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Week 8 Physics 5b Notes

by: Shanee Dinay

Week 8 Physics 5b Notes PHYS 5B

Shanee Dinay
GPA 3.94

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Week 8 notes for Physics 5b or Intro to Physics 2. Topics include Intensity of Light, Diffraction Grating, Single Slit Diffraction, Location of Dark Fringes and Bright Fringes, Polarization, Double...
Intro to Physics II
Class Notes
Physics 5b, Physics II, light, Intensity, diffraction, Polarization
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This 11 page Class Notes was uploaded by Shanee Dinay on Monday February 29, 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 24 views. For similar materials see Intro to Physics II in Physics 2 at University of California - Santa Cruz.


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Date Created: 02/29/16
Day 20 ­ 2/22/2016  Phys 5b    Intensity at random point on screen  I(y) or I(θ )  Δθ 2π A = | 2acos( 2 ) | Δϕ = hΔr =  λ Δ r Δr = dsinθ   1 2π 2 2 A = | 2acos [ 2 λ d sinθ] I ~ A​I​ o​= Ca​   2​ 2​ 2​πd I(θ) = CA​  = 4Ca​ cos​ [λ sinθ]  2​πd πd I(θ) = 4I​o​os​ [ λ sinθ ]α =  λ sinθ   2 Double Slit Pattern: ​ I(α) = 4I​o​os​ α    2 πd N­Slits: I(θ ) = I​(r) = sin [πdsinθ] α =  πd sinθ   o​ sin λ sinθ] λ sin [Nα]  I(θ ) = o​r) =  sin α   Max: dsinθ = nλ → n = 0, 1, 2, 3,...     sin [λsinθ] πd I(θ) = I​o​r) = sin [λsinθ]α =  λ sinθ   We are trying to build the limit from α → nπ  2 2sin[Nα]cos(Nα) I(α) = I​o​lim sin 2Nα = NI​o 2   α→nπ sin α sin αcosα         = derivative using quotient rule  Summary:  N slits diffraction grating of bright angles: sinθ​ n​=  nλ  d locations on the screen y​ n​= Ltanθ ​n  n = 0,1, 2, …  (2n+1)λ angles of dark fringes: sinθ ​ n​=  2d   Diffraction Grating  w width of grating w, N total numbers of slits d =  N nλ λ 500nm sinθ​ n​  d d  =  1 x 10 m = 0.5  λ red λ violet nλ nλ sinθ​ n​  d = w   λ r λ  v Single Slit Diffraction  Intensity of Individual Source  2 I(θ) = I​ si2 η β = πasinθ  o β λ Minima: β = nπ    n = 0, 1, 2, … dark fringes  πa sinθ​  = nπ  λ n​ Location of Dark Fringes: sinθ   =  nλ   n a CAnnot determine max, only dark fringes  Condition for dark fringes  λ sinθ n= n d n = 1, 2, 3, …   λ θ n= n d yn​= Ltanθ  n Width of Central Max  w = y​  ­ y’ = 2Lλ  1​ 1​ a   Day 21 ­ 2/24/2016    Polarization Chapters 34.11, 35.12    Double Slit:  Example  d = 0.4mm λ1= 589nm λ2= 589.6nm  λL yn​= n d   λ1L yn1​= n d n=1  λ2L yn2​= n d L = 1m  y1 1​  y​ =   1 2​ We want to see how for apart they are:  10 m Δy = y​12​­ 11​= 0.4•1030.6)  = 1.5 μm  Now in diffraction grating:  w = 1mm N = 1000  −3 ­6​ d =  N  11000 = 10​ m  yn​= L tan θ n λ sinθ n= n  d nλ θn = arcsin( d )  589•10  m o n = 1 → θ  1 arcsin( 10 m ) = 36.086​   o n = 2 → θ  1 36.128​   y11​= 1m tan θ  110.7288 m  y12​= 0.72996 m  Δy = y​  ­ y​= 1.16 mm  12​ 11 ​   For a given set up, how many orders can we see on the screen:  λv= 397nm λr = 656.3nm  Diffraction grating with N = 500 lines, w = 1mm  −3 d =  10 m   500 sinθ  =  2nλ= nλ =  5−3nλ = 5•10​ 5 (nλ )m​­1  n 2d d 10 n = 1  5​ ­1​ ­9​ violet: sinθ 1V = 5•10​  m​  397•10​ m = 0.199  o θ1V  = 11.48​ y​ 1V​tan θ 1V  = 20.31cm  o red: θ 1R 19.15​ yi1R​34.73cm  Δy = y​1R​y​1V​(34.73 ­ 20.31)cm = 14.42cm  When you continue up the screen to second order and third order:    sinθ n =  nd < 1 !  