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Reflection and Refraction of Light IB 12 A laser pointer is aimed at the surface of a plane mirror. Use a protractor and straight-edge to construct the laser beam after it reflects from the mirror. Plane Mirror: Normal: Angle of incidence: Angle of reflection: Law of Reflection Properties of Images formed by Plane Mirrors Each object below is in front of a plane mirror (seen on edge). Sketch the image that you would see in each case if you were looking into the mirror. Then, check your result by placing a plane mirror on top of this page at each location and looking into it. What are some properties of images formed by plane mirrors? 1. 4. 2. 3. 5. 1 IB 12 Locating Images using the Law of Reflection Locate the image of this dot by means of two lines of sight. 1. 2. How much of this 2.0 meter tall mirror is actually needed for the man to see the reflection of his entire body? Virtual Image: 2 Spherical Mirrors IB 12 Source of Parallel (paraxial) Rays: Method of locating focal point: Principal Axis: Relationship between radius of curvature and focal length Center of Curvature (C): Radius of Curvature (R): Focal Point (F): Focal Length (f): Power (P): Converging Mirror Shape: Focal Point: Focal Length: Images: Examples: Diverging Mirror Shape: Focal Point: Focal Length: Images: Examples: 3 Real and Virtual Images Real image: Properties: IB 12 Virtual image: 1. 2. Properties: 1. 2. Ray Tracing to Locate Images Ray #1: Your Turn Ray #2: Your Turn Ray #3: Your Turn 4 IB 12 Image Properties Case 1: Your Turn Image Properties Case 2: Image Properties Case 3: Image Properties Case 4: General Trend: As the object moves closer to the mirror . . . 5 IB 12 Image Properties Case 5: Application: Your Turn Summary Chart for Converging (Convex) Mirror Object Position Image Position Image Properties Convex Mirror Image Properties: Application: 6 Calculating Locations and Sizes of Images IB 12 p= q= ho = hi = Mirror Equation Linear Magnification Sign Convention 1. A 2.0-cm-high object is placed 7.10 cm from a concave mirror whose radius of curvature is 10.20 cm. a) Locate the image by means of a ray diagram. c) Calculate the magnification of the mirror. d) Calculate the size of the image. b) Calculate the location of the image. e) Describe the image. 7 IB 12 2. An object is placed 6.00 cm in front of a concave mirror that has a 10.0-cm focal length. a) Determine the location of the image. b) The object is 1.2 cm high. Find the height of the image. 3. A convex mirror is used to reflect light from an object placed 66 centimeters in front of the mirror. The focal point is 46 centimeters from the mirror. Find the location of the image. In reality, why is a parabolic mirror preferred over a concave or convex mirror? 8 Refraction of Light IB 12 Refraction: Complete the path of each light ray shown below. Angle of refraction: In which substance does light travel faster – air or water? Snell’s Law for Refraction Snell’s Law Use a protractor and Snell’s law to construct the refracted ray on the diagram at right. air flint glass 9 Total Internal Reflection IB 12 Total Internal Reflection: Conditions for Total Internal Reflection: 1. 2. Critical Angle (θc): Formula: 1. What is the critical angle as light exits from water into air? 2. What is the critical angle as light exits from water into crown glass? Applications of Total Internal Reflection Fiber Optic Cables How do fiber optic cables (optical fibres) work? Cladding: 10 IB 12 Dispersion Natural Dispersion: since the refractive index of a medium depends on the wavelength of the light travelling through it, light of different wavelengths will travel through the glass core of an optical fibre at different speeds Result: a set of light rays of different wavelengths will reach the end of a fibre at different times, even when following the same path High-order-mode paths: rays that undergo many internal reflections over a given distance Low-order-mode paths: rays that undergo few internal reflections over a given distance Waveguide dispersion: the result of rays of the same wavelength following different paths (having different order modes), thus reaching the end in different amounts of time Multimode fibre vs Monomode fibre Multimode fibres: Light of different wavelengths follow different paths, being subjected to both material and waveguide dispersion; diameter of the core is about 100 µm Monomode fibers: All light propagates along the same path; diameter of the core is very small, about 10 µm, with