Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Page 1 of 46 AP Physics – Electromagnetic Waves, Reflection and Refraction of Light – Chapter 23-24 Chapter 23 - Reading pp. 719-739 - text HW #17, 18, 24, 27 Chapter 24 - Reading pp. 748-771 – text HW #6, 7, 8, 9, 12, 13, 17, 18, 19, 20, 24, 28, 29, 32, 33, 36, 37, 38, 39, 42, 30(E.C.) The story so far… In the last chapter, we studied mechanical waves such as sound waves which required a matter medium for propagation. And now… 23.1- Electromagnetic Waves- waves resulting from the vibration of charged particles. Such waves are the result of: 1) the changing electric field around the oscillating particle and; 2) the changing magnetic field induced by the changing electric field. Since these vibrations manifest themselves in the field, they do not require a matter medium for transmission. James Clerk Maxwell – 1865 – developed a theoretical mathematical theory that showed the above relationship between electrical and mathematical phenomena. Maxwell’s four equations allowed him to derive the speed of propagation for EM Waves in a vacuum, which is: c=2.99792458 x 108 m/s. Wow. Maxwell’s theoretical EM waves would travel at the speed of light! This means: Light is an Electromagnetic wave! Page 2 of 46 23.2 - This led to the discoveries of Heinrich Hertz in 1887. He used a variable AC circuit that could alternate the charge on the two metal spheres on his transmitter circuit (left), causing a spark to jump at different frequencies. As he adjusted the frequency of the receiver by changing the spacing between the brass spheres, he found that he could get a spark to jump across the gap in the receiver. Hertz invented the first Radio. Transmitter Receiver Note: 1) In air, light travels at approximately c=3 x 108 m/s (for calculations in air, use speed of light in a vacuum) 2) In denser optical media (water, glass, diamond) Light and EM waves move SLOWER than c. Page 3 of 46 23.5 - The Electromagnetic Spectrum Which has the highest frequency? Gamma rays Which has the lowest frequency? Radio waves Which has the largest wavelength? Since c=fλ, radio waves Which has the smallest wavelength? Gamma rays Which has the most energy? Gamma rays Why? E = hf Which has the least energy? Radio waves Note: for mechanical waves (and harmonic oscillators like the pendulum and mass spring system), energy affects the amplitude of the vibration, only. Whereas a higher frequency for a mechanical wave yields the same energy per oscillation, a higher frequency for EM waves means MORE ENERGY per oscillation, if you think of light as a photon of course, but I digress… Back to light as a wave… Fluoroscope Videos Page 4 of 46 The Visible Light Spectrum- Some important facts: 1) ROYGBIV 2) Wavelength range of visible light in air/vacuum: In nanometers (1x10-9m): 400-700 nm (approximate) Note: 400nm = Violet, 700 nm=Red In Angstroms (1x10-10m): 4000-7000 Å 3) Frequency range of visible light in air/vacuum: 4.3 x 1014 - 7.5 x 1014 Hz (approximate) Note: What is the wavelength and frequency range of visible light in water? 300-500 nm (approximate) 4.3 x 1014 - 7.5 x 1014 Hz (approximate) Therefore, the color/characteristic of a wave is determined by the wave’s FREQUENCY. Now go home and do HW Chapter 23 -- #17, 18, 24, 27 Page 5 of 46 AP Physics - Reflection and Refraction of light - Chapter 24 Chapter 24 - Reading pp. 748-771 – text HW #6, 7, 8, 9, 12, 13, 17, 18, 19, 20, 24, 28, 29, 32, 33, 36, 37, 38, 39, 42, 30(E.C.) 24.2 Measuring the Speed of Light in Air – Precisely measured in 1926 by Albert Michelson, after the predecessor to the US Geologic Survey painstakingly measured the baseline distance from the Mount Wilson Observatory in Los Angeles to Old Baldy, 22 mi away (35 km) and he was still off by about 48000 m/s. How did Michelson use this apparatus? The rotating mirror had to be in the exact orientation above to properly reflect light from the light bulb to the telescope. The frequency of rotation was increased until the point when he could first see the reflection of the light through the telescope. (That frequency was 32000 rpm, or 533.33 Hz, a period of .001875 sec/rotation, or a time of .001875/8=.000234 sec for 1/8 a rotation. This yields a v=70000m/.000234sec=2.9866x108 m/s) Page 6 of 46 (using 3x108m/s, t=2.3933x10-4s, ω=3281.65Rad/s or 522Hz or 31320rpm) (this would be a quarter turn in same time, or