This Week v Waves v Standing waves Musical

This Week v Waves v Standing waves Musical instruments, guitars, pianos, organs v Interference of two waves tuning a piano, color of oil films v Polarisation Why have polaroid sun glasses? v Electromagnetic waves and telescopes v How do we see color 10/31/2020 Physics 214 Fall 2011 1

Periodic waves One can propagate waves which are a single complicated pulse e. g. an explosion or a complicated continuous wave e. g. the wind. We will focus on regular repetitive waves λ These waves have a pattern which repeats and the length of one pattern is called the wavelength λ The number of patterns which pass a point/second is called the frequency f and if the time for one pattern to pass is T then f = 1/T v = λ/T = fλ 10/31/2020 Physics 214 Fall 2011 2

Waves on a string If we shake the end of a rope we can send a wave along the rope. The rope must be under tension in order for the wave to propagate v = √(F/μ) F = TENSION μ = MASS/UNIT LENGTH 10/31/2020 Physics 214 Fall 2011 3

Standing waves If two identical waves exist on the same string but traveling in opposite directions the result can be standing waves in which some points never have a deflection, these are called nodes Some points oscillate between plus and minus the maximum amplitude, these are called antinodes. Standing waves provide the notes on musical instruments. When a string is secured at both ends and plucked or hit the generated waves will travel along the string and be reflected and set up standing waves. 10/31/2020 Physics 214 Fall 2011 4

Musical notes Each end of the string must be a node so the possible standing waves must be multiples of λ/2 Fundamental f = v/λ = v/2 L 2 nd Harmonic f = v/λ = v/L 3 rd Harmonic f = v/λ = 3 v/2 L Musical sound is a mixture of harmonics modified by the body of the instrument. v = √(F/μ) so a piano or a violin is tuned by changing the tension in the string 10/31/2020 Physics 214 Fall 2011 5

Closed organ pipe Node at one end an antinode at the other Fundamental f = v/4 L 2 nd Harmonic f = 3 v/4 L 3 rd Harmonic f = 5 v/4 L The velocity of sound in air is v = 340 m/s 10/31/2020 Physics 214 Fall 2011 6

Beats If two waves have slightly different frequencies then the sum has a frequency which is f 1 – f 2. The human ear can detect beats and this is used to tune an instrument. For example using a tuning fork at a known frequency and adjusting a piano string until no beats heard f = f 1 – f 2 10/31/2020 Physics 214 Fall 2011 7

Doppler effect The Doppler effect is the change in frequency of a wave when the source and observer are moving. As the source is approaching the frequency increases and if it is receding the frequency decreases 10/31/2020 h t t p http: //www. physics. purdue. edu/class/applets/phe/dopplereff. htm Physics 214 Fall 2011 8

Sonic boom At each point on the path the sound wave expands radially and they all combine along a single wave front which is a pressure wave causing us to hear the boom http: //www. phy. ntnu. edu. tw/java/airplane. html 10/31/2020 Physics 214 Fall 2011 9

Summary of Chapter 15 v = λ/T = fλ Transverse or longitudinal Standing waves Fundamental f = v/λ = v/2 L 2 nd Harmonic f = v/λ = v/L 3 rd Harmonic f = v/λ = 3 v/2 L Fundamental f = v/4 L 2 nd Harmonic f = 3 v/4 L 3 rd Harmonic f = 5 v/4 L 10/31/2020 Physics 214 Fall 2011 10

Sound effects Doppler effect f increases as sound approaches Beats f = f 1 – f 2 Sonic boom 10/31/2020 Physics 214 Fall 2011 11

1 S-13 Slinky on Stand Creating longitudinal compression waves in a slinky What happens when you pull back and release one end of the slinky ? THE SLINKY TRIES TO RETURN TO EQUILIBRIUM. BUT INERTIA CAUSE THE LINKS TO PASS BEYOND THE EQUILIBRIUM POINT AND THIS CREATES A COMPRESSION. THUS, BOTH A RESTORING FORCE (BACK TO EQUILIBRIUM) AND INERTIA PLAY KEY ROLES IN THE TRANSMISSION OF THE WAVE. 10/31/2020 Physics 214 Fall 2011 12