considerably thicker cladding 11 IB 12 Step-index fibre vs Graded-index fibre Step-index fibres: the refractive index of the core is constant, but is slightly higher than the constant refractive index of the cladding. Graded-index fibres: the refractive index decreases smoothly from the center of the core to the outer edge of the core, with a constant value in the cladding Attenuation Attenuation: A loss of power as a signal is traveling through a medium Causes in an optical fibre: the scattering of light and the impurities in the glass core Amount of attenuation depends on: the wavelength of the light being transmitted Formula: units: General rule of thumb: Specific Attenuation: The power loss in decibels per unit length travelled Formula: units: 12 IB 12 1. An amplifier amplifies an incoming signal of power 0.38 mW to 2.7 mW. Calculate the power gain of the amplifier in decibels. 2. A source produces an initial power output of 12 mW. What is the final power output after it experiences a power loss of 18 dB while traveling along a cable? 3. A signal of power 15 mW is input into a cable of specific attenuation 3.0 dB km-1. Calculate the power of the signal after it has travelled 8.0 km Advantages of Optical Fibres Advantages of optical fibres over coaxial cables (and also over twisted copper wires): 1) Fast (travels at “c”) and cheap transfer of information in digital form 2) Low attenuation 3) No interference from stray EM signals 4) Greater capacity (bandwidth), making possible the transmission of many signals 5) Security against “tapping”, i.e. unauthorized extraction of information from the signal 13 Lenses IB 12 Why do lenses converge or diverge light? How can the focal length of a converging lens be found? Shape: Converging Lens Focal Point: Focal Length: Images: Converging lenses are . . . Diverging Lens Shape: Focal Point: Focal Length: Images: Diverging lenses are . . . Factors Affecting Focal Length a) Thickness c) Frequency b) Refractive Index n = 1.52 n = 1.68 14 Image Formation by Converging Lenses IB 12 Sample of a ray diagram used to locate and describe image Three Principal Rays used in Ray Diagramming 1. Ray parallel to principal axis . . . 2. Ray through F . . . 3. Ray through the center of the lens . . . Case 1 Properties of image: Case 2 Properties of image: Case 3 Properties of image: Case 4 Properties of image: As the object moves closer to the lens . . . 15 IB 12 Virtual Images Case 5 Properties of image: Case 6 Properties of image: Applications State the type of lens, locate the object and image, and describe the image for each device below. a) Camera b) Projector c) Magnifying Glass Lens: Lens: Lens: Object: Object: Object: Image: Image: Image: d) Security “Peephole” Lens: Object: Image: 16 The Thin-Lens Equation and Linear Magnification IB 12 f = focal length p = object distance q = image distance ho = height of object hi = height of image Thin-Lens Equation Linear Magnification Sign Conventions positive = negative = 1. A 3.0 cm high object is placed 15 cm from a converging lens whose focal length is 6 cm. Determine the location of the image and describe its properties. Determine the magnification of the lens and the height of the image. 2. A 20 cm high object is placed 10 cm in front of a convex lens whose focal length is 30 cm. Determine the location and properties of the image. 3. A 10 cm high object is placed 20 cm in front of a diverging lens whose focal length is 60 cm. Determine the location and properties of the image. 17 Comparisons – Mirrors and Lenses Converging IB 12 Diverging 18 IB 12 Lens Combinations: Virtual Objects By combining multiple lenses, the image produced by the first lens can serve as the object for the second lens. Overall Magnification: 4. An object lies on a table. A converging lens of focal length 8.0 cm is placed 6.0 cm above the object. a) Determine the image formed by this lens. b) A second converging lens of focal length 5.0 cm is now placed 3.0 cm above the first lens. Determine the image formed by this combination of lenses. 5. An object is placed 10.0 cm to the left of a converging lens of focal length 5.0 cm. A second diverging lens of focal length 7.0 cm is placed 4.0 cm to the right of the converging lens. Determine the image of the object in the two-lens system, AND verify your results with a scaled ray diagram. 