, 2 mm, 1 mm – left to right) at a distance of 1 m from the disc. The point source has a wavelength of 633 nm (e.g. He-Ne ) and is located 1 m from the disc. The image width corresponds to 16 mm. Airy Disk Due to diffraction, the smallest point to which a lens or mirror can focus a beam of light is the size of the Airy disk. Even if one were able to make a perfect lens, there is still a limit to the resolution of an image created by such a lens. An optical system in which the resolution is no longer limited by imperfections in the lenses but only by diffraction is said to be diffraction limited. Diffraction Resolution

Fig 38-13, p.1215 Resolution

• The ability of optical systems to distinguish between closely spaced objects is limited because of the wave nature of light • If two sources are far enough apart to keep their central maxima from overlapping, their images can be distinguished – The images are said to be resolved • If the two sources are close together, the two central maxima overlap and the images are not resolved Resolving Power The diffraction limit: two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other. Rayleigh Criterion    1.22 min D

D: Aperture Diameter Space Shuttle Resolution

What, approximately is the distance between objects on Earth that the astronauts can resolve by eye at 200 km height from the space shuttle? Assume l = 500 nm light and a pupil diameter D = 0.50 cm. Ignore the index of refraction from vitreous eye fluid.

Rayleigh Criterion    1.22 min D http://www.smithsonianmag.com/science- nature/prehistoric-pointillism-long-seurat-ancient- artists-chiseled-mammoths-out-dots-180962306/ Georges Seurat’s 1884 pointillist masterpiece A Sunday on La Grand Jatte

Standing back from a Georges Seurat painting, you can cannot resolve the dots but a camera, at the same distance can. Assume that light enters your eyes through pupils that have diameters of 2.5 mm and enters the camera through an aperture with diameter of 25 mm. Assume the dots in the painting are separated by 1.5 mm and that the wavelength of the light is 550 nm in vacuum. Find the distance at which the dots can just be resolved by a) the camera b) the eye. Thin Film Interference RGB Color Theory What is color? What is light? “Light” is what we see. An optical Tuning Fork I am Watching YOU!! Human Retina Sharp Spot: Fovea Blind Spot: Optic Nerve Optical Antennae: Rods & Cones Rods: Intensity Cones: Color Visible Light

• Different wavelengths correspond to different colors • 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 How is Light Made? Atoms are EM Tuning Forks

They are ‘tuned’ to particular frequencies of light energy. Visible light ( and UV and IR) is emitted when an electron in an atom jumps between energy levels either by excitation or collisions. Light Emission Atomic Emission of Light

Each chemical element produces its own unique set of spectral lines when it burns Dispersion with a Diffraction Grating Dispersion via Diffraction constructive: d sin m , m 0,1,2,3 Hydrogen Spectra Incandescent Light Bulb Full Spectrum of Light All frequencies excited! If you pass white light through a prism, it separates into its component colors.

long wavelengths short wavelengths R.O.Y. G. B.I.V Dispersion via Diffraction constructive: d sin m , m 0,1,2,3 Diffraction Grating Problem

constructive: d sin m , m 0,1,2,3 White light is spread out into spectral hues by a diffraction grating. If the grating has 1000 lines per cm, at what angle will red light (l = 640 nm) appear in first order? The ResolvingDiffraction PowerGrating of a Diffraction Grating

. Diffraction gratings are used for measuring the wavelengths of light. . If the incident light consists of two slightly different wavelengths, each wavelength will be diffracted at a slightly different angle.

Slide 22-56 Resolving Power of a Diffraction Grating

• For two nearly equal wavelengths, λ1 and λ2, between which a diffraction grating can just barely distinguish, the resolving power, R, of the grating is defined as

λλ Rave  ave  Nm λ21 λ λ

• Therefore, a grating with a high resolution can distinguish between small differences in wavelength • The resolving power in the mth-order diffraction is • R = Nm – N is the number of slits – m is the order number • Resolving power increases with increasing order number and with increasing number of illuminated slits Grating Resolution λλ Rave  ave  Nm λ21 λ λ Determine the number of grating lines necessary to resolve the 589.59 nm and 589.00 nm sodium lines in second order. Intensity of a Diffraction Grating

Because intensity depends on the square of the amplitude, the intensities of the bright fringes are:

Slide 22-55 QuickCheck 22.8

In a laboratory experiment, a diffraction grating produces an interference pattern on a screen. If the number of slits in the grating is increased, with everything else (including the slit spacing) the same, then

A. The fringes stay the same brightness and get closer together. B. The fringes stay the same brightness and get farther apart. C. The fringes stay in the same positions but get brighter and narrower. D. The fringes stay in the same positions but get dimmer and wider. E. The fringes get brighter, narrower, and closer together.

Slide 22-57 QuickCheck 22.8

In a laboratory experiment, a diffraction grating produces an interference pattern on a screen. If the number of slits in the grating is increased, with everything else (including the slit spacing) the same, then

A. The fringes stay the same brightness and get closer together. B. The fringes stay the same brightness and get farther apart. C. The fringes stay in the same positions but get brighter and narrower. D. The fringes stay in the same positions but get dimmer and wider. E. The fringes get brighter, narrower, and closer together.

Slide 22-58 Reflection Gratings in Nature

. Naturally occurring reflection gratings are responsible for some forms of color in nature. . A peacock feather consists of nearly parallel rods of melanin, which act as a reflection grating.

