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Total Internal Reflection Lecture 24, ACT 2: Critical Angle… Total Internal Reflection – Consider light moving from glass (n1=1.5) to air (n2=1.0) incident reflected sinθ n ray ray 2 = 1 >1 n1 θ1 θr θ2 > θ1 GLASS sinθ1 n2 θi =θr AIR n2 θ2 I.e., light is bent away from the normal. refracted as θ1 gets bigger, θ2 gets bigger, but θ2 ray can never get bigger than 90°!! n1 sinθ1 = n2 sinθ 2 2 In general, if sin θ1 > (n2 / n1), we have NO refracted ray; we have TOTAL INTERNAL REFLECTION. sin θc = n2/n1 For example, light in water which is incident on an air surface -1 with angle θ1 > θc = sin (1.0/1.33) = 48.8° will be totally reflected. This property is the basis for the optical fiber communication. Total Internal Reflection Fiber Optic Network Maps Total internal reflection occurs when θ>θc and provides 100% reflection. This has better efficiency than silvered mirror. Examples of devices using Critical Angle • Prism Binoculars • Fiber Optics Fiber optics has replaced copper for Fiber optics is extremely important for high speed long distance communication and is Internet and digital data transfer at long distances. becoming increasingly used for local Many companies (Lucent) have laid fiber over long communications. Distances to provide internet service. Lecture 24, ACT 2: Critical Angle… Lecture 24, ACT 2: Critical Angle… An optical fiber is An optical fiber is Case I water n =1.33 Case I water n =1.33 surrounded by another surrounded by another dielectric. In case I this is dielectric. In case I this is θ glass n =1.5 θ glass n =1.5 water, with an index of c water, with an index of c refraction of 1.33, while in water n =1.33 refraction of 1.33, while in water n =1.33 case II this is air with an case II this is air with an index of refraction of 1.00. index of refraction of 1.00. Case II air n =1.00 Case II air n =1.00 Compare the critical angles Compare the critical angles for total internal reflection for total internal reflection θ glass n =1.5 θ glass n =1.5 in these two cases c in these two cases c air n =1.00 air n =1.00 a) θcI>θcII a) θcI>θcII n2 Since n1>n2 TIR will occur for θ > critical angle. b) θcI=θcII b) θcI=θcII n1 Snell’s law says sinθc=n2/n1. c) θ <θ c) θ <θ n1>n2 cI cII cI cII If n2=1.0, then θc is as small as it can be. So θcI >θcII . 1 Polarization Consider our EM plane wave. The E field is polarized in the Y-direction. We say this is “linearly polarized light”. E y = E 0 sin( kx − ωt) TA E y ο 0 45 B z = sin( kx − ωt) x c E fast Most light sources are not polarized ow UP sl in a particular direction. This is called Es λ/4 unpolarized light or radiation. x z polaroid (sunglasses) Ef RCP Long molecules absorb E- TA field parallel to molecule. (transmission axis) EM wave; energy and intensity EM wave; Intensity Previously, we demonstrated the energy density existed in E fields in The intensity, I, is the energy flow per unit area. Capacitor and in B fields in inductors. We can sum these energies, In Y&F, section 32.4, the average for a sinusoidal plane wave 1 2 1 2 (squared) is worked out, u = ε 0 E + B 2 2μ0 ε c Since, E=cB and c = 1 / μ ε then u in terms of E, 0 2 0 0 I = Emax Y&F section 32.4 2 2 1 2 1 ⎛ E ⎞ 1 2 1 2 2 u = ε 0 E + ⎜ ⎟ = ε 0 E + ε 0 E = ε 0 E 2 2μ0 ⎝ c ⎠ 2 2 => EM wave has energy and can transport energy at speed c Suppose we have transverse E and B fields moving, ct at velocity2 c, to the right. If the energy density is u = ε0E and the field is moving through area A A at velocity c, the energy in volume Act, uAct, passes through. Hence the amount of energy flow per unit time per unit surface area is, 1 dU 1 = ()uAct = uc = ε cE 2 A dt At 0 LP Intensity Reduction Cell Phone Problem TA 32.18) A sinusoidal electromagnetic wave emitted by a cell nˆ phone has a wavelength of 35.4 cm and an electric field UP TA -2 E1 θ amplitude of 5.4 x 10 V/m at a distance of 250 m from the (unpolarized light) antenna. (a) The intensity? (b) The total average power? I=I 