19 Optical Instruments IB 12 PURPOSE: To increase the angular size of an image in order to aid the eye to see it more clearly Linear Size (ho or hi): height of object or image – measured in meters Angular Size (θo or θi): angle subtended at the eye by the object or image – measured in radians An object has one linear size but different angular sizes depending on how far it is from the observer. Far Point: distance between the eye and the furthest object that can be comfortably brought into focus 1. Taken to be infinity – approx. > few meters in practice 2. Most comfortable viewing arrangement Near Point: distance between the eye and the nearest object that can be comfortably brought into focus 1. “least distance of distinct vision” 2. D = 25 cm for normal eye– depends on age Standard Case – When the object is at the near point, the it occupies the greatest angular size. The magnification of any optical instrument is measured against this case. 20 The purpose of an optical instrument is to magnify the angular size of an image. Without an optical instrument to magnify image IB 12 With an optical instrument to magnify image Angular Magnification (M): ratio of the angle subtended at the eye by the image to the angle subtended at the eye by the object Formula: Possible Angular Magnifications of a Magnifying Glass Extreme Case 1: Image is formed “at infinity” 1. object placed at focal point (u = f) 2. image formed at infinity (v →∞) – at far point 3. smallest possible angular magnification for mag. glass Angular Magnification Derivation 1. A stamp collector wishes to look at a 1.20 cm square stamp for awhile comfortably with a magnifying glass whose focal length is 10.0 cm. a) Where should she place the stamp for comfortable prolonged viewing? b) Where will the image appear? c) What is the angular magnification of the lens for this situation? 21 IB 12 Extreme Case 2: Image is formed at the near point 1. object placed in front of focal point 2. image formed at near point (v = -25) 3. largest possible angular magnification for mag. glass Angular Magnification Derivation 2. The stamp collector now wishes to examine a particular detail on the stamp more closely so she wishes to have the maximum possible angular magnification. a) Where will the image be formed? b) Where should the stamp be placed? c) What is the angular magnification of the lens for this situation? 22 The Compound Microscope IB 12 Purpose: Objective Lens: Eyepiece Lens: lens close to object, lens very near eye Short focal length (f < 1 cm) Longer focal length (f ≈ few cm) Forms real image to be used for second lens Forms virtual larger image Focal Lengths: 1. 2. Principles of Operation 1. The object is placed just outside the objective lens’ focal point in order to form a real image for the eyepiece lens. 2. A real, inverted, larger image is formed more than twice the objective lens’ focal length away on the other side of this lens. 3. The eyepiece lens is placed so that it acts as a simple magnifying glass, that is, so that the real image falls within its focal length. This lens then forms the final image – a virtual, non-inverted, larger image somewhere behind the real image. 4. The eye is placed very near to the eyepiece lens and the position of this lens is adjusted so that the final image is located at the near point for maximum angular size. This is called “normal adjustment.” Linear Magnifications: (pg 21 book packet) Overall Angular Magnification at normal adjustment of the microscope: (pg 20/21 of book packet) 23 1. a) Locate the final image in the diagram below by means of ray tracing. IB 12 \ b) The objective and eyepiece of the compound microscope above have focal lengths of fo = 15.0 mm and fe = 25.5 mm. A distance of 61.0 mm separates the lenses. The microscope is being used to examine an object placed 24.1 mm in front of the objective. Find the final image distance and linear magnification. c) Where would the eyepiece need to be placed for the microscope to yield the maximum possible angular magnification? 24 The Astronomical Telescope IB 12 Purpose: Objective Lens: large diameter – to collect more light, reduce diffraction, increase resolution Long focal length Object approximately at infinity Forms real image in focal plane of lens Eyepiece Lens: Focal Lengths: short focal length 1. Acts as magnifying glass with object at focal point 2. Forms large virtual image at infinity Principles of Operation 1. Object (star, planet, etc.) is very far away – approximately at infinity. As a result, wavefronts arriving from object are approx. parallel as are the rays. 2. Real, inverted, smaller image is formed in the focal plane of the objective lens. 3. The eyepiece lens is placed so that it acts as a simple magnifying glass - the real image is at the focal point of the eyepiece lens and acts as an object for this lens. The eyepiece lens’ focal point then coincides with the focal point of the objective lens. The real image now falls in the mutual focal plane of the two lenses. 