Slide 22-63 Diffraction & Interference Iridescence Thin Film Interference Radiation of Visible Sunlight Additive Primary Colors Red, Green, Blue 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 = ?? FYI: Mixing Colored Pigments Subtractive Colors Pigments subtract colors from white light.

Yellow + Cyan = Green

Cyan + Magenta = Purple

Yellow + Magenta = Red

Yellow + Cyan + Magenta = Black Why are some materials colored? Why is a Rose Red?

Colored materials absorb certain colors that resonate with their electron energy levels and reject & reflect those that do not. Shine cyan light on a red rose and what color do you see? Shine cyan light on a red rose and what color do you see? Why is the Ocean Cyan?

White light minus cyan is red. Ocean water absorbs red. Thin Film Interference 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 • For a vacuum, n = 1 speedof lightin a vacuum c λ – We assume n = 1 for air n    also speedof light in a medium v λ n • For other media, n > 1 λλin vacuum • n is a dimensionless number n  greater than unity, not λλn in a medium necessarily an integer Some Indices of Refraction Variation of Index of Refraction with Wavelength speedof light ina vacuum c λ n    speedof light inamedium v λn

• This dependence of n on λ is called dispersion • The index of refraction for a material generally decreases with increasing wavelength • Violet light bends more than red light when passing into a refracting material Interference in Thin Films • Light will be cancelled when reflecting off a thin film due to phase changes. • Find the total path difference due to wave 2 travelling 2t and any phase changes of waves. • The wavelength of ray 1 in the film is /n • Apply interference conditions Application: Thin-Film Optical Coatings

. Thin transparent films, placed on glass surfaces, such as lenses, can control reflections from the glass. . Antireflection coatings on the lenses in cameras, microscopes, and other optical equipment are examples of thin-film coatings.

© 2013 Pearson Education, Inc. Slide 21-96 Interference in Thin Films Depends on Medium and index of refraction! When reflecting off a medium of greater refractive index, a light wave undergoes a phase shift of ½ a wavelength. When reflection off a medium of lesser refractive index, a light wave undergoes NO phase shift. Reflections Phase Shift Chant From Low to High, a phase change of pi! From High to Low, a phase change? NO! Thin Film Interference

The light reflected from a soap bubble (n = 1.40) appears red ( = 640 nm). What is the minimum thickness (in nm)?

Constructive or destructive?

constructive interference

2t = (m + ½) /n (m = 0, 1, 2 …) Interference in Thin Films with the thin film on air

• The wavelength of ray 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 optical path length for the two rays and the 180° phase change • For destructive interference

2t = m/n (m = 0, 1, 2 …) Example 21.9 Designing an Antireflection Coating

© 2013 Pearson Education, Inc. Slide 21-98 Example 21.9 Designing an Antireflection Coating

© 2013 Pearson Education, Inc. Slide 21-100 Application: Thin-Film Optical Coatings

. The phase difference between the two reflected waves is:

where n is the index of refraction of the coating, d is the thickness, and  is the wavelength of the light in vacuum or air. . For a particular thin-film, constructive or destructive interference depends on the wavelength of the light:

© 2013 Pearson Education, Inc. Slide 21-97 The equations in the text is for ONLY the first case where the coating lies on the glass with a higher index of refraction. BUT… What if the film floats on water? Or in air?

Just don’t use those equations!! DERIVE THEM!!!!!! MORE Applications of Interference: Interference: Holography

. The diffracted reference beam reconstructs the original scattered wave. . As you look at this diffracted wave, from the far side of the hologram, you “see” the object exactly as if it were there. . The view is three dimensional because, by moving your head with respect to the hologram, you can see different portions of the wave front. Slide 22-104 https://www.youtube.com/watch?v=0ics3RVSn9w 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 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 interference changes from a max to a min and back to a max every time L increases by half a wavelength. The number of maxima that appear as the Mirror shifts a length

https://www.youtube.com/watch?v=j-u3IEgcTiQ https://www.youtube.com/watch?v=j-u3IEgcTiQ Interferometer Problem

Monochromatic light is beamed into a Michelson interferometer. The movable mirror is displaced 0.382 mm, causing the interferometer pattern to produce 1 700 fringes. Determine the wavelength of the light. What color is it? James Clerk Maxwell 1860s Light is wave. The medium is the Ether.

1 c3.0 x 108 m / s 0 o Rotate arms to produce interference fringes and find different speeds of light caused by the Ether Wind, due to Galilean Relativity: light should travel slower against the Ether Wind. From that you can find the speed of the wind. https://www.youtube.com/watch?v=7qJoRNseyLQ http://www.youtube.com/watch?v=XavC4w_Y9b8&feature=related http://www.youtube.com/watch?v=4KFMeKJySwA&feature=related http://www.youtube.com/watch?v=ETLG5SLFMZo http://www.youtube.com/watch?v=Z8K3gcHQiqk&feature=related 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! On the Electrodynamics of Moving Bodies 1905 LIGO: Gravitational Interferometry

https://www.youtube.com/watch?v=HwC5IYw5uAE

http://www.youtube.com/watch?v=RzZgFKoIfQI&feature=related LISA The Laser Interferometer Space Antenna

http://www.youtube.com/watch?v=DrWwWcA_Hgw&feature=related http://www.youtube.com/watch?v=tUpiohbBv6o Physics Diffraction Fun on an Airplane Always sit on the side opposite the sun when traveling north-south!! Interference of Light Wave Optics