LC circuit: Energy stored. This lecture reviews some but not all of the material that will be on the final exam that covers in Chapters

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Disclaimer: Chapter 29 Alternating-Current Circuits (1) This lecture reviews some but not all of the material that will be on the final exam that covers in Chapters 29-33. LC circuit: Energy stored LC circuit: Oscillation frequency RLC circuit: With time varying emf Chaper 29 Alternating-Current Circuits (2) Time-varying emf Chapter 29 Alternating-Current Circuits (3) Summary: Phase and Phasors Time-varying emf V R with resistor Resistance R Time-varying emf V C with capacitor Capacitive Reactance X C Time-varying emf V L with inductor Inductive Reactance X L 1

Chapter 29 Alternating-Current Circuits (4) Transformers Consider the second coil with N S turns The time-varying emf in the primary coil induces a timevarying magnetic field in the iron core. This core passes through the secondary coil Thus a time-varying voltage is induced in the secondary coil described by Faraday s Law Because both the primary and secondary coils experience the same changing magnetic field, we can write Chaper 29 Alternating-Current Circuits (5) Single loop RLC Circuit with time-varying emf: Phase constant: The average power step-up step-down Chaper 29 Alternating-Current Circuits (6) Resonance For given peak, R, L, and C, the current amplitude I peak will be at the maximum when the impedance Z is at the minimum. Maxwell s Equations Integral Form Differential Form i.e., load purely resistive ε and I in phase This is called resonance. Resonance angular frequency: Wave solutions: where k = 2 / and = 2 f with wavelength and frequency f 2

Characteristics 1. E B 2. If E = E 0 sin(kx ωt), B = B 0 sin(kx ωt). 3. The Poynting vector S = 4. E and B are both S. 5. c = 6. E = cb 1 ε 0 µ 0 E B is in the direction of propagation. µ 0 2.99792458 10 8 m/s (exact) Intensity of an electromagnetic wave: Energy density: z x S 4/25/12 9 Chapter 30 Total Absorption P r = I c Pressure P = Force/Area Total Reflection P r = 2I c Chapter 31 Properties of Light (5) Intensity after Polarizer The intensity of the light I 0 before the polarizer is given by After the light passes through the polarizer, the intensity I is given by *Polarization The transmitted intensity in terms of the initial intensity is 4/25/12 11 This result is called the Law of Malus This equation only applies to the case of polarized light incident on a polarizer 3

Chapter 31 Properties of Light (1) Law of Reflection: Focal length of a spherical miorr: Mirror Equation: Chapter 31 Properties of Light (2) Laws of Reflection and Refraction: Summary Law of Reflection A reflected ray lies in the plane of incidence The angle of reflection is equal to the angle of incidence Medium 1 Magnification: Law of Refraction (Snell s Law): The critical angle for a total reflection: Law of refraction Indenx of refraction n: A refracted ray lies in the plane of incidence The angle of refraction is related to the angle of incidence by Snell s Law where is the wavelength in vacuum Medium i=2 Total Internal Reflection The critical angle, c, at which total internal reflection takes place is given by Which we can rewrite as For n 2 = 1 (air) At angles less than c, some light is reflected and some transmitted At angles greater than c, all the light is reflected and none is transmitted Chapter 31 Properties of Light (3) air = 2 and n air =n 2 2 > 1 % is Chapter 31 Properties of Light (4) Chromatic Dispersion The index of refraction of a medium is usually a function of the wavelength of the light. It is larger at shorter wavelengths. This causes spreading of light which is called chromatic dispersion. White light consists of components of nearly all the colors in the visible spectrum with approximately uniform intensities. The component of a beam of white light with shorter wavelength tends to be bent more. Spectrometer (such as a prism) 4