nλ d d  < 1 → n < n​ max​ λ Magnetic Field  Light is a Transverse Wave  Electromagnetic Wave  Electric Field Oscillates like this in 2­dimensions    The wave travels in the x direction at the speed of light  In 3­dimensions:  blue ink is x direction  Equations we have derived:  2 2 2 2 d y =  1 d y d ρ  =  1 d ρ  dx2 vw dt2 dx2 cw dt2 d E 1 d E Magnetic Field:  dx2  =  c2 dt  → E(x, t) = E​ o​in(kx ­ wt + ϕ )  o Our electric field can only swing this way:    We can have another wave that does this in the z­direction:    Unpolarized:    Polarizers:  Oriented in the y­direction:  The beam is coming through the polarizer:   The are coming in as represented by the blue lines  We turned light that was randomly polarized in to light that is linearly polarized in the  x­direction    Day 22 ­ 2/26/2016  Phys 5b    Unpolarized Beam passed through a Polarizer:    Example 1:     Now we will pass it through the second polarizer:    The cross polarizer gives us nothing out of it.  Example 2:    We inserted a third polarizer between the two:  E is the electric field:    We have an electric field in the second polarizer with an angle:      We can substitute I​ 1​rom the first equation:  Now for the right­most polarizer:    2 I3​ I2​in​θ2  I = I​cos​θ  sin​2θ  = Icos​2θ cos​ 2θ sin​θ   3​ 1​ 2 2 o​ 1 2 2 θ =60​ o θ  = 30​o  1 2 I = I​ •  3 1 =  3 I​ 3​ o4 4 4 64 o  What would be the state of polarization whose electric fields are perpendicular to each other:  Δϕ o =   ± 2nπ n = 0, 1, 2, …  we will choose n =0    this is the first wave    By adding the two perpendicular waves to each other, that have phases that differ from  each other by 2π we get a linearly polarized wave.  We will rotate the wave to see the plane and the sum of polarization:  Ex​(z,t) = iox​n(kz ­ wt)  Ey​(z,t) = oy​n(kz ­ wt)  Adding the two together → E(z,t) = ( iEox ​E​oy​in(kz­wt)  Plane of Oscillation:    (2n + 1)π 2) Eox​= Eoy​ Eo Δϕo = − 2   E​x​= Eo​cos(kz ­ wt)  E​  = Ej cos(kz ­ wt ­  ) = Ejsin(k ­ wt)  y​ o​ 2 o ​ + E = E​o​ icos(kz ­wt) + jsin(kz ­ wt)] ← looks like a circle  Pick x up, y to the right, z is the direction of propagation:    Now we do the sign wave:    the amplitude of the wave changes from pointing  along x and y axes    We can see how the state of polarization changes via a circle:    When light passes through a medium:  ­ speed of propagation decreases  vn​< c n =  c Index of Refraction  vn E​p​= ?  What is the difference when we place a glass screen?  ­ the glass forces the light to oscillates  Ep​= Es​+ Echarges    wave front propagates this way:    Now the wave length as it passes through the medium is shorter  c v c λ < λ →oλ = o f λ =   f λ = o →  f v​n​= c  c λo n =   n  λ   Law of Refraction: another way    What does an electromagnetic wave look like at point P:  E​  = E​ cos(kz ­ wt) = E​ cos[w(t­ )]  s​ o​ o​ c k =  w This is what the wave would look like if we did not have a glass plate.  c   z1+2 Δt The traveling time of the wave: t =  c + c   Now we have the glass:    z +z z +z The new traveling time: t’ =  1 2+ Δz =  1 2 + nΔz   c √ n c c t’ = t ­ Δz  + nΔz  = t + (n­1) Δz  c c c t = t’ ­ (n­1) Δz  c E​ = E​ [w(t’ ­   ­ (n­1) Δz )]  PSG ​ o​ c c E​= E​ cos{w(t­ )}  SP ​ o​ c Inserting a glass plate allows us to slow the travel time  How does the elactric field get altered when we add a glass plate with respect to when there  was no glass plate?  Conclusion: The new Field differs from the old one by a phase ϕ =− (n − 1) ω  Δz o c   Scattered Light:  ­ incoming electric field forces light particles to go up and down  ­ oscillating in the y­direction:     


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