waves_05 ELECTROMAGNETIC WAVES

Transcription

1 waves_05 ELECTROMAGNETIC WAVES 1

2 waves_05: MINDMAP SUMMARY - ELECTROMAGNETIC WAVES Electromagnetic waves, electromagnetic radiation, speed of light, electromagnetic spectrum, electric field, magnetic field, reflection, refraction, refractive index, Snell s Law, superposition principle, constructive interference, destructive interference, thin film interference, conditions for constructive & destructive interferecne, path length difference, phase difference, phase changes upon reflection, wave particle models, photons, E = hf, energy [ J and ev], absorption and emission of light, ground state, excited states c = m.s -1 c n n sin v n sin c f E h f f o n f 2d 2d n 2 2 f f o 2

3 ELECTROMAGNETIC WAVES Transverse wave Oscillating quantities: electric and magnetic fields, in phase, perpendicular to each other and direction of propagation Can travel through vacuum, speed is c = 3.0 x 10 8 m.s -1 In transparent medium, speed is v = c / n where n is a property of the medium called refractive index When electromagnetic waves are emitted or absorbed by an atom, done so in quanta of energy: E = h f CP 551 3

4 Increasing frequency Electromagnetic Spectrum c = f E = h f visible Gamma rays X-rays Ultraviolet Visible Infrared Microwaves Radio Increasing wavelength CP 588 4

5 REFRACTION How can we see? What produces an image in an optical microscope? What happens when light passes through a transparent material? dispersion CP 588 5

6 The ratio of the speed of an electromagnetic wave, c in vacuum to that in the medium, v is defined as the refractive index, n n c v When a wave enters a new medium (different wave speed) at an angle the wavefront must bend. This bending is called refraction. The amount of bending is described by Snell's Law (Law of Refaction) n sin n sin CP 588 6

7 n 1 < n 2 normal 1 n n 2 n sin n sin CP 588 7

8 n 1 normal n 1 normal reflection 1 1 reflection n 2 n n sin n 1 < n 2 sin 1 > sin 2 refracted ray bent towards normal n sin n 1 > n 2 sin 1 < sin 2 refracted ray bent away from normal Critical angle sin 2 = 1 sin 1 = sin C = n 2 / n 1 total internal reflection CP 8588

9 9 1 2 medium 1 medium n n v v f f Plane wave incident upon an interface x I +X 1 1 sin 2 2 x t y A T sin 2 2 I I I x x x t if x x y A T

10 Plane wave incident upon an interface medium n v f n 1 2 v f 1 2 medium 2 normal 1 2 Snell s Law n sin n sin

11 Refraction When light enters the eye, most of the light is bent at the cornea and fine adjusts by the lens to focus the light onto the retina. Glasses correct for eye defects to produce a focussed image. In a transmission optical microscope light passes through a sample and the path of the light is bend by the various lens. The image gives a "map" of the refractive index variation throughout the sample. Fiberoptics bending of light through a glass fibre (total internal reflection). n sin n sin CP

12 THIN FILM INTERFERENCE CP

13 THIN FILM INTERFERENCE Thin films are responsible for colours of soap bubbles, oil sticks, iridescence of peacock feathers, blooming of camera lenses. When light impinges on the first surface of a transparent film, a portion of the incident wave is partially reflected and partially transmitted. The transmitted portion is then reflected from a second surface and emerges back out of the film. Thus, emerging from the thin film are two waves (1) wave reflected from front surface and (2) wave reflected from rear surface. The two waves have different path optical lengths that is determined by the width of the film. The two waves will eventually interfere and the interference pattern observed will depend upon the thickness of the film. Consider a thin oil film with varying thickness. Whenever the film is exactly the right thickness for the two waves of emerging red light to undergo constructive interference, the film will appear red at this location and the same for the other colours for different thicknesses in different locations. Thus, the oil film will show its characteristic with multicoloured fringe pattern when viewed under white light. CP

14 Review: Interference We have seen examples already: beats and standing waves When superimposing (adding) waves we sometimes get constructive interference (max amplitude), destructive interference (zero amplitude), and in between situations. Consider two sources which are in phase (coherent), emitting waves of the same amplitude At any location the resultant disturbance is the sum of the individual waves at that point. In some places there will be maxima (constructive interference) In other places there will be zeros (destructive interference) CP

15 Review: Interference For every wavelength path difference there is 2 phase difference What determines the type of interference? Constructive interference - why? Two sources, in phase (coherent) 15 CP 555

16 Interference of waves - water waves Interference of waves - light fringes CP

17 n 1 n f incident light light reflected at front surface light reflected from rear surface THIN FILM INTERFERENCE n 1 < n f 1 = n 1 > n f 1 = 0 thickness d optical path length 2dn f Phase changes upon reflection n f < n 2 2 = n f > n 2 2 = 0 n 2 CP