1 S-41 Standing Waves in Rope Creating transverse standing waves on heavy cable How might we create different wave patterns on the cable ? How is this related to tuning a guitar ? BY ADJUSTING THE SPEED OF THE JIGSAW OR THE TENSION IN THE CORD, DIFFERENT STANDING WAVE PATTERNS CAN BE GENERATED BETWEEN THE FIXED ENDS. THOUGH THE WAVELENGTH OF THE STANDING WAVE IS FIXED BY THE LENGTH BETWEEN THE FIXED POINTS, THE FREQUENCY DEPENDS ON THE TENSION IN THE CABLE. YOU TUNE A GUITAR BY CHANGING THIS TENSION. 10/31/2020 Physics 214 Fall 2011 13

4 B-01 Standing Waves in a Gas What happens when an acoustic standing wave is introduced in the tube ? Effects of acoustic standing wave on air pressure The wave pattern indicates a pressure non-uniformity within the tube. STANDING WAVES ARE PRODUCED WHENEVER TWO WAVES OF IDENTICAL FREQUENCY INTERFERE WITH ONE ANOTHER WHILE TRAVELING OPPOSITE DIRECTIONS ALONG THE SAME MEDIUM. 10/31/2020 Physics 214 Fall 2011 14

4 B-13 Hoot Tubes (Resonance in Pipes) Creating acoustic resonances in glass tubes with hot air If the same heated grid is used, why do the different tubes give off different sounds ? Why does horizontal tube not emit sound ? L’ L 1 st Harmonic: λ = 2 L , f = v/λ Length of tube determines resonant frequency THE HOT AIR FROM THE HEATED GRID GENERATES A DISTURBANCE THAT CAN BE THOUGHT OF AS “NOISE. ” THE RESONANT FREQUENCY OF THE PARTICULAR TUBE DETERMINES WHICH COMPONENTS OF THIS NOISE ARE AMPLIFIED. 10/31/2020 Physics 214 Fall 2011 15

4 B-10 MONOCHORD What is the purpose of tightening or loosening the string ? What role do the frets play ? Chinese Zither Real Musical Instrument CHANGING TENSION OF THE STRING AFFECTS THE SPEED OF WAVE PROPAGATION AND CHANGES THE FUNDAMENTAL FREQUENCY THE BRIDGE ACTS AS A FRET THAT EFFECTIVELY CHANGES THE LENGTH OF THE WIRE AND THE FUNDAMENTAL FREQUENCY 10/31/2020 Physics 214 Fall 2011 16

4 B-14 Whirly Tubes Forcing air thru a tube to create acoustic resonances Why do different tubes give off different sounds ? How can we increase the pitch emitted from any one whirly tube ? AIR FLOWS UP THE TUBE DUE TO THE “CENTRIFUGAL” EFFECT FROM ROTATION. THE SOUND RESULTS FROM THE AIR PASSING OVER THE CORRUGATIONS IN THE TUBE. FASTER WHIRLING RESULTS IN HIGHER FREQUENCY OF SOUND (HIGHER PITCH). 10/31/2020 Physics 214 Fall 2011 17

4 B-11 Resonance Two identical wooden have identical tuning forks attached at the center of the top of the box. When the tuning fork is struck, the sound is amplified by the resonance occurring in the box. When the one box is placed such that its opening is a few centimeters from the other’s opening, striking one tuning fork sets up a sympathetic vibration in the other, which can be shown by having a ping-pong ball, attached to a thread, lightly touch the second fork. The ping-pong ball will bounce back and forth indicating that the fork is vibrating. 10/31/2020 Physics 214 Fall 2011 18

4 A-02 Sound Production with Rotating Slat Making acoustic compressions with the Bull Roarer How does the Bull Roarer make a sound ? Is it a high frequency or low frequency acoustic wave ? Why could we THE ROTATING SLAT WHEN SWUNG IN A CIRCLE WILL PRODUCE COMPRESSIONS AND RAREFACTIONS, CREATING A LOW FREQUENCY SOUND. 10/31/2020 Physics 214 Fall 2011 19