double the frequency) Also, read about the first successful terrestrial (non astronomical) method (1849) FIZEAU’S technique for calculating the speed of light in air on page 752. Page 7 of 46 24.4-24.5 Reflection of Light When a light ray is incident upon a boundary leading into a second medium, it is reflected back into the first medium, governed by the law of reflection. θi=θr (measured to a line normal to the boundary) Regular reflection and diffuse reflection? DEMO EACH (Laser on paper and on mirror and smoke machine) #1) Find the angle of reflection off of mirror M2 Now go home and do HW Chapter 24 #6,7,8 Page 8 of 46 24.5 REFRACTION I. What happens when a ray of light traveling through a transparent medium encounters the boundary leading into another transparent medium? 1. Partial Reflection 2. Partial Absorption 3. Partial Transmission – this part may be refracted. Why do materials look darker when they are wet? If your shirt gets wet, the light that is illuminating the shirt now has to pass through a layer of water. Since some of the light is reflected and some gets absorbed, less light illuminates the shirt, less light reflects off the shirt, and therefore the shirt looks darker. Refraction – the bending of a wave as it travels from one medium to a different medium density at an angle. (as it bends, its speed also changes) DEMO: Blackboard optics or beaker of Water / Laser The Index of Refraction of a transparent medium is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium c n v Indices of refraction measured with λ = 589 nm (Yellow line in bright-line spectrum of sodium) (Text Ref. Table 24-1) Air n = 1.000293 Water n = 1.333 Crown Glass n = 1.52 Flint Glass n = 1.66 Diamond n = 2.419 Zircon n = 1.923 Page 9 of 46 #2) What is the velocity of light (λ = 589 nm) in A) air? B) crown glass? C) diamond? #3) The speed of light in a material is 1.807 x 108 m/s. That material may be ___. #4) What material may have an index of refraction of 0.6? Therefore, how is the index of refraction “n” related to the speed in that medium? #5) In the table of indices of refraction, why was in necessary to mention that the values were those measured with λ = 589 nm? Some media are DISPERSIVE, which means waves of different wavelengths travel at different speeds in that medium. This means that: Index of refraction is a function of the wavelength. Hmmm…. Why is index of refraction not a function of frequency? Try to derive an equation with n and f in it. v1 n c v f f o o 1 or 2 n1 1 n2 2 v2 f f v1 v2 c c n1 n2 n2 n1 Page 10 of 46 #6)Draw the path as the ray enters the water Compare the following: v1 > v2 f1 = f2 λ1 > λ2 *Remember: when a ray is refracted, the wave’s frequency and phase remain the same. Refraction – Which way does it bend? When a light ray goes from low n to high n, its speed ________ and the ray bends TOWARDS the normal. When a light ray goes from high n to low n, its speed ________ and the ray bends AWAY from the normal. Gruber says: remember these two important memory tricks: HIGH to LOW, AWAY we go! LOW to HIGH, Stay close BY! Page 11 of 46 Recall, Willebrord Snellius’ Law (1621) n1sinθ1 = n2sinθ2 #7) For each light ray (λ=589 nm), draw the path and find the angle of refraction. Page 12 of 46 #8) Trace the path Page 13 of 46 #9) A monochromatic ray with wavelength of 6.5 x 10-7 m in air is incident upon a flint glass boundary. The angle of incidence is 40o. A) Sketch the path and find the angle of refraction. B) Find the speed of light in the flint glass (n=1.66). C) Find the frequency in the flint glass. D) Find the wavelength in the flint glass. E) What color is the light in air? F) What color is the light in the flint glass? Note: We know that n c v n o n Page 14 of 46 What other ratio is n equal to? Now go home and do HW Ch. 24 # 9,12,13,17,18,19,20,24,28,29 #10) A ray of light enters and exits a glass of water. θ = 20° A) Trace the path as light enters and exits the glass of water. Show all angles. B) What is the time it takes for the light to pass through the entire glass of water? Page 15 of 46 24.6 Dispersion and Prisms What is white light? White light is what the human eye perceives when all three cones in the eye are hit by a combination of all visible wavelengths. Review: Which color in the visible spectrum has the Longest wavelength ______ Shortest wavelength ______ Recall that the index of refraction is a function