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- (12) Faraday’s Law Faraday’s Flux Rule
- OPTIC, WAVES, THERMO AND STAT : (13) Properties of an EM plane wave and the Poynting Vector
- The N slit pattern constructive interference condition
- Intensity pattern = single slit * multi-slit pattern The single slit pattern zeros: a sin = m m 0 The multi-slit bright
- The figures below show the intensity pattern of six narrow (width
- Birefringence (double refraction)
- Polazation, polarizers
- Natural (unpolarized)
- (16) Basic Wave Properties
By knowledge: The field grows linearly inside a uniformly charge sphere so the field has 1/3 of the strength at r = R or  . The direction is radially away by symmetry. A
By Gauss:  directed away. For the point R/3 from the center, the Gaussian surface is a sphere surface of radius R/3 concentric with the sphere. (The point at which the field is to be found must lie on the Gaussian surface. The surface has the symmetry of the problem.)
Qenc = (1/3)3 Q;  = (1/3)2 (4 R2)
(12) Faraday’s Law 
Faraday’s Flux Rule = Emf = - Schematically: pure induction motional motional transformer slidewire gen. rotating generator In some cases, one needs to multiply by the number of turns. Ex. 12) Four meters of wire form a square that is placed perpendicular to a uniform magnetic field of strength 0.1 Tesla. The wire is reduced in length by 4.0 centimeters per second while still maintaining its square shape. Which one of the following gives the initial induced emf across the ends of the wire? (A) 1 mV (B) 2 mV (C) 4 mV (D) 8 mV (E) 16 mV The magnetic field is not varying in time, so we will use the flux rule form: Emf = -  .
The easy way: The side s is the perimeter divided by 4. The side at time zero is s(0) = 1 m. At one second, s(1 sec) = 0.99 m. Using Emf = - As a function of time. Use: Area = side2 = (perimeter/4)2  = B side2 = B (perimeter/4)2 = (0.1 ) (4.0 – 0.04 t )2/16 Tm2
d/dt B =(0.1 ) ( [2 (4.0 – 0.04 t) (- 0.04)]/16) Tm2s-1 At time zero, the magnitude of the emf is |(0.1) ( [8 (- 0.04)]/16)| Tm2s-1 = 2 mV. B The final technique to compute a motional emf is as  . In this problem you have 4 meters of wire moving inward at ½ cm/s at t = 0. Bv (0.1T)(4m)(0.005 m/s) = 2 mV.
OPTIC, WAVES, THERMO AND STAT: (13) Properties of an E&M plane wave and the Poynting Vector EM waves are transverse which means that the electric and magnetic fields are perpendicular to the direction of propagation of the wave. The electric and magnetic fields are mutually perpendicular. The wave propagates in the direction of the Poynting vector, Energy Densities: E = ½ o E2 ; B = ½ 2/o
At that position and time , the electric field points along the (A) positive y-axis (B) negative y-axis (C) positive z-axis (D) negative z-axis (E) negative x-axis
EM waves are transverse which means that the electric and magnetic fields are perpendicular to the direction of propagation of the wave. The electric and magnetic fields are mutually perpendicular. As the wave propagates in the z direction and at that point and time has the magnetic field in the x direction, the electric field must be in the y direction. The Poynting vector 
The electric field is in the negative y direction at that point and time. B (14) Diffraction and Interference Standing waves As a first guess, diffraction limits angular resolution to min = /d where d is the width of the aperture. (If the aperture is circular, use min = 1.22 /D where D is the diameter of the aperture.) The N slit pattern constructive interference condition: d sinbright = m where m is the order of the interference. Array of equally spaced finite slits. A array of 100 identical slits spaced by 32 m and of width a yields a diffraction pattern in which the first three orders of the multiple slit pattern appear, but the fourth m = 4 peaks are missing (dark). What is the width a of the individual slits? Intensity pattern = single slit * multi-slit pattern The single slit pattern zeros: a sin = m m 0 The multi-slit bright: d sin = m m = integer The fourth bright is missing: d sin = 4 anda sin = m m = 1, because this is the first missing peak. 