4 A-03 Sound Production in Bell Jar Investigating the medium through which sound waves propagate Where does the sound go when the bell jar is evacuated ? We know waves carry energy and can do work, so what happens to the energy emitted by the tone generator ? AIR MOLECULES PROVIDE THE MEDIUM THROUGH WHICH ACOUSTIC WAVES PROPAGATE. WHEN THAT MEDIUM IS NOT PRESENT, SOUND CANNOT PROPAGATE. 10/31/2020 Physics 214 Fall 2011 20

4 B-02 Standing Waves in a Vertical Air Column Finding resonances of acoustic vibration in tube open at one end How do we know that we have found a resonance ? How is the length of the tube that gives resonance determined ? L = n(λ/4), n is odd STRONG RESONANCES OCCUR WHEN THE LENGTH OF THE AIR COLUMN FORMS A STANDING WAVE SUCH THAT THERE IS A NODE AT THE CLOSED (WATER)-END AN ANTINODE AT THE OPEN-END. 10/31/2020 Physics 214 Fall 2011 21

4 C-01 Doppler Effect Investigating change in sound due to the Doppler effect At what point in circular movement does sound change ? What is relative motion between source and listener at these points ? WHEN THE SOURCE MOVES TOWARD (AWAY FROM) LISTENER, THE FREQUENCY OF SOUND, OR PITCH, INCREASES (DECREASES). 10/31/2020 Physics 214 Fall 2011 22

4 C-02 Doppler Effect II Beat frequency, Doppler Effect and a virtual source Wall Virtual source, due to reflection off wall Why do you hear beats (interference between two sound waves of different frequencies) ? v v Because the speed of sound is finite, the virtual source can be thought of as a time-delayed (out of phase) replica of actual source. AN OBSERVER AT THE WALL WOULD NOTICE A FREQUENCY INCREASE AND DECREASE AS THE ACTUAL SOURCE MOVED TOWARD AND AWAY FROM THE WALL. THE CHANGES IN FREQUENCY CAUSE THE INTERFERENCE THAT CREATE THE BEATS. 10/31/2020 Physics 214 Fall 2011 23

Questions Chapter 15 Q 2 Waves are traveling in an eastward direction on a lake. Is the water in the lake necessarily moving in that direction? No. The main motion is just up and down Q 3 A slowly moving engine bumps into a string of coupled railroad cars standing on a siding. A wave pulse is transmitted down the string of cars as each one bumps into the next one. Is this wave transverse or longitudinal? Longitudinal. The motion is back and forward in the direction of the wave 10/31/2020 Physics 214 Fall 2011 24

Q 4 A wave can be propagated on a blanket by holding adjacent corners in your hands and moving the end of the blanket up and down. Is this wave transverse or longitudinal? It is transverse the motion of the blanket is up and down Q 13 Suppose that we increase the tension in a rope, keeping the frequency of oscillation of the end of the rope the same. What effect does this have on the wavelength of the wave produced? The increase in tension cause an increase in velocity so the wavelength increases 10/31/2020 Physics 214 Fall 2011 25

Q 20 If we increase the tension of a guitar string, what effect does this have on the frequency and wavelength of the fundamental standing wave formed on that string? The fundamental wavelength is fixed by the length of the string. Increasing the tension increases the wave speed so the frequency increases. Q 23 Is it possible for sound to travel through a steel bar? Yes. It is just a compression wave that travels through the steel. Putting your ear to the rails is a good way of hearing a train coming a long distance away 10/31/2020 Physics 214 Fall 2011 26

Q 24 Suppose that we increase the temperature of the air through which a sound wave is traveling. A. What effect does this have on the speed of the sound wave? Explain. B. For a given frequency, what effect does increasing the temperature have on the wavelength of the sound wave? Explain. The speed increases and the wavelength increases Q 25 If the temperature in a organ pipe increases above room temperature, thereby increasing the speed of sound waves in the pipe but not affecting the length of pipe significantly, what effect does this have on the frequency of the standing waves produced by this pipe? The length is unchanged so the wavelength is unchanged so the frequency increases 10/31/2020 Physics 214 Fall 2011 27