of wavelength. This means that light of different wavelengths will be bent at different angles when entering a dispersive refracting material. So… What happens when white light enters a glass prism? It undergoes DISPERSION. Dispersion – The breaking up of polychromatic light into separate monochromatic rays as it passes into a dispersive medium or prism. Page 16 of 46 According to the above graph, Index of refraction decreases with increasing wavelength. This means that VIOLET light bends the MOST VIOLENTLY, and Red light bends the least. Trace the path as the white light enters and exits the prism. In the prism, which color travels the fastest? slowest? 24.7 How are rainbows made? Double Rainbows? Video – how rainbows are made. Video – Whoa… Double Rainbow… Why? Video – How double rainbows are made http://www.youtube.com/watch?v=FJVvtOy-ukE 22:50-40 min Now go home and do HW ch. 24 #32,33 Page 17 of 46 24.9 Total Internal Reflection Total internal reflection occurs when a ray, moving from a medium of high n to low n, is incident at an angle greater than the critical angle. The critical angle, θc, of a medium is the angle that yields an angle of refraction = 90º #11) Monochromatic light travels from air into water (n=1.33). Complete the chart for the following angles and draw each refracted ray. θincident 0° 25° 48.7534664° 89° Θrefracted Hence, when light is traveling from n1 to n2 where n1 < n2 (low to high), refraction occurs for 0 < θi < 90. Monochromatic light travels from water into air. Complete the chart for the following angles and draw each refracted ray. θincident 0° 25° 48.7534664° 89° Θrefracted Hence, when light is traveling from n1 to n2 where n1 > n2 (high to low), refraction occurs only for 0 < θi < θc. Page 18 of 46 Critical angle (θc): When a ray moves from high n to low n at θi = θc, the refracted ray moves parallel to the boundary. Thus: If θi < θc then refraction occurs. If θi = θc then refraction occurs and θr = 90°. If θi > θc then total internal reflection occurs. Using Willebrord Snellius’ Law, derive the formula for θc in terms of n1 and n2, and the special case equation when n2 is air. http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/totintrefl/index.html DEMO: Show total internal reflection with laser and block. DEMO: Fiber-optic cables DEMO: Water/laser through hole in cylinder (Use Torricelli’s Law apparatus) DEMO: Beer mug filled with beer in air and water. Discussion on next page. Page 19 of 46 Explain this one: a frosty mug (soda-lime glass, n = 1.5) filled with your favorite beer, when viewed in air, looks fuller than when viewed in a medium such as water. In water, the very thick glass of the beer mug is noticeable. View from above: Hints: #12(a) For a θi =40°, find θr going from air into glass and from water into glass. (b) Find the critical angle for the glass-air boundary. Repeat for the glasswater boundary. Putting the mug in water changes the outer glass interface from an air-glass interface to a water-glass interface. This means there is ________ of a bend as light rays travel from water into the glass and travels through the wall. This also means that there is a _________ critical angle at the glass-water interface. When the ray hits the other side, there is a ________ chance of the incident light ray undergoing total internal reflection, so refraction ________, and then light that passed through the glass hits your eye. DEMO: Fiber-optic cables in water? Page 20 of 46 #13) Find the critical angle for a diamond–air boundary. Diamond ring and laser in and out of water. Glycerine? Now go home and do HW Ch. 24 #36,37,38,39,42,30(E.C.) Page 21 of 46 #14 (A) Trace the path as a beam of monochromatic light in air enters and exits the glass prism. (Show all angles) B) Repeat when the prism is submerged in water (n = 1.33) C) What would be the index of refraction of an unknown liquid that the prism would need to be dropped into such that the ray would refract at 900 at the hypotenuse boundary? Page 22 of 46 Here’s another interesting question: If a fish is underwater, and I see it, does that mean it can see me? Conversely, if a fish is underwater, and it sees me, does that mean I can see it? #15) (A) A fish is 6 m below the water surface. Where does a person in the air have to look to see the fish? The critical angle makes a “cone of visibility” for seeing the fish. The person in the boat