The figures below show the intensity pattern of six narrow (width << ). The blue line is the intensity pattern of a single slit with width of ¼ of the slit to slit spacing. The next figure shows the intensity pattern of six slits each with width of one-fourth of the slit to slit spacing. The final figure is the pattern of 20 equally spaced slits each with width of one-fourth of the slit to slit spacing. The individual intensity spikes are narrower. A diffraction grating has a great many slits and so produces very narrow spikes. The single slit pattern multiplies the N slit pattern.   
Six Narrow Slits + finite Single Slit Pattern of Six Finite Slits Pattern of 20 Finite Slits (Ex. 14) An observer looks through a slit of width 5 x 10-4 meters at two lanterns a distance of 1 kilometer from the slit. The lantern emits light of wavelength 5 x 10-7 meters. The minimum separation of the lanterns at which the observer can resolve the lantern lights is most nearly (A) 0.01 m (B) 0.1 m (C) 1 m (D) 10 m (E) 100 m In this case, = 5 x 10-7 meters and d = 5 x 10-4 meters so min = 10-3 radian or one milli-radian. One should prepare a sketch, but one can guess that the answer is 10-3 of one kilometer or one meter. C Polarization Methods: Selective Absorption: Light can be polarized by selective absorption which is the method used in most sheet polarizers. The material absorbs light polarized along one of the transverse directions and passes light polarized along the pass axis (which is perpendicular to the absorption axis). Reflection: The fraction of the light of each of the two polarizations that is reflected depends on the angle of incidence. At Brewster’s angle, the refracted ray make a 900 angle with respect to the reflected ray, and the reflected light is 100% polarized perpendicular to the plane of incidence. n1 sinp = n2 sin2 and p + 2 = 900 so sinp = cos2 tanp = n2/n1. 
Birefringence (double refraction): Certain materials have different indices of refraction for the two polarization so the two polarizations are refracted at different angles leading to a separation of the incident beam into two beams of pure linear polarization (that are polarized perpendicular to one another).  
Scattering: The electric field accelerates the charges in material along its direction. The scattered light has electric field (polarization) components in the direction of the charges acceleration that are perpendicular to the direction of propagation.
Polazation, polarizers: The interaction of E-M waves with matter is described by the Lorentz force law,  . For a wave, |B| c-1 E so Fmagnetic (vq/c) Felectric. The electric interaction dominates so the polarization of the wave is associated with the direction of the electric field. That field is transverse to the direction of propagation and there are two independent transverse directions so there are two independent polarizations for a given propagation direction. These polarizations might be chosen to be vertical and horizontal linear polarizations or left- and right-handed circular polarizations. Circular polarizations carry angular momentum along the direction of propagation. Linear polarizations carry zero net angular momentum. Natural (unpolarized light) can be assumed to be an equal mix of the two independent polarizations.
The wave immediately after a polarizer has a polarization parallel to the pass axis and an average electric field strength of Eincident cos. As the intensity of a wave is proportional to the square of its amplitude, Itransmitted = Iincident cos2. (The Law of Malus.) Natural (unpolarized) light is an equal mix of the two polarizations. If natural light is incident on a polarizer, light of one-half the intensity that is 100% polarized along the pass axis is transmitted. Io natural ½ Io vertical ½ Io cos2 along last pass axis ½ Io cos2cos2 along last pass axis One half the intensity of the incident intensity of natural light passes through the first polarizer. All the light that passes is polarized along the pass axis direction. Incident polarized light obeys the Law of Malus, and the transmitted light is 100% polarized along the pass axis of the last polarizer. (15) Lens Problems 
Ray Tracing a Positive (Converging) Mirror Virtual for object inside focal point The parallel ray: In parallel; out through the focus The focal ray: In (as if) through the focal point; out parallel The chief ray: In at optical axis point in the interface plane. r = i. Center Ray: In through the center of curvature; out through the center of curvature
Ray Tracing a Convex (Diverging/Negative) Mirror Virtual Image The parallel ray: In parallel; out as if from the back focal point The focal ray: In (as if) toward the focal point; out parallel The chief ray: In at optical axis point in the interface plane. r = i. Center Ray: In through the center of curvature; out through the center of curvature You must bend rays at the plane of the optic (optical interface plane) in order to get accurate ray trace answers.