Q 26 Is the wavelength of the fundamental standing wave in a tube open at both ends greater than, equal to, or less than the wavelength for the fundamental wave in a tube open at just one end? A tube open at both ends half a wavelength whereas the tube open at one end has one quarter of a wavelength. Q 27 A band playing on a flat-bed truck is approaching you rapidly near the end of a parade. Will you hear the same pitch for the various instruments as someone down the street who has already been passed by the truck? The pitch increases as the band approaches and decreases as it is going away 10/31/2020 Physics 214 Fall 2011 28

Ch 15 E 2 Water waves have a wavelength = 1. 4 m and Period T = 0. 8 s. What is the velocity of the waves? v = / T = 1. 4/0. 8 = 1. 75 m/s 10/31/2020 Physics 214 Fall 2011 29

Ch 15 E 8 String of length 0. 8 m is fixed at both ends. The story is plucked so that there are two nodes along the string in addition to those at either end. What is the wavelength of the interfering waves? fixed head node λ ½λ v snapshot T R v At each node wave T and wave R cancel. From the picture a total of 3/2 fit on 0. 8 m string. 3/2 = 0. 8 m → = (2/3) 0. 8 = 0. 53 m 10/31/2020 Physics 214 Fall 2011 30

Ch 15 E 10 What is the frequency of a sound wave with wavelength = 0. 68 m travelling in room temperature air (v=340 m/s)? v = f → f = v/ = 340/0. 68 m = 500 Hz 10/31/2020 Physics 214 Fall 2011 31

Ch 15 E 12 Suppose we start a major scale at 440 Hz (concert A). If we call this frequency do, what is the ideal ratio frequency of a) mi ? b) sol ? a) mi = 5/4 do = 5/4 (440 Hz) = 550 Hz b) sol = 3/2 do = 3/2 (440) = 660 Hz 10/31/2020 Physics 214 Fall 2011 do re mi 5/4 4/3 3/2 fa sol la ti do 32

Ch 15 E 14 In just tuning, major third ratio = 5/4. In equally tempered tuning the ratio = 1. 260. If we start a scale at do = 440 Hz, what is the difference in frequency of a major third in each style of tuning? Just tuning mi = 5/4 (440 Hz) = 550 Hz = fj Equally tempered tuning mi = 1. 260 (440 Hz) = 554. 4 Hz = fet – fj = 4. 4 Hz 10/31/2020 Physics 214 Fall 2011 33

Ch 15 CP 2 A guitar string has length 1. 25 m and mass 40 g. After stringing the guitar, string has 64 cm between fixed points. It is tightened to tension = 720 N. a) What is mass per unit length of string? b) What is wavespeed on tightened string? c) What is wavelength of traveling waves that interfere to form the fundamental standing wave? d) What is frequency of the fundamental standing wave? e) What are wavelength and frequency of the next harmonic? 10/31/2020 Physics 214 Fall 2011 34

Ch 15 CP 2 cont. a) =M/L=0. 04/1. 25=0. 032 kg/m b) = F/ = 720/0. 032 = 22500 = 150 m/s c) v R = T v node antinode Snapshot of fundamental standing wave 0. 64 m = ½ = 1. 28 m node d) =1. 28 m ; f = / = 150/1. 28 = 117 Hz e) = 0. 64 m f = / = 150/0. 64 = 234 Hz node anti-node Snapshot of second harmonic 10/31/2020 Physics 214 Fall 2011 35

Ch 15 CP 4 Concert A = 440 Hz. A is nine half steps above C (12 half steps in full octave). In equally-tempered-tuning each half step has a ratio 1. 0595 above preceding step. a) b) c) d) What is the frequency of A-flat, one half step below A? Find the frequencies at each half-step until C. In just-tuning, middle C = 264 Hz. How does b) compare to this? Now, work up from concert A to high C in equal temperament. Is this frequency twice than in b) (one octave high than middle c)? a) A = 440 Hz, Ab = A/1. 0595 = 415. 2902 Hz b) C, C#, D, D#, E, F, F#, G, G#, A → up the scale A, Ab, G, Gb, F, E, Eb, D, Db, C → down the scale 10/31/2020 Physics 214 Fall 2011 36

