Week 9 Physics 5b Notes
Week 9 Physics 5b Notes PHYS 5B
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This 14 page Class Notes was uploaded by Shanee Dinay on Tuesday March 8, 2016. The Class Notes belongs to PHYS 5B at University of California - Santa Cruz taught by A.Steinacker in Fall 2015. Since its upload, it has received 13 views. For similar materials see Intro to Physics II in Physics 2 at University of California - Santa Cruz.
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Date Created: 03/08/16
Day 23 2/29/2016 Physics 5b Homework 8 is up on geometric objects reflection and refraction This week we will use lenses before we understand them Friday we will hold an additional lecture: 5:106:30pm in the Earth & Marine Classroom the lecture will be recorded Sunlight as it passes through the atmosphere is polarized: the wave travels in the direction perpendicular to the electric field there is no energy transported in the direction of the electric field If the wave strikes an atom: the energy by the wave that is generated cannot travel in the direction of the electric field but in the perpendicular direction we can measure the wave as a source If we measure a beam, we will see a wave that is polarized like the incoming wave Incoming wave that strikes a surface: we have the reflection plane, and the incidence place and we want to figure out how polarization occurs in this case: we will refer to the angles of incidence angle of incidence = angle of refraction We will see some intensity / radiation ni= 1 nr n o together i + r (refracting) = 90 if we want to create a situation where our incoming incident beam is very large o o i + r = 90→ r = 90 i nsin i = n sin r n = 1, n = n i r 0 i r sin i = n sin(90 i) = n cos i tan i = n i = β Brewster’s angle Example nr= n = 1.5 → β = tan (1.5) = 56.3 Particles will osciallate: Substances with Optical Activity: substances who contain molecules who contain no mirror symmetry if the current is moving upward, then the molecule is moving upward also if we have an extra + at the top and a minus at the bottom we generate an extra electric field: We measured a different light, just like when we had the corn syrup: Dark and Bright Fringes Review: sinθ = λ 1 a λ << a θ 1= a y 1= La 2λL w = a Circular opening: w = 2.44 aL w ≃ a λL 2 if we ignore the optics of 2.44, then we will have a ≃ a → a = λL a is going to be about 1mm Geometric Optics light travels in straight lines → rays if two rays intersect, nothing happens. Rays can intersect without consequences Pinhole camera: m magnification = = di hθ dθ If we take a mirror, a reflective surface if a beam makes an angle with a surface, the wave leaves the surface equals the angle of incidence How does light know to do that? Light takes the path of least time Now we have an incidence angle and a refraction angle θ i= θ r this is not so trivial! Next: take a surface and investigate how does light go through that surface… Day 24 3/2/2016 PHYS 5b Reading 32.8, 33.2, 33.4 Homework correct #4: means 75% not .75% Optical Path Length t = PO +OP′ c vn 2 2 PO = h √ x OP’ = h + (a − x) √ n 2 n = vn = vn= n 12 2 √h +x2 √ h +(a−x) t(x) = c + c 2x n2(a−x) t’ = 2 2 2 2 = 0 2c√h′+x c2√h′ +(a−x) sini = sinθ ← Snell’s Law c c r Total Internal Reflection key n1> n2! n1inθ1= n2nθ r θ i → θ r↓ o when θ r 90 ← → θ i= θ ccritial angle o n1inθc= n2n90 sinθ = n2 c n1 Ex θ > θ i c if θ = θ → nsinθ = 1 • sin90o i c c sinθ = c n sinθ > sinθ = i c n θ = 45o i √2> → n > 1.4 2 n Ex o θi= 60 nsinθ = nsinθ 1 i 2 r θ = 60 i θr= 35.3 Optical Devices Slab of Glass Choosing different angles on a planar piece of glass: o o Ex. Two beams that make 15 and 30 on glass n = 1.5 Apply Snell’s law to find index of refraction o o o sin15 = 1.5sinθ r θ i 15 θr= 9.9 Then we draw how the ray is going through the glass: Second Ray: 30 o θi= 30 o o sin30 = 1.5sinθ r → θ r1= 19.47 Then ray would travel up light this: We know that the resulting ray would come out parallel to the incoming beam: The two arrows show that the first image and the second image show that the glass is useless. A planar surface is no good for creating sharp images of objects The line in the photo: Optical Axis if the angles of incidence are small enough, the rays are paraxial rays, so it is very close to the optical axis. Then P1≃ P2≃ …. as long as we keep the angles small. We are going to use paraxial rays o o We will use small angles meaning less than 5 → θ << θ When we chose rays closer to the optical axis we did manage to get the two rays to come together to one point Now, we want the shortest distance from P to P o A lensmaker can find the closest path from P to P o curved path Lens Magic Converging Lens Diverging Lens and Converging Lens (symbolically) Thin Lens Lens has two focal points: distance between F and F’ are equal only for thin lenses F = F’ we place an object. s = object distance s’ = image distance First array we draw: parallel to optic axis and then gets refracted as it hits the yaxis Second array: goes through center and is unbent Third one: aray going to near focus and refracts parallel to the optic axis on the right side You can use any two arrays Then we can find the location of the image, s’ 1 + = 1 Lensmaker’s Equation s s′ f If we put a screen at the image distance, then we can project the image on the screen 1) s > 0, if object is located on the left of the lens 2) s’ > 0 if the image is on the right of the lens 3) the radius of curvature of the refracting surface is positive if the center of the refracting surface is to the right of the surface Day 25 3/4/2016 Physics 5b Reading: 32.8, 32.2, 33.4 Homework 4 take problem 10 off homework 8 Unbold Problem 10 we have not talked about Diverging Lens number 4 not worked out properly we want to find θ ind θ rin image b Two images of object P from last class: Converging Lens Where does the image form? When the object is located at the focus… image does not show up and s = f so we get s’ = ∞ Object Located Inside Focus… we get a virtual image because the image does not come together on the right Remember F = F’ Outside Focal Point… image size h’ distance of object: s distance of image: s’ Image that we have a medium that is stratified: light take the path of least time from one point to the other path from P to P’ calculate the total travel time: s1 2 sn n i t = v + v+ … + ∑vn v 1 2 i = 1i c c ni vi→ v i= ni n t = c∑ ns i i optical path length OPL i=1 n OPL = ∑ ns shortest path = shortest optical path length i=1 i i General way to finding minimum time: P1 t = 1 n(s)ds c ∫ P 2 Single Refracting Surface OPL = l = n 1o+ n2i 2 2 2 ΔPAC : lo = R + (So+ R) 2R(So+ R)cosϕ 2 2 ½ lo= [ R + (So+ R) 2R(So+ R)cosϕ ] 2 2 2 ΔACP: l i= R + (Si R) 2R(Si R)cos(180 ϕ ) 2 2 ½ Ii= [ R + (Si R) 2R(Si R)cos(180 ϕ) ] 2 2 ½ 2 2 ½ l = n1R + (So R) 2R(So+ R)cosϕ ] + n2 R + (Si R) 2R(Si R)cos(180 ϕ) ] Spherically Curved Surface Single Refracting Surface
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