Chapter 31 Properties of Light (6) Wave-Particle Duality of Light Wave nature of light Refraction, diffraction (Huygens s Principle) Interference (Young s double slit interference) Electromagnetic waves (Maxwell s Equations) Light and other EM radiation often come together (e.g., in Black-body radiation) Doesn t require a medium Particle nature of light Quantum Mechanics: Duality for all particles Collisions, scattering as in photoelectric effect (Albert Einstein) Chapter 32 Optical Images (1) Thin-Lens Equation: Lens-Maker s Formula: Energy is quantized: massless photons Planck s constant Chapter 32 Optical Images (2) Mirror Equation and Magnification (f = r/2) Chapter 32 Optical Images (3) Lens Equation Using these sign conventions we can express the mirror equation in terms of the object distance, d o, and the image distance, d i, and the focal length f of the mirror (f = r/2) Images Formed with Concave Lenses The image formed is virtual, upright, and reduced The magnification m of the mirror is defined to be 5

Chapter 32 Optical Images (4) Systems of Lenses Images formed by systems of lenses can be obtained using a set of equations Chapter 33 Interference and Diffraction (1) Huygens s Principle: every point on a propagating wave front serves as a source of spherical secondary wavelets The requirement for constructive interference of two coherent waves with : The requirement for destructive interference of two coherent waves with : Two slits with separation d: bright interference fringes On a screen a long distance L away: Chapter 33 Interference and Diffraction (2) Two slits with separation d: dark interference fringes On a screen a long distance L away: For a thin film with thickness t, the condition for constructive interference is The radii of the bright circles in Newton s rings are given by Chapter 33 Interference and Diffraction (3) The angle of the first minimum from a circular aperture with diameter D illuminated with light of the wave length : Rayleigh s Criterion: The angle of the maxima from a diffraction grating illuminated with light of the wave length (d is the distance between rulings of the grating): The angle of dark diffraction fringes from a single slit of width a : Because diffraction gratings produce widely spaced narrow maxima, they can be used to determine the wavelength of monochromatic light 6

A capacitor in an LC circuit oscillator has a maximum potential difference of 15 V and a maximum energy 360 µj. At a certain instant, the energy in the capacitor is 40 µj. At that instant what is the potential difference across the capacitor? An AC generator producing 10 V (rms) at 200 rad/s is connected in series with a 50 ohm resistor, 400 mh inductor and a 200 µf capacitor. The rms voltage (in volts) across the resistor is: E x = E 0 sin(kz - wt) Consider a point (x,y,z) at -me t when E x is nega-ve and has its maximum value. At (x,y,z) at -me t, what is B y? A) B y is posi-ve and has its maximum value B) B y is nega-ve and has its maximum value C) B y is zero D) We do not have enough informa-on 7

An unpolarized beam of light has intensity I o. It is incident on two ideal polarizing sheets. The angle between the axes of polarization of these sheets is. Find if the emerging light has intensity I o /4. UP (unpolarized light) I=I 0 LP TA E 1 % I 1 =?? nˆ TA % E 2 Problem LP I 2 = I o/4 A laser beam of power 4.60 W and diameter 2.60 mm is directed upward at one circular face (of diameter less than 2.60 mm) of perfectly reflecting cylinder, which is made to hover by the beam s radiation pressure. The cylinder s density is 1.20 g/cm 3. What is the height (in meters) of the cylinder? 8

: Snell s Law Chapter 24 Geometric Optics (b) Calculate the critical angle for a diamond under water. (1) 30.61 (2) 33.44 : Convex Mirror In a subway station, a convex mirror allows the attendant to view activity on the platform. A woman 1.64 m tall is standing 4.5 m from the mirror. The image formed of the woman is 0.500 m tall. (a) What is the radius of curvature of the mirror? A concave spherical mirror has a focal point of 12 cm. If an object is placed 6 cm in front of the mirror, the image position is (1) 3.9 m; (2) 4.5 m; 9

How thick is an anti-reflection coating for a glass lens which is chosen for light of wavelength 500 nm if the index of refraction of the coating is 1.29? How thick is an anti-reflection coating for a glass lens which is chosen for light of wavelength 500 nm if the index of refraction of the coating is 1.29? : Double Slit : Thin Film d sinθ = mλ sinθ ~ θ = λ d = 0.00562 tanθ = y L y = L tanθ = 1.68mm 10