Ch 15 CP 4 cont. d) A, Bb, B, C → up to scale b) continued: A = 440 Hz Bb = A 1. 0595 = 466. 18 Hz B = Bb 1. 0595 = 493. 9177 Hz C = B 1. 0595 = 523. 3058 Hz A = 440 Hz Ab = A/1. 0595 = 415. 2902 Hz G = Ab/1. 0595 = 391. 9681 Hz Gb= G/1. 0595 = 369. 9557 Hz F = Gb/1. 0595 = 344. 1796 Hz E = F/1. 0595 = 329. 5701 Hz Eb = E/1. 0595 = 311. 0620 Hz D = Eb/1. 0595 = 293. 5932 Hz Db = D/1. 0595 = 277. 1054 Hz C = Db/1. 0595 = 261. 5435 Hz middle C = 261. 5435 Hz high C = 523. 3058 Hz high C/ middle C = 2. 00; Yes, high C is twice the frequency of middle C!! c) Cj = 264 Hz; Cet = 261. 5435 Hz Cj – Cet = 2. 46 Hz 10/31/2020 Physics 214 Fall 2011 37

Electromagnetic waves EM waves consist of oscillating magnetic and electric fields transmitted through vacuum at a constant speed of c = 3 x 108 m/s. They are produced whenever there is a changing electric or magnetic field. The acceleration of electric charge produces EM waves such as a broadcast antenna, AC wiring and lightning. A simple EM wave in vacuum has f = c/λ where c = 3 x 108 m/s In a medium like glass the velocity decreases, f stays the same so λ also decreases and f = v/λ The factor for the decrease is the refractive index which for glass is typically 1. 4 to 2. 0 http: //www. physics. purdue. edu/class/applets/phe/emwave. htm 10/31/2020 Physics 214 Fall 2011 38

Electromagnetic spectrum Almost all the information we receive from outside the earth is in the form of EM radiation. Different parts of the spectrum correspond to different physical processes We can understand what is going on in the Universe and back in time to near the beginning of the Universe using a variety of earth and space based telescopes. 10/31/2020 Physics 214 Fall 2011 39

The telescopes 10/31/2020 Physics 214 Fall 2011 40

How do we see color An image is formed at the back of the eye like a camera and there are receptors called cones that respond to different wavelengths. The brain interprets the mixture of the three signals as color. If we look at an object the color we see is the reflected light so the other colors were absorbed. If we are looking through a colored object then the object lets that color be transmitted and the other colors are absorbed 10/31/2020 Physics 214 Fall 2011 41

Refractive Index The speed of electromagnetic waves in vacuum is 3 x 108 m/sec and no energy can be transferred faster than this speed. When an electromagnetic wave passes through a transparent medium such as glass the speed is reduced by a factor n that we call the refractive index. If the speed in glass is v then n = c/v. The frequency of the wave does not change so that the speed is reduced by the factor n and therefore since c = fλvacuum and v = fλglass then c/v = n and λvacuum/λglass = n and λglass is smaller than λvacuum 10/31/2020 Physics 214 Fall 2011 42

Interference When two coherent beams of light are brought together they will add just like two waves on a string. So when two peaks coincide the light will be a maximum and when a peak coincides with a valley there will be no light. If we use two slits then as we move on the screen the path difference between the two beams will vary and we will get bright and dark fringes. 10/31/2020 Bright fringes are located at positions given by dy/x = nλ where n is an integer Dark fringes occur when dy/x = nλ/2 where n is an odd integer Physics 214 Fall 2011 43

Thin film interference occurs when light is reflected from the top surface and the underneath surface. This provides the two beams of coherent light that interfere. Since we normally observe this with white light we see colors because the path difference varies depending on the angle of observation So different wavelengths (colors) have constructive and destructive interference at different places on the film. 10/31/2020 Physics 214 Fall 2011 44

Diffraction Interference occurs even for a single aperture and this is called diffraction. The pattern shown is from a square aperture and the pattern can be thought of as light from different parts of the aperture interfering. As the aperture is made smaller the pattern actually expands. This effect can limit our ability to see detail in small objects or to resolve two stars nearby to one another. 10/31/2020 Physics 214 Fall 2011 The position of the first dark fringe is given by y = λx/w. This means the central bright fringe has a width ~ 2 y and increases as w gets smaller 45