0 θ I =?? E2 1 LP 1 a.) I = ε cE 2 ; 2 0 1 −12 2 2 8 −2 I =?? I = (8.85×10 C / Nm )(3×10 m / s)(5.4×10 V / m) LP 2 2 −6 2 • This set of two linear polarizers produces LP light. What is I = 3.87 ×10 W / m the final intensity? – First LP transmits 1/2 of the unpolarized light: I = 1/2 I b.) Assume isotropic: I=P/A; P=4πr2 I = 4 π (250m)2 I 1 0 P = 3 W – Second LP projects out the E-field component parallel to the TA: E2 = E1 cosθ 2 I2 = I1 cos θ Law of Malus I ∝ E2 2 Preflight 23 y y An EM wave polarized along the y-axis, Yes … is incident on two orthogonal polarizers. TA TA TA adding an intermediate TA polarizer will restore some light !!! z x z x 5) What percentage of the intensity gets through both polarizers? • A polarizer at an intermediate angle can change the direction of polarization a) 50% so that some light is able to get through the last filter. b) 25% c) 0% 6) Is it possible to increase this percentage by inserting another polarizer between the original two? Explain. Lecture 23, ACT 1 Lecture 23, ACT 1 nˆ 1 TA TA • Light of intensity I0, polarized along the x • Light of intensity I0, polarized along x θ the x direction is incident on a set of θ x direction is incident on a set of 2 E0 E0 linear polarizers as shown. 2 linear polarizers as shown. E1 y y nˆ 2 – Assuming θ = 45°,what is I45, – Assuming θ = 45°,what is I , 1A I=I0 TA 1A 45 I=I0 TA the intensity at the exit of the 2 E the intensity at the exit of the 2 polarizers, in terms of I ? 0 z z I =? 2 polarizers, in terms of I0? I =? 1 1 45 1 1 45 (a) I = I (b) I = I (c) I 45 = 0 (a)I = I (b) I = I (c) I 45 = 0 45 2 0 45 4 0 45 2 0 45 4 0 • What is the relation between I and I , the final intensities in the • We proceed through each polarizer in turn. 1B 45 30 2 I situation above when the angle θ = 45° and 30°, respectively? • The intensity after the first polarizer is: II=−=cos() 45 0 0 10 2 I45 > I30 (a) I45 < I30 (b) I45 = I30 (c) • The electric field after the first polarizer is LP at θ1 = 45. 30 TA x 45 TA x • The intensity after the second polarizer is: E0 E0 y TA y 2 I1 1 I=I0 I=I 0 TA II45= 1 cos() 90−= 45 I = I 2 454 0 z z I45=? I30=? Lecture 23, ACT 1 nˆ y 1 Polarization by reflection TA E0 • Light of intensity I0, polarized along the x x direction is incident on a set of 2 E θ 0 E0y linear polarizers as shown. E1 y ˆ unpolarized – Assuming θ = 45°, what is the n 2 I=I TA E 1A relation between the I , the intensity 0 0x x 45 E at the exit of the 2 polarizers, in 2 Can always describe z θ θ E in terms of terms of I0? i i I45 =? n 1 1 a components in two (a) I 45 = I 0 (b) I45 = I0 (c) I = 0 arbitrary directions. 2 4 45 nb The components are 1B • What is the relation between I45 and I30 , the final intensities in the equal for unpolarized situation above when the angle θ =45° and 30°, respectively? Medium b light. I > I θ (a) I45 < I30 (b) I45 =I30 (c) 45 30 r • In general, the first polarizer reduces the intensity by cos2θ, while the second polarizer reduces it by an additional factor of cos2(90 – θ). The reflected rays are partially • Thus, the final output intensity is given by: polarized in the horizontal plane. 22 22 2 The transmitted rays are also IIout =−=∝00cos()θ cos ( 90θθθθ ) I cos () sin () sin ( 2 ) partially polarized. This has a maximum when θ = 45°. 3 Polarization by reflection Light reflected on dashboard to the windshield will be polarized in the horizontal plane. Using polaroid dark For a certain angle, glasses with a vertical axis will remove most of reflected the Brewster angle, light θp θp the reflected light is na completely polarized nb in the horizontal plane. This occurs when the angle between the refl. and o Medium b θr refr. rays is 90 . From Maxwell’s eqn. it can be shown that Brewster’s angle is given by nb tanθ p = na Polarization by Scattering Electric field lines from oscillating dipole • Suppose unpolarized light encounters an atom and scatters (energy absorbed & reradiated).
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