has to be in the area of radius R above the fish. (B) Determine R. Page 23 of 46 So what does the fish see? Swimming pool vid 1– Scuba diver surfacing vid 2 #16) motivated by his neighbor’s fake rock hide-a-key, Mr. Gruber decides he is going to build one using a coaster of diameter _____, hang the key from the center of it, and let it float around the pool. What is the maximum length that the key can hang from the coaster so that no one can see it from above the surface of the pool? Page 24 of 46 Inverse Square Law of Illumination Luminous Intensity – The Luminous strength of a light source in a particular direction. Roughly speaking, it’s how bright the source is. (Units: candela (Cd) or CandlePower). A 40-watt tungsten filament ≈ 35 Cd A 40-Watt fluorescent lamp ≈ 200 Cd Why? Fluorescent lamps are more efficient (heating a low pressure gas requires less thermal energy than a solid tungsten filament Note: Luminous intensity is not to be confused with intensity of illumination. Intensity of illumination (Unit – lumen (lm)) 1 lm is defined as the amount of light falling upon a 1-m2 area from a 1-cd light source a distance 1 m away. Note: I 1 2 r #17) (A) (B) At a distance of 2 m from the 1-cd light source, what is the intensity of illumination on a 1-m2 area? 0.25 lumen The intensity of radiation from the Sun is 9126 W/m2 at the distance of Mercury (0.387 Astronomical Units); Determine the approximate intensity of radiation at the distance of Earth (1 AU)? 1367 W/m2 Page 25 of 46 (C) Graph the intensity of a light source vs. the distance from the source. INTENSITY = POWER/AREA (SI units: Watt/m2) In the Modern Physics section, we will talk about Intensity as photons falling upon a unit area per unit time. Page 26 of 46 AP Physics – Wave Optics – Diffraction, Interference and Polarization – Chapter 26 Chapter 26 - Reading pp. 809-833- text HW #1,5,6,9,17,18,27,28,38,40 26.2 Diffraction and Young’s Double Slit Interference What is Diffraction? Diffraction is the spreading out of a wave when it passes through an opening (slit) or around a barrier Draw what happens to the parallel wave fronts when they encounter the break in the sea wall as shown: Draw what happens to the parallel wave fronts when they encounter two breaks in the sea wall as shown: Page 27 of 46 Remember interference? Interference: The resulting wave pattern that occurs when two waves pass through each other in the same medium What are the conditions in which 2 waves have maximum constructive interference? In phase, same frequency (and same amplitude helps too.) What are the conditions in which 2 waves have maximum destructive interference? 180° phase difference, same frequency, same amplitude. For points A, B and C in the interference pattern below, determine the following: Type of Interference Point A Constructive Point B Destructive Point C Constructive Distance from S1 3λ 3.5λ 4λ Distance from S2 4λ 4λ 4λ Path difference δ 1λ ½λ 0λ DEMO: Ripple Tank Simulator http://www.falstad.com/ripple/ Page 28 of 46 And now… Thomas Young’s Double Slit interference experiment (1801) There was much debate over the nature of light in the 1600s. Hooke, Huygens, and Euler proposed a wave theory of light, while Newton rejected wave theory and proposed a corpuscular (particle) theory of light (interesting). Of course, everyone accepted Newton’s Theory – duh – it was Newton – SMH. Anyway, when Young revisited the debate in the late 1700s, going against Newton, this was akin to heresy. But his double slit experiment shut everyone up, at least until 1901 when Max Planck resurrected Newton’s particle theory under the name “Quantum Theory”. (DEMO: Laser/Diffraction Grating) λ = wavelength d = distance between the 2 slits L = distance between slits and screen Y = distance from central maximum to first maximum Page 29 of 46 Derivation of the equation for path difference: δ = r2 – r1 = d sinθ = mλ For constructive interference m = 0, +/-1, +/-2, +/- 3… For destructive interference m = +/- (0+ ½), +/-(1 + ½)… Determine the m values: Page 30 of 46 Note: The textbook shows 2 formulas for the path difference; one formula for the constructive maxima, and another for the destructive maxima. It is easier to combine the cases using one formula. Book: constructive δ = dsinθ = mλ for m = 1,2,3,4... destructive δ = dsinθ = (m + ½)λ for n = 0,1,2,3,4… What does the intensity graph look like? Page 31 of 46 Derivation of the APPROXIMATION equation relating λ, d, x, L: This equation works as long as L>>d and d>>λ, because this keeps θ really small, and thus: sin θ ≈ tan θ m d xm L So, assuming all else stays constant: 1) As L decreases, x _____________ (move the screen, move the slits move laser) 2) As d decreases, x _____________ CD-R vs DVD-R reflection interference pattern. 