The individual electric and magnetic waves are in phase. The fields peak at the same position at the same time.

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Stanley Carroll
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1 1 Part 3: Otics 3.1: Electromagnetic Waves An electromagnetic wave (light wave) consists of oscillating electric and magnetic fields. The directions of the electric and magnetic fields are erendicular. The wave travels in the third erendicular direction. It is a transverse wave. The fields store energy and transort the energy in the wave. For examle, the icture below shows the wave traveling in the +x direction. The electric field oscillates in the y direction and the magnetic field oscillates in the z direction. The individual electric and magnetic waves are in hase. The fields eak at the same osition at the same time. In a harmonic light wave, the fields can be written using a cosine or sine function. E(x, t) = E cos M sin (kx ± ωt + φ) and B(x, t) = B cos M (kx ± ωt + φ) sin E M = electric field amlitude of wave [V/m] B M = magnetic field amlitude of wave [T] k = angular wave number [rad/m] = angular frequency [rad/s] = hase angle [rad]

2 2 = wavelength = distance between consecutive crests [m] k = 2 / [rad/m] T = eriod = time for one oscillation or cycle [s] f = 1 / T = frequency = number of cycles er time [cycles/s = Hz] = 2 / T = 2f [rad/s] The sign between kx and t determines the direction the wave travels along the x-axis. + wave travels to left (in the direction of decreasing x) - wave travels to right (in the direction of increasing x) The hase angle shifts the cosine or sine function left or right. This can be used to match some initial condition for the wave function. (We won t have to worry about it.) Wave Seed and Index of Refraction The seed of the light wave in a medium is v = E M B M = f = ω k = 1 εμ where and are the electrical ermittivity and magnetic ermeability of the medium, resectively. In free sace (vacuum), v = E M B M = o f = ω k o = 1 ε o μ o = c = 3x10 8 m/s This is the fastest that light can travel and only light can travel at this seed. In any other medium, the seed of light is v = c n where n is the refractive index or index of refraction of the medium. For vacuum, n = 1. For any other medium, n > 1. Air s refractive index is aroximately 1.

3 3 Power and Intensity The ower is the rate that energy is carried in the wave. P E M B M f 2 v E M 2 f 2 The ower is sread out over an area A. The intensity is ower er area. It determines the brightness of the light. I = P A Examle: The Sun emits ower in all directions. At Earth, the received intensity is aroximately 1370 W/m 2. This is known as the solar constant. Electromagnetic Sectrum Frequency of light determines the art of the sectrum. You should know the order of the sectrum and of the colors in the visible art of the sectrum (ROYGBV). Light in ifferent Media When light is traveling in Medium 1 and enters a different media (Medium 2) the frequency is constant. The seed and wavelength change. n 1 λ 1 = n 2 λ 2 Light in vacuum has a free-sace wavelength of o. Wavelength in medium with index n is λ = λ o n Examle: A laser ointer delivers a red beam in air with a wavelength of 650 nm. What is the wavelength and color of the beam underwater (n ~ 1.33)? Ans. 488 nm and the beam remains red because the frequency is the same

4 4 3.2 Reflection & Refraction Law of Reflection θ 1 = θ r n1 1 r Law of Refraction (Snell s Law) n 2 2 n 1 sin θ 1 = n 2 sin θ 2 Case 1: n 1 < n 2 (lo index into hi index) Case 2: n 1 > n 2 (hi index into lo index) 1 > 2 (light bends towards normal like in icture) 1 < 2 (light bends away from normal) Critical Angle (only for Case 2) sin θ c = n 2 n 1 If 1 > c then all light reflects for Total Internal Reflection (TIR) Normal isersion Index decreases as wavelength increases. For light going from air into medium, violet bends most, red least.

5 5 3.3 Image Formation Plane Mirror = object distance (from object to element) q = image distance (from element to image) h = object height (from axis to ti) h = image height (from axis to ti) Lateral Magnification M = h h q = h = h M = 1 virtual, non-inverted image eye h h q Single Refracting Surface (Aarent eth) q < h = h M = 1 virtual, non-inverted image closer to surface for small angles, q n 2 n 1 h h n 1 q n 2 eye

6 6 Thin Lens Positive (Converging) Lens Negative (iverging) Lens f F F f Lens Maker s Equation 1 f = (n lens n sur n sur ) ( 1 R 1 1 R 2 ) F = focal oint f = focal length n lens = index of lens material n sur = index of medium surrounding lens R 1 = radius of curvature of left surface (front surface) R 2 = radius of curvature of right surface (back surface) R > 0 for ( -shaed surface R < 0 for ) -shaed surface Thin Lens Equation 1 f = q h q Lateral Magnification h M = h h = q