4. The eye is placed very near to the eyepiece lens. The final image is a virtual, non-inverted, larger image at infinity. 25 IB 12 Your Turn Magnification of the Astronomical Telescope Angular Magnification: Length of telescope tube: 1. A telescope has the following specifications: fo = 985 mm and fe = 5.00 mm. From these data, find (a) the angular magnification of the telescope (b) the approximate length of the telescope. 26 IB 12 Reflecting Telescopes Reflecting telescopes use mirrors rather than the lenses of refracting telescopes. The largest telescopes used are reflecting. Advantages: 1) To see distant faint objects requires large lenses (to collect more light); but large lenses are hard to make and can only be supported along their rim; large lenses may collapse under their own weight By contrast, large mirrors can be supported along the rim and at the back. 2) Mirrors do not suffer from chromatic aberration 3) Only 1 side has to be ground, as opposed to 2 for lenses Two General types of reflecting telescopes: 1) 2) Single-dish Radio Telescopes A radio telescope receives and detects EM waves in the radiofrequency region. What types of objects are known to radiate in this region? Properties of single-dish radio telescopes: 1) Since frequencies are small, their wavelengths are large, so the diameter of the radio telescope has to be large as well in order to achieve reasonable resolution (Arecibo radio telescope =300 m diameter , while Hubble Space Telescope = 2.4 m diameter) 2) Large radio telescopes are very heavy steel structures 3) Can be difficult to steer (if it can be steered at all) 4) Parabolic shape with a detector placed at the focus of the mirror Note: Diffraction places limits on resolution, i.e. that ability of an instrument to see two nearby objects as distinct. Resolution: Where: θA = λ= b= 27 IB 12 Radio Interferometry Telescopes Radio interferometry: How does it work? Use a very large array of radio telescopes very far apart and appropriately combine the signals from the individual dishes. This can achieve the same resolution as a single dish with a diameter equal to the length of the entire array. Satellite-borne Telescopes Earthbound telescopes are limited for a number of reasons: 1) Light pollution (excess light in the atmosphere) 2) Atmospheric turbulence (mainly due to convection currents and temperature differences) 3) Absorption of various wavelengths (especially X-rays and UV) by the atmosphere. This makes observation at these wavelengths impossible. These problems do not exist for satellite-based telescopes in orbit around the Earth. 28 Lens Aberrations IB 12 Aberration: A single large converging lens is unable to form a perfectly sharp image for two reasons. Spherical Aberration: Result: Method to reduce or eliminate aberration: use a “stop” to reduce the size of the aperture– an aperture with a disk with an adjustable hole to block some rays far from the principal axis Tradeoff: also reduces amount of light so image is sharper but fainter Chromatic Aberration: Result: Method to reduce or eliminate aberration: “chromatic doublet” – a compound lens formed by adding a second lens that is diverging, usually of a different material, so that the dispersion is reduced Make a sketch to show how the image of this object would be distorted due to spherical aberration. aberration object fix Make a sketch to show how the image of this object would be distorted due to chromatic aberration. 29 Medical Imaging and Diagnostic Techniques IB 12 Various imaging techniques are used for diagnostic purposes: 1) X-rays: use high energy radiation 2) Computed Tomography (CT): 3D x-ray (also known as a CAT scan) 3) Ultrasound: uses high frequency sound waves 4) Nuclear Magnetic Resonance (NMR): uses magnetic fields (also known as MRI) 5) Lasers Production of X-rays: a) Electrons are boiled off a filament due to heating. b) They are accelerated through a high potential difference. c) They strike a metal target, often tungsten. d) The inner shell electrons of the target (tungsten) jump to high energy level and emit x-ray photons when they relax. Imaging technique: a) X-rays pass through person and fall on a photographic film which darkens when x-rays hit it. b) Bone absorbs more of