18 What physics is displayed in the animation? Can you explain what is happening? 18

19 The reflected wave from the rear surface travels an extra distance 2d (path difference) before it is superimposed and interferes with the wave reflected from the front surface. The wavelength f of the wave in the film is different from that in a vacuum o f c v c 0 0 f f nf f nf f o n f The optical path length is (2d) n f The phase difference between the two waves is 2d 2d n 2 2 f f o The 's are determined from the reflections at the interfaces. Remember a pulse travelling down a thin string is reflected with a phase shift of rad (inverted) at the interface with a heavy string. So a reflected light wave has a change of phase when it is incident upon a material that has a greater refractive index (optically more dense). CP

20 2dnf o determine the value of the 1 and 2 at each interface Path length difference Phase changes on reflection For constructive interference = m ( 2 ) rad m = 0, 1, 2, 3,. For destructive interference = (2m+1)( ) rad Thin film destructive interference is the principle behind coating optical surfaces such as lenses to reduce reflections so more light then can be transmitted. An uncoated air-glass interface reflects ~ 4% of the incident light. In a multiple lens system, the loss of transmitted light can be significant. Often lens are coated with magnetism fluoride to reduce reflections to produce a gain in the transmission of the lens system. For thick films it becomes possible for many different colours to have their maxima at the same locations. Where many wavelengths can interfere constructively at the same time, the reflected colour becomes increasing unsaturated and the fringe contrast disappears. CP

21 Soap film fringes illuminated by white light Why does the film look dark at the top (where it is thinnest)? 2dnf o air water air CP 21563

22 Reflection from soap films Film has higher refractive index than medium on either side (air) At first interface refractive index increases, at second refractive index decreases In this case there is (180) phase difference between the reflected beams, in addition to phase difference due to path difference. 1 = 2 = 0 What is the thickness of the soap film? constructive interference m(2) = 2 (2d n f / o ) - d destructive interference (2m+1) = 2 (2d f / o ) - d 1 ( m 2) 0 n m0 m1, 2,3, n 2 f 2 f m 0,1, 2,3, 2dnf o CP 22563

23 Anti-reflection coating Thin film of material with refractive index between that of air and glass on a lens. Choose thickness so there is destructive interference between light reflected from each interface Why do such lens look purple? Anti-reflection coating: destructive interference The film thickness d is equal to an odd number of quarter wavelengths f Waves reflected from the two interfaces are (180) out of phase CP 23563

24 Anti-reflective coating You want minimum reflection 2dnf o air n 1 = 1 1 = 2-1 = 0 2 = Destructive interference thinnest film film n f d = 2 (2d n f / o ) + 0 glass n 2 d = o / (4 n f ) = f / 4 n 1 < n f < n

25 Problem solving strategy: I S E E Identity: What is the question asking (target variables)? What type of problem, relevant concepts, approach? Set up: Diagrams Equations Data (units) Physical principals PRACTICE ONLY MAKES PERMANENT Execute: Answer question Rearrange equations then substitute numbers Evaluate: Check your answer look at limiting cases sensible? units? significant figures? 25

26 Problem 1 PHYS 1002 Exam, 2002 In air (n = 1.00) light is incident normally on a thin film with an index of refraction n = The film covers a glass lens of refractive index What is the minimum thickness of the film to minimise reflection of blue light (400 nm)? n n n 1 f 2 n 1 = 1 thin film nf 1.25 d n 2 = 1.45 [Ans: d =80 nm] 26

27 Solution 1 o = 400 nm n n n 1 f 2 n 1 = 1 thin film n 1.25 f d n 2 = 1.45 Min reflection 2dn f o d o 4 n f 80 nm 27

28 Problem 2 Light is incident normally on a thin film with an index of reflection n =1.35. The film covers a glass lens of refractive index 1.5. What is the thickness of the film to maximise the reflection of red light (633 nm)? 28

29 Solution 2 I S E E n 1 = 1 n f = 1.35 n 2 = 1.5 o = 633 nm = m you need to add diagram To maximise reflections we need the reflected waves to interfere constructively. 2dnf rad o reflection at front interface n 1 < n f 1 = reflection at rear interface n f < n 2 2 = 2-1 = 0 can ignore the total phase change due to interface reflections 2dnf 2 2 rad o f 9 o d = m 234 nm 2 n (2)(1.35) 29