3) As λ decreases, x _____________ Demo: Green Laser, Red Laser with Diffraction Grating Page 32 of 46 #18) Two slits are illuminated with coherent light of wavelength 560 nm. The first order interference maximum appears at 6o. (A)What is the distance separating the slits? (B)Also, what is the path difference? (C) If a diffraction grating were used, what is the number of lines per mm? #19) (A) Two slits separated by 1 x 10-5 m. If the distance between the slits and the screen is 1.2 m, and the distance from the central maximum to the first order maximum is 2.9 x 10-2 m, what is the wavelength of the light? (B) What is the distance from the central maximum to the 3rd order minimum? Page 33 of 46 #20) Two loudspeakers separated by 0.5 m and operating in phase send out identical sound waves. An observer who is 10 m from the speakers finds out the first sound maximum at a distance of 12.5 m as shown. (A) What is the wavelength of the sound using path difference equation? (.39 m) (B) What is the wavelength of the sound using the approximation equation. (.625 m) (C) Give justification for the discrepancy. For sin θ ≈ tan θ, both opposite sides must be small compared to their respective adjacent side/hypotenuse (X<<L, λ << d). Since both x and λ are relatively large, the small angle approximation breaks down. Page 34 of 46 #21) Light of wavelength 440 nm falls upon two slits spaced 0.31 mm apart. What is the angle between the first and second dark fringes? #22) Light of wavelength 700 nm is incident upon a diffraction grating that has 1000 lines/cm. If the diffraction grating is 3 m from the screen, find the distance from the central maximum to the 5th dark fringe. Now go home and do HW #1,5,6,9 Page 35 of 46 26.4 Interference in Thin Films I remember as a kid growing up in the 70s, walking around the neighborhood after a rain storm, puddles and wet ground everywhere looked like the picture to the right. Only now do I realize it was a sign of the times: old cars back then were notorious for leaking oil. In fact my 1973 Buick Le Sabre (in 1988) had three leaks I had to keep up with at once: motor oil, transmission fluid power steering fluid. Today I see less oil films on the road, but is it due to cars leaking less, or to New Yorkers buying new and keeping them for fewer years? Now, to finally get to the point: what’s up with those rainbows in the street, and in soap bubbles? Reflection Rule: i. When a ray of light reflects off of a surface, the reflected ray undergoes a 180o phase change at the boundary if n1 < n2 such as when n1 = air and n2 = water. ii. If n1 > n2 no phase change occurs Analogy: show reflected pulses Thick to thin string/Thin to thick string n1 n2 n1 n2 Page 36 of 46 THIN FILMS CASE I – phase inversion occurs at BOTH surfaces (So think of it as NO PHASE SHIFT): (A) Derive a general expression for the possible thicknesses t of the oil film such that the resultant intensity of the light reflected back into the air is a maximum (constructive interference) At minimum, to be in phase, make t= λfilm/2, then the wave in the oil has to travel 2t (1λfilm) to meet the wave reflected off the top surface. Thus: c f 0 m 0 m film 0 n t t film m=1, n 2n v f 2 film 2… (B) Derive a general expression for the possible thicknesses t of the oil film such that the resultant intensity of the light reflected back into the air is a minimum. (destructive interference) At minimum, to be 180° out of phase, make t= λfilm/4, so the wave in oil has to travel 2t (1/2 λfilm) to meet the wave reflected off the top surface. Thus: t m film 4 t m 0 4n m= 1, 3, 5… Page 37 of 46 (C) Calculate the minimum thickness t for the example above such that the resultant intensity of the light reflected back into the air produces: i. constructive interference ii. destructive interference THIN FILMS CASE II (Soap Bubble)-phase inversion occurs at one surface (So think of it as PHASE SHIFT) (A) Derive a general expression for the possible thicknesses t of the soap bubble such that the resultant