x-rays than soft tissue so on the film tissue appears darker and bone appears lighter. Attenuation: a) b) Therefore, Contrast: 30 IB 12 1. X-rays are incident on the muscle of a patient as shown at right. The intensity of the transmitted x-rays is measured and plotted as a function of the thickness of the muscle. I0 = I= x= a) What is the relationship between the transmitted intensity and the thickness? Symbol: Half-value Thickness: Units: b) What is the half-value thickness of the muscle shown above? Intensity Equation: where μ = Relationship between Half-value thickness and attenuation coefficient Attenuation coefficient: Symbol: Units: Mass absorption coefficient: Symbol: Units: 31 IB 12 1. The graph at right shows the transmitted intensity of a parallel beam of X-rays versus the thickness of a certain type of tissue. Determine the: a) half-value thickness of this tissue b) attenuation coefficient of this tissue 2. The half-value thickness for soft tissue (muscle) is about 20 cm and for bone is 150 times less than this. Determine the attenuation coefficient for bone and for soft tissue. 3. Compare the half-value thickness and attenuation coefficients for bone and muscle. 4. A parallel beam of X-rays passes through muscle and reduces to one-eighth of its initial intensity. Determine the fraction of the intensity of this beam when it is transmitted through the same thickness of fatty tissue. The half-value thickness of muscle is 4.0 mm and the half-value thickness of fatty tissue is 6.0 mm. 32 IB 12 5. The half-value thickness of a 30 keV X-ray photon in aluminum is 2.4 mm. The initial intensity of the X-ray beam is 4.0 x 102 kW/m2. a) What is the beam intensity after passing through 9.6 mm of aluminum? b) What is the beam intensity after passing through 1.5 mm of aluminum? 6. Analyze the intensity graph at right by straightening it. 33 Computed Tomography (CT) IB 12 CT or CAT scan: Computer-Assisted Tomography or Computerized Axial Tomography Outline the basis of CT scanning – X-ray image of target taken at different angles (many different directions) - computer produces detailed image of slice (these images are combined using computers to form a two-dimensional image of section) – images of many sections/slices can be obtained – combined to build up a 3D image so image can be rotated for viewing from any angle Compare X-ray imaging and CT scans X-ray imaging 1) uses ionizing radiation (X-rays) 2) short duration of exposure 3) smaller dose of radiation 4) two-dimensional shadow image CT scanning 1) uses ionizing radiation (X-rays) 2) long duration of exposure (hard for kids) 3) larger dose of radiation 4) three-dimensional image Why/how are barium meals used to assist X-ray imaging of stomach or intestinal tract? a) All tissues in the abdominal cavity have approximately the same attenuation coefficient so there is little to no contrast on photographic film. b) The attenuation coefficient for barium is greater than for the tissues in the abdominal cavity. c) Barium meal lines the stomach. d) Outline of stomach becomes clearer (greater contrast). 34 Ultrasound Imaging IB 12 Type of energy used: Advantage: What is ultrasound? Typical operating frequencies of ultrasound waves: Operating principles: High frequency sound waves are transmitted from a transducer (a probe) into the patient’s body and are reflected at each boundary between different types of tissue and bone. The same probe both transmits and receives the ultrasound waves. By measuring the time between transmission and reception, the distance to each boundary can be calculated using the speed of sound and thus the location and surface of each organ can be mapped. Production and detection of ultrasound waves Piezoelectric crystal: a quartz crystal that changes shape when a potential difference is applied across it Production: apply an AC voltage to generate a vibration at desired frequency Detection: received sound wave causes it to vibrate and generate an AC voltage that can be measured Factors affecting choice of diagnostic frequency: a) Resolution: size of smallest object that can be imaged. Ultrasound is a wave so diffraction effects must be minimized. Use smallest wavelength possible. Favors: b) Attenuation: absorption of signal. Attenuation increases as frequency increases. High frequency ultrasound will be absorbed and not reflected back to receiver. High frequency can’t scan organs deep in body – poor depth penetration Favors: Compromise: 35 IB 12 1. The time delay for an ultrasound pulse going through body fat to reach and be reflected from the liver is 0.133 ms. The speed of the ultrasound through fat at the chosen frequency is 1450 m/s. a) Calculate how far from the probe the liver is. b) What are some assumptions made in this calculation? Acoustic Impedance: Formula: Where Units: Z= ρ= c= Density (kg m-3) Speed of sound (m s-1) Air (200 C) 1.21 344 Muscle 1075 1580 Bone 1900 3780 Fat 900 1480 Soft Tissue (skin) 1060 1540 Material Acoustic Impedance (kg m-2 s-1) 2. Complete the table with the acoustic impedance of each of the above substances. 