Diffraction grating If we have multiple slits very close together maybe several 1000/centimeter the diffraction from each slit adds in a way that produces very narrow and bright fringes. For small angles the location of the bright fringes is given by dy/x = mλ where m is an odd or even integer 10/31/2020 Physics 214 Fall 2011 46

Polarization In a normal beam of light the electric field vector points in all directions. There are materials like polaroid that will only transmit light with the electric field vector along a specific direction. Light reflected by water also is polarized so polaroid sun glasses at the right orientation will block the reflected light. Transparent objects under stress can also cause transmitted light to be polarized. Car windshields are one example where patterns can be seen when wearing polaroid sun glasses 10/31/2020 Physics 214 Fall 2011 47

Summary of Chapter 16 f = c/λ c = 3 x 108 m/s Refractive index c = fλvacuum and v = fλglass then c/v = n and λvacuum/λglass = n and λglass is smaller than λvacuum Polarization Thin film interference 10/31/2020 Physics 214 Fall 2011 48

Interference Bright fringes are located at positions given by dy/x = nλ n is an integer Dark fringes occur when dy/x = nλ/2 where n is an odd integer Diffraction grating dy/x = mλ where m is an odd or even integer The position of the first dark fringe is given by y = λx/w. This means the central bright fringe has a width ~ 2 y and increases as w gets smaller 10/31/2020 Physics 214 Fall 2011 49

7 B-11 Color in Thin Films Studying thin-film interference effects What kind of patterns appear on the screen ? Why are the effects on white light and monochromatic light different ? LIGHT REFLECTED FROM THE FRONT AND THE BACK OF A THIN SOAP FILM INTERFERES WITH ITSELF. INTERFERENCE OF MONOCHROMATIC LIGHT PRODUCES BRIGHT AND DARK FRINGES WHILE INTERFERENCE OF WHITE LIGHT PRODUCES COLORED BANDS (DIFFERENT FREQUENCIES OF LIGHT INTERFERE DIFFERENTLY). 10/31/2020 Physics 214 Fall 2011 50

7 B-15 Line Spectrum Identifying the emission line spectrum of specific elements What do you expect to see with the diffraction grating ? THE COLOR WE SEE IN THE TUBE (WITHOUT PASSING THROUGH THE GRATING) IS A COMPOSITE OF COLORS. THE COLORS ARE SEPARATED (DIFFERENT COLORS INTERFERE AT DIFFERENT POSITIONS) BY GRATING AND PROVIDE UNIQUE IDENTIFICATION OF ELEMENT. 10/31/2020 Physics 214 Fall 2011 51

Coatings for lenses As light passes from one transparent medium to another a few percent of the light will be reflected. This is a particular problem in optical systems like lenses where there may be many glass elements. For example if 96% of the light is transmitted at a surface after 8 surfaces only 72% of the light remains and the other 28% will be scattered everywhere. Thin coatings are put on glass surfaces so that for particular wavelengths the light reflected from the top surface is exactly cancelled by the light from the bottom surface. This is only true for a single wavelength and to reduce the reflections for a range of wavelengths requires multiple thin film layers very often just λ/4 thick. This is why camera lenses usually have a blue green color. Often binocular lenses are ruby red. 10/31/2020 Physics 214 Fall 2011 52

Questions Chapter 16 Q 3 Is it possible for an electromagnetic wave to travel through a vacuum? Yes Q 11 Skylight is produced by scattering of light from the direct beam coming from the sun. Why is the color of the sky different from the color of the light of the sun itself? Blue light is scattered more strongly than red light 10/31/2020 Physics 214 Fall 2011 53

Q 12 Two waves interfere to form fringes in Young’s double-slit experiment. Do these two waves come from the same light source? Yes. They must be coherent Q 13 If two waves start out in phase with one another, but one wave travels half a wavelength farther than the other before they come together, will the waves be in phase or out of phase when they combine? They will be exactly out of phase 10/31/2020 Physics 214 Fall 2011 54

Q 15 When light is reflected from a thin film of oil on a water puddle, the colors we see are produced by interference. What two waves are interfering in this situation? One ray is reflected by the top of the oil, the second from the oil/water interface 10/31/2020 Physics 214 Fall 2011 55