intensity of the light reflected back into the air is a maximum (constructive interference) Because of the phase shift, to be in phase, make t= λfilm/4, then the wave in the soap bubble has to travel 2t (1/2 λfilm) to meet the wave reflected off the top surface. Thus: m 0 m film t t 4n m=1, 3, 5… 4 Page 38 of 46 (B) Derive a general expression for the possible thicknesses t of the soap bubble such that the resultant intensity of the light reflected back into the air is a minimum. (destructive interference) Because of the phase shift, to be 180° out of phase, make t= λfilm/2, so the wave in oil has to travel 2t (1λfilm) to meet the wave reflected off the top surface. Thus: t m film t m 0 2n m= 1, 3, 5… 2 (C) Calculate the minimum thickness t for the example above such that the resultant intensity of the light reflected back into the air produces: i. constructive interference ii. destructive interference #23) A thin film of oil (n = 1.26) coats the water in a marina (n = 1.33). If light in air with wavelength 500 nm strikes the oil, what are the three smallest thicknesses of the oil film such that the light reflected back in the air interferes constructively. Now go home and do Ch 26 HW # 17,18,27,28 Page 39 of 46 #24) 1984 AP Exam – Free Response #5 The surface of a glass plate is coated with a transparent thin film. A beam of monochromatic light of wavelength 6 x 10-7 m in air is incident normally on surface S1. The beam is partially transmitted and partially reflected. (A) Calculate the frequency of the light in the air. (B) Calculate the wavelength of the light in the thin film. (C) Calculate the minimum thickness d1 of the film such that the light reflected into the air is a minimum. (D) Calculate the minimum thickness d2 of the film such that the light reflected into the air is a maximum. Page 40 of 46 Application of thin films: Antireflective (AR) coating on eyeglasses. AR coatings serve two purposes: i. They reduce the amount of reflected glare, making the glasses look better on the wearer. ii. They increase the amount of light that passes through the lens. This is because the amount of reflection is related to the difference between indices of refraction. Going from air (n=1.00) into crown glass (n=1.52) is a big jump and leads to more reflection than going from air into a coating of MgF2 (n=1.38), or a fluoropolymer (n=1.30). MgF2 on a crown glass surface gives a reflectance of about 1%, compared to 4% for bare glass. MgF2 coatings perform much better on higher-index glasses, especially those with index of refraction close to 1.9. MgF2 coatings are commonly used because they are cheap, and when they are designed for a wavelength in the middle of the visible band they give reasonably good antireflection over the entire band. Page 41 of 46 26.6 Single Slit Diffraction of light When light is sent through a small single slit and the transmitted wave is projected on a screen, a diffraction pattern is established that is similar to the double slit interference pattern with one big difference: The single slit pattern has a larger central maximum (about twice the size of the next weaker bright band). This can be done in two ways: Fraunhofer Diffraction – lens used when screen is far away to focus the parallel rays. Fresnel Diffraction – no lens is used when the screen is close. DEMOS: Laser through single slit or through a pin hole in a piece of paper. Form Single Slit with 2 fingers! Why does this happen? Huygen theorized that as a wave front passes through a medium, each point hit by the wave produces a circular wave front that interferes with the other circular fronts produced in its vicinity. When the resulting wave front hits a barrier with an opening or slit in it, only the points inident on the slit can produce circular wave fronts. It is these points that act as individual point sources of waves. Hence, light from one portion of the slits can interfere with light from another portion. The light interferes with…itself. Whoa. Video demonstrating Huygens Principle Page 42 of 46 Derivation of Single Slit equation that relates slit length d with λ If we break the slit into 2 equal parts and analyze three coherent rays: o Rays 1 and 2: 180 phase difference o Rays 1 and 3: 0 phase difference i. If we choose a θ that relates d with λ, we get: sinθ = 2 d 2 Thus the condition for destructive interference is: dsinθ = 1λ ii. If we break the