36 IB 12 Reflection at a boundary: 3. From what boundary will most of the energy of the ultrasound waves be reflected? 4. What is the purpose of putting gel on probe and the patient’s skin? Ultrasound of intensity Io is traveling in a medium of impedance Z1 and is incident on a medium of impedance Z2. The intensity of the reflected ultrasound is IR and is given by the formula: 5. Calculate the fraction of ultrasound that would be reflected from a patient’s skin if gel were not applied to the tip of the ultrasound wand. Types of ultrasound scans I. B-scan: a) b) Presentation: Uses the signal strength to affect the brightness of a dot on the screen which can be displayed as a real-time video. Many B-scans are combined to give an image of the internal organs or baby. Use: 37 IB 12 II. A-scan: a) b) Presentation: Uses: The diagram above shows an ultrasound transducer (probe) in contact with the skin in an effort to determine the depth and width of the organ shown. An A-scan of the results is shown also. The average speed of the ultrasound used in tissue and muscle is 2.0 x 103 m/s. a) Identify the source of each reflection labeled A, B, C, and D. b) Calculate the depth of the organ. c) Calculate the width of the organ. 38 NMR and MRI IB 12 NMR: MRI: Type of energy used: 1) 2) Uses: 1) imaging blood flow and soft tissue in the body 2) Preferred for the brain and central nervous system 3) Detecting tumors, strokes, infections in brain, spine, joints Operating Principles: a) Large constant uniform magnetic field causes hydrogen atoms to line up (align their spin axes) – act like tiny magnets b) Small non-uniform magnetic field is superimposed on top of larger field – localized magnetic field – weak oscillating field in the form of pulses of radio waves c) If frequency of radio waves matches that of the hydrogen atoms (resonance) then the smaller field makes some hydrogen atoms realign d) When small non-uniform field is removed, atoms relax back to original alignment e) As they relax they emit radio-waves f) Time it takes to relax is measured g) Frequency of emitted radio waves and relaxation times are processed to produce the NMR image Advantages: 1) Any situation where detailed tomography/slicing/imaging is required 2) Large scale investigations where dose of ionizing radiation would be too great 39 Comparisons of Diagnostic Imaging Techniques IB 12 1. What are the advantages and disadvantages of using ultrasound instead of X-rays? Advantage: not as harmful since no ionizing radiation And so can be used for pregnant women Disadvantages: Small depth of penetration Limit to size of objects that can be imaged Blurring of images due to reflection at boundaries 2. Why are X-rays preferred over ultrasound for bone fractures? Nearly all ultrasound is reflected by bone (at bone/tissue boundary) but x-rays can penetrate bone therefore X-rays show up internal structures. 3. What are the main advantages of each of the following imaging techniques? X-rays: To detect broken bones because bone and tissue show different attenuation/good distinction between bones and flesh Ultrasound: Any soft tissue analysis – takes advantage of reflections off organ boundaries Pre-natal scans because there is no risk from ionizing radiation NMR: Any situation where detailed tomography/slicing/imaging is required Large scale investigations where dose of ionizing radiation would be too great 4. Apart from health hazards, why are different means of diagnosis needed? a) b) c) d) e) f) g) Different types of tissues and bone have different absorption/attenuation properties Some are better at distinguishing boundaries of organs Some provide two dimensional slice imaging – some provide complete three dimensional images Some are better at monitoring static or dynamic conditions Some are better to investigate at large or small scales Some can be used to study concentrations of specific types of tissue or pharmaceuticals Some are better at monitoring specific organ functions 40