7 7 Simle Otical Systems (You should know how to do the ray traces for these systems.) Camera ositive lens, object outside focal oint and far from lens real, inverted image image smaller than object Projector ositive lens, object outside focal oint but close to lens real, inverted image image bigger than object Magnifer (Magnifying Glass) ositive lens, object inside focal oint virtual, non-inverted image image bigger than object Angular Magnification m = 25 cm Largest useful angular mag occurs when q = -25 cm. Then, and only then, m = M. m max = 25 cm f + 1 Examle: You use a lens with a focal length of +5 cm as a magnifying glass to look at a 1 cm-long bug. (a) How far is the lens from the bug if you get the maximum useful angular mag? (b) What is the maximum useful mag? (c) What is the lateral mag? (d) How long does the bug look to you? Suose you now move the lens so that it is 4.8 cm from the bug. (e) How far is the image of the bug from you? (f) What is the angular mag now? (g) What is the lateral mag now? (h) How long does the bug look to you? Ans. (a) 4.17 cm (b) 6x (c) 6x (d) 6 cm (e) 120 cm (f) 5.2x (g) 25x (h) 5.2 cm

8 8 Eye: Correcting Vision Problems with Lenses Myoia (Near-Sightedness) can t see far-away objects eye has an unhealthy far oint (F.P.) of less than infinity need negative lens to fix f = F.P. Hyeroia (Far-Sightedness) can t see near objects eye has an unhealthy near oint (N.P.) greater than 25 cm need ositive lens to fix with = 25 cm and q = -N.P. Lens Power (L.P.) 1 f = 1 25 cm 1 N.P. L. P. = 1 f mks units [1/m = dioter]

9 Interference Young s ouble Slit and if L >> d mλ d sin θ = { (m + 1 2)λ m λl d y = { (m + 1 2) λl d maxima minima maxima minima d θ L y 0 where m = 0, ±1, ±2, Sacing between consecutive maxima = Sacing between consecutive minima = y = L / d iffraction Grating Grating Equation d sin θ = mλ where order number m = 0, ±1, ±2, d θ The grating constant is the recirocal of the line sacing d. [# lines / length] Thin Films If reflection is lo n off of hi n, then hase shift is π. If reflection is hi n off of lo n, then hase shift is 0. Find hase shifts for surfaces 1 and 2. Subtract the two hase shifts to get Δ. 1 2 cover n c film n substrate n s t For near-normal incidence, mλ 2nt = { (m + 1 2)λ Δ = 0 maxima minima Δ = minima maxima

10 10 Examle: A lens with an index of 1.5 has an anti-reflection coating with an index of 1.4 that minimizes the reflection of 580 nm-light. (a) What is the thinnest that this layer can be? (b) What is the longest wavelength that is strongly reflected? Ans. (a) nm (b) 290 nm so the transmission is enhnaced for all visible light

$iffraction Single Slit Intensity minima given by L y sin θ = mλ θ 0 and if L >> y = m λl where m = ±1, ±2, Width of central diffraction eak (CP) W CP = 2 λl$

11 iffraction Single Slit Intensity minima given by L y sin θ = mλ θ 0 and if L >> y = m λl where m = ±1, ±2, Width of central diffraction eak (CP) W CP = 2 λl Circular Aerture First minimum given by and if L >> sin θ 1 = 1.22λ y = 1.22 λl θ L y 0 iameter of Airy disk (CP) W CP = 2.44 λl W CP where m = 1, 2,

12 12 iffraction-limited Resolution Two oints on the object are just resolvable as two distinct oints on the image when sin θ 1 = 1.22λ d min θ 1 q d min The angle θ 1 is the resolution angle and since it is very small, sinθ 1 ~ θ 1 so the above equation can be written as θ 1 = 1.22λ [rads] The minimum searations of these two oints on the object and image are given by θ 1 [rads] = d min = d min q Useful angle conversions: π rad = = 60 = 60 1 arcminute = 60 arcseconds = 60 Examle: The uil of a erson s eye is oened to a diameter of 2 mm. Find (a), the resolution angle of this eye, and (b), the closest that two oints can be searated bye on the age of a book that is 25 cm from the eye. Use a free-sace wavelength of 560 nm which is in the middle of the visible sectrum. (c) These two oints are imaged on the retina which is 20 mm from the uil. How close are the two imaged oints on the retina? Ans. (a) 2.57x10-4 rad = 53 (b) 64 μm (c) 5 μm

13 Polarization Linear Polarization of Light by Linear Polarizer transmission axis incident electric field This arallel comonent gets through. The energy from this erendicular comonent is absorbed. Law of Malus I = I o cos 2 θ Intensity I o Intensity I

14 14 Linear Polarization by Reflection (Brewster s Law) As the incident angle θ gets closer to the olarization angle (Brewster angle) θ P, more of the reflected light is linearly olarized with the electric field axis arallel to the surface. When θ = θ P, all of the reflected light is linearly olarized with the electric field axis arallel to the surface. tan θ P = n 2 n 1 *Review the Poweroint slides on olarization located on the course age: htt://facstaff.cbu.edu/~jvarrian/252/252polweb/252polabs.t 3.7 Scattering and Absortion *Review the Poweroint slides on olarization located on the course age: htt://facstaff.cbu.edu/~jvarrian/252/252polweb/252polabs.t There are a few slides at the end of the resentation that discuss scattering and absortion.

Transverse wave - the disturbance is perpendicular to the propagation direction (e.g., wave on a string)

1 Part 5: Waves 5.1: Harmonic Waves Wave a disturbance in a medium that propagates Transverse wave - the disturbance is perpendicular to the propagation direction (e.g., wave on a string) Longitudinal