Ch 16 E 4 X-rays often have wavelength of about =10 -10 m. What is the frequency of such waves? f = c/ = 3 1018 Hz 10/31/2020 Physics 214 Fall 2011 56

Ch 16 E 8 Light of 500 nm is reflected from a thin film of air between two glass plates. The thickness of film: d=1 m=1000 nm. a) How much farther does light reflected from bottom plate travel than that reflected from top plate? b) How many wavelengths of light does this represent? top bottom a) Path difference = 2 d = 2000 nm b) 2000 nm/ = 4 4 wavelengths of light 10/31/2020 Physics 214 Fall 2011 d 57

Ch 16 E 12 Diffraction grating has 1000 slits ruled in space of 1. 4 cm. What is the distance d between adjacent slits? ∙∙∙ d d 1. 4 cm 1000 d = 1. 4 cm (see picture) d = 1. 4 cm/1000 = 0. 014 m/1000 = 1. 4 10 -5 m 10/31/2020 Physics 214 Fall 2011 58

Ch 16 E 14 When passed through a diffraction grating of slit spacing 4 10 -6 m, the first-order fringe, for light of a single wavelength , lies 29. 0 cm from center of screen 2. 0 m away from grating. What is ? y = 0. 24 m X = 2. 0 m grating d = 4 10 -6 m d y/x = m , m=1 because this is first order fringe = dy/x = (4 10 -6)(0. 29)/2. 0 = 5. 8 10 -7 m = 580 nm 10/31/2020 Physics 214 Fall 2011 59

Ch 16 CP 2 Light of wavelength =600 nm passes through a double slit with d = 0. 03 mm. The resulting fringe pattern is observed on a screen 1. 2 m from the double slit. a) b) c) d) How far from the center of the screen is the first bright fringe? Second bright fringe? First dark fringe? Sketch a picture of the central seven bright fringes (central fringe & 3 on either side). Clearly indicate the distance from each fringe to the center of the screen. d = 0. 03 mm 10/31/2020 y 1 y 2 z 1 x = 1. 2 m Physics 214 Fall 2011 z 1 - first dark fringe y 1, y 2 - first and second bright fringes 60

Ch 16 CP 2 cont. a) d y/x=n or yn= n x/d = n (1. 2)/(0. 03 10 -3) (600 10 -9) yn=n(0. 024)m y 1 = 1(0. 024) = 0. 024 m = 2. 4 cm b) y 2 = 2 (0. 024) m = 4. 8 cm y 3 = +7. 2 cm c) z is spacing nth dark fringe dz/x = (n-1/2) or zn = (n-1/2) x/d zn = (n-1/2) (0. 024 m) z 1 = (1 -1/2)(0. 024 m) = 1. 2 cm y 2 = 4. 8 cm y 1 = +2. 4 cm y=0 y-1= -2. 4 cm d) y 1 = 2. 4 cm y 2 = 4. 8 cm y 3 = 3(0. 024 m) = 7. 2 cm +4 -4 y-2 = -4. 8 cm y-3 = -7. 2 10/31/2020 Physics 214 Fall 2011 61

Ch 16 CP 4 Soap film has index of refraction n=1. 333. This means that the wavelength of light in the film is reduced by a factor of 1/n. Index of refraction for air = 1. 0. a) If light has = 600 nm in air enters the soap film, what is the new ? b) Film is 900 nm thick. How many wavelengths farther does light reflected off bottomyou of film travel than off top? c) Would be surprised tothat findreflected the thickness produces destructive interference for reflected light? 10/31/2020 Physics 214 Fall 2011 62

CH 16 CP 4 cont top air n = 1. 0 a) f = 600 nm/1. 333 = 450 nm b) Light reflected from bottom surface travels 2 (900 nm) farther. 2(900 nm)/ f = 4. = 600 nm bottom film n = 1. 333 f = 450 nm In the film this is a 4 full wavelengths. 900 nm c) The above analysis determines that the reflected light should constructively interfere. That is, the light reflected from the bottom surface is in-phase with that reflected from the top. However the actual reflection process introduces a phase change between the two waves so the waves are in fact out of phase. 10/31/2020 Physics 214 Fall 2011 63