slit into 4 equal parts, sinθ = 2 d 4 or: dsinθ = 2λ (destructive) iii. If we break the slit into 6 equal parts, sinθ = 2 d 6 hence dsinθ = 3λ Formula for finding angle to 1st dark fringe using single slit: dsinθ = mλ m = +/-1, +/-2, +/- 3… Page 43 of 46 What does the intensity graph look like for the single slit. Remember: the single slit pattern has a larger central maximum (about twice the size of the next weaker nth order maximum). #25) Monochromatic light with a wavelength of 750 nm passes through a slit 1 µm wide. The distance from the screen to the slit is 20 cm. The first dark band appears at point p. (A) Calculate angle θ (48.6°) (B) What is the width of the central maximum? (θ gets you to the minima, so use tanθ = x/L) (0.23 m) Now go home and do Ch 26 HW – #38, 40 Page 44 of 46 26.7 Polarization – Polarization is the process of turning unpolarized light into polarized light. Polarization can ONLY BE DONE with transverse waves. Therefore, only electromagnetic radiation and light can be polarized. SOUND CANNOT BE POLARIZED, EVER, EVER, EVER!!!! What do electromagnetic waves really look like? http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35.0 Unpolarized light – Light with EM fields oscillating in more than one plane of vibration. Define: Polarized light – Light with with EM fields oscillating in a single plane of vibration http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=692.0 (show two waves, move around polarizers. Read directions to adjust parameters) What are polarizing filters made of? In 1938, Edwin H. Land discovered a material, which he called Polaroid, that could polarize light through selective absorption. He made a solid sheets of long-chained hydrocarbons, such as polyvinyl alcohol (PVA used in making slime), that are stretched during the manufacture process to align the molecules in one direction. The sheets are then dipped in a solution containing iodine; the iodine atoms impregnate the PVA molecules and provide electrons that can move along the length of the chain. Thus when light with an electric field vector E parallel to the hydrocarbon chains is incident on the sheet, the field causes electrons to move, thus absorbing the energy of the incident wave. Light with E perpendicular to the chains cannot be absorbed and passes through the sheet. Land is the founder of the Polaroid Corporation, which I remember as a kid because he invented the instamatic camera, which could take a picture, spit out the photographic sheet, and then we would all watch for sixty seconds as the picture developed in front of our eyes. Page 45 of 46 What happens when unpolarized light enters a vertical polarizer? Repeat for a horizontal polarizer. What happens when a horizontal and vertical polarizer are 90o apart? Practical applications 1. In polarized sunglasses- These types of sunglasses reduce glare. Glare is reflected light off of a surface that interferes with you actually seeing the surface. It is important to note that glare is polarized as it is reflected off a surface from above some extreme angle that is surface dependent. If glare is reflected off of a horizontal surface, it is horizontally polarized, and off of a vertical surface, it is vertically polarized. For drivers since most glare is road glare, how would the polarizers in the glasses need to be oriented ? Page 46 of 46 3d-movie technology- Anyone who has seen an Imax 3d movie or has been to Universal’s Islands of Adventure to take a ride on “the Amazing Adventures of Spiderman” knows that modern 3d technology seems so lifelike compared to the old red/blue 3d technology from the early 1900s. The recording system is expensive because it employs 2 cameras, each filming the same image through separate polarizers The projector has a rotating polarizer that is frequency matched with one moving film that has both sets of camera images on it alternating left and right. The rotating polarizing filter is there to allows for every other picture to be plane polarized 90º to each other and offset a few centimeters. The viewer wears 3d glasses also fitted with polarizers. How would the polarizers in the glasses need to be oriented now? What is the problem with plane-polarized 3d technology? You cannot turn your head or lie down, as the 3d effect would be lost. Thus Circular Polarization is employed in current movies and 3d LCD TVs Chapter 23, 24 and 26: Done.