30 Problem 3 PHYS 1002 Exam, Question 12 (a) (b) (c) (d) Explain the meaning of the concepts of constructive and destructive interference when applied to two monochromatic waves. You can draw diagrams showing the waves produced by two sources as part of your answer. When light reflects off a surface it can have zero change in phase or a change in phase. What is the significance of the refractive index in determining this change upon reflection? A fused silica lens of refractive index 1.46 is surrounded by air. Explain why a quarter wavelength thick magnesium fluoride (refractive index of 1.38) coating over the front surface of the lens can reduce reflections and hence increase the amount of light transmitted through the lens. Light from a helium neon laser (wavelength 633 nm) is normally incident upon the coated lens described in part (c). What is the minimum thickness of the film that will result in minimum reflected intensity [Ans: d = 115 nm] 30

31 Problem 4 Soap film fringes illuminated by white light Why does the film look dark at the top (where it is thinnest) 31

32 Electromagnetic waves A progressive electromagnetic wave is a self-supporting, energy-carrying disturbance that travels free of its source. The light from the Sun travels through space (no medium) for only 8.3 minutes before arriving at Earth. Each form of electromagnetic radiation (radiowaves, microwaves, infrared, light, ultraviolet, x-rays and rays) is a web of oscillating electric and magnetic fields inducing one another. A fluctuating electric field (electric charges experience forces) creates a magnetic field (moving charges experience forces) perpendicular too itself, surrounding and extending beyond it. That magnetic field sweeping off to a point further in space is varying there, and so generates a perpendicular electric fields that spreads out. Nothing is actually displaced in space like a water wave where the water oscillates up and down and side-ways. 32

33 All electromagnetic waves propagate in vacuum at exactly speed of light c = m.s -1 This is a tremendous speed, light travels 1 m in only s. "There are only two fundamental mechanisms for transporting energy and momentum: a streaming of particles and a flowing of waves. And even these two seemingly opposite conceptions are subtly intertwined there are no waves without particles and no particles without waves " from Hecht. The particles associated with electromagnetic waves are called photons. The energy of a single photon is E = h f E energy of photon (J) electron volt 1 ev = J f frequency of electromagnetic wave (Hz) h = J.s Planck's constant (J.s) 33

34 Absorption and emission of light The emission and absorption of light (transfer of energy & momentum) takes place in a particle manner. All forms of electromagnetic radiation, interact with matter in the process of emission and absorption. The radiation propagates in a wavelike fashion but in an interaction the radiation behaves as a concentration of energy (photons) moving at the speed of light. Each photon carries very little energy. However, even an ordinary torch beam is a torrent of ~10 17 photons.s -1. When we see light what we observe by eye or on film is the average energy per unit `area per time arriving at some surface. 34

35 Most of the chemical and optical properties of a substance are dependent upon the outer most bound electrons in atoms. Each electron is usually in its ground state the lowest energy state. Only when an atom absorbs specific and sufficient energy can the atom move to a well-defined higher energy state (excited state). energy excited states ground state 35

36 Resonant Absorption E 2 E 1 h f = E 2 - E 1 E 2 E 1 h f E 2 - E 1 36

37 Spontaneous emission E 2 h f = E 2 - E 1 E 1 Excited state Oscillating mixed state Ground state ~ 10-9 s later 37

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40 Radiation f (Hz) (m) Photon Energy (ev) Comments power lines Currents. Radiation emitted from 50 Hz power lines are they dangerous? radio AM FM TV < 10 9 > 0.3 < Currents, electronic circuits. AM FM radio, TV. microwaves Magnetron. ~ 10 mm to 30 m can penetrate atmosphere satellite communications, mobile phones, police radar, microwave cooking (polar water molecule), medical diathermy. 40

41 Radiation f (Hz) (m) Photon Energy (ev) infrared light ultraviolet Comments Molecular vibrations & rotations, all objects. Why do you get warm standing in the sun? Thermography, IR satellite images, detecting tumors & cancers. Outer electrons, incandescent lamps, lasers, arcs. Eye sensitive to this narrow band. White light mixture of all colours of visible spectrum. TV (red blue green). If a an uranium atom enlarged to a size of a pea, red ~ 20 m long. Sunlight ~ photons.s -1.m -2. Blue light photons sufficient energy to disrupt chemical bonds. Inner & outer electrons. Ionising (4 ev photons break C-C bonds). Skin cancers, tans the skin, vitamin D synthesis, damage to eyes. O 3 layer absorbs UV protection layer. Can kill microorganisms. 41

42 Radiation f (Hz) (m) Photon Energy (ev) x-rays Comments Inner electrons, x-ray tubes. Diagnostic medical-rays. Computed tomography CT scan. Crystal & molecular structures. -rays > > Nuclei, accelerators. Sterlisation, food preservation, cancer treatment, medical diagnosis, flaw detection. Very penetrating can pass through large thickness of concrete. 42