Positive lens. Three rays: (1) in parallel, out through focus; (2) in through focus, out parallel; (3) Chief Ray: straight through the lens center. Shows the image point if there were no second lens. Ray Tracing Negative lenses The parallel ray: In parallel; out as if from the front side focal point The focal ray: In at the backside focal point; out parallel The chief ray: Straight through the lens center
(Ex. 15) Two thin converging lenses A and B, each having a focal length of 6 centimeters. are placed 10 centimeters apart, as shown in the figure above. If an object is placed 10 centimeters to the left of the lens A, the final image is (A) 30 cm to the right of lens B (B) 30/11 cm to the right of lens B (C) 30/10 cm to the right of lens B (D) 30/11 cm to the left of lens B (E) 30/10 cm to the left of lens B A positive lens increases the convergence of the light that passes through it. Light diverges from a real object. Real objects have positive object distances. Light converges to a real image with a positive image distance. The focal length of the lenses in this problem is f = 6 cm. The object is 10 cm before the left lens and so it has an object distance + 10 cm. Using the lens equation,  
Lens A would form an image 15 cm to its right if there were no second lens. That would be 5 cm to the right of lens B. Lens B is converging so the rays should converge closer to lens B than 5 cm. (Lens B is converging so we are focusing on responses (B) and (C).) The appearance of 5 cm and 6 cm makes the 30/11 cm value look attractive. The light arriving at lens B is converging. That case is called a virtual object and corresponds to a negative object distance. (Light diverges from a real object and converges to form real image. These cases correspond to positive object and image distances. Converging light incident on a lens and diverging light leaving the lens corresponds to virtual objects and virtual images negative distances for the lens formula.) Rays are converging as they reach the second optic lens two has a virtual object negative object distance. The rays as converging to a point 5 cm after the lens dObj,2 = - 5 cm. Applying the formula to lens B, 
The image distance is positive so the light converges to a point on the side of the lens opposite to that from which the light was incident. right side at 30/11 cm. B
First lens. Three rays: (1) in parallel, out through focus; (2) in through focus, out parallel; (3) straight through the lens center. Shows the image point if there were no second lens. To trace the second lens, you generate construction rays that are directed to converge at the position of the image that would have been formed by the first lens in the absence of the second lens. New image.
The blue rays shows the final image after the action of the second lens. Real image: the rays actually converge to and pass through a real image point. Second lens: Three rays: (1) in parallel directed at the image point of the first lens, out through focus; (2) in through focus directed at the image point of the first lens, out parallel; (3) straight through the lens center directed at the image point of the first lens. Shows the intermediate and final images. Rays are converging as they reach the second optic lens two has a virtual object negative object distance. The rays as converging to a point 5 cm after the lens dobj,2 = - 5 cm.  
(16) Basic Wave Properties Basic Wave Plug*: v = f The wave-speed is the frequency of the wave times its wavelength. THE BASIC WAVE PLUG (BWP) is the most important wave property relation. Wave Speed: v = Traveling Wave: y(x,t) = A sin[ k x - t + ] k = 2; T = 2
Standing Wave: y(x,t) = A cos[t + ] sin[ k x + ] node to node spacing: ½ node to anti-node spacing: ¼ Wave types: longitudinal and transverse There are two transverse directions so transverse waves have polarizations Waves reflect with a sign change when they reflect from a region with lower wave speed. Waves reflect with a sign change when they reflect from a fixed end. Waves reflect with no sign change when they reflect from a region with higher wave speed. Waves reflect with no sign change when they reflect from a free end.
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= 2 mV
. The E and B fields oscillate in phase with one another and B/E = v, the wave speed (= c in vacuum). The Poynting vector has units W/m2 and represents the power flow of the wave.
and time t, the magnetic field points along the positive x-axis, as shown in the figure to the right.
represents the intensity of the wave times its propagation direction. Find the direction of 
