θ =θ i r n sinθ = n sinθ
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1 θ i = θ r n = 1 sinθ1 n sin θ
2 Total Internal Reflection Consider light moving from glass (n 1 =1.5) to air (n =1.0) n 1 incident ray θ 1 θ r reflected ray GLASS sinθ sinθ 1 > 1 = n n 1 θ > θ 1 n θ refracted ray AIR I.e., light is bent away from the normal. as θ 1 gets bigger, θ gets bigger, but θ can never get bigger than 90!! In general, if sin θ 1 > (n / n 1 ), we have NO refracted ray; we have TOL INTERNAL REFLECTION. sin θ c = n /n 1 For eample, light in water which is incident on an air surface with angle θ 1 > θ c = sin -1 (1.0/1.33) = 48.8 will be totally reflected. This property is the basis for the optical fiber communication.
3 Preflight 4 4) The path of light is bent as it passes from medium 1 to medium. Compare the indees of refraction in the two mediums. a) n 1 > n b) n 1 = n Snell s Law: n 1 sinθ 1 = n sinθ Here, θ >θ 1 implies n < n 1 c) n 1 < n 5) A light ray travels in a medium with n 1 and completely reflects from the surface of a medium with n. The critical angle depends on: a) n 1 only b) n only c) n 1 and n Critical angle occurs when θ = 90 o Therefore, sinθ critical = n /n 1
4 Total Internal Reflection Total internal reflection occurs when θ>θ c and provides 100% reflection. This has better efficiency than silvered mirror. Eamples of devices using Critical Angle Prism Binoculars Fiber Optics Fiber optics is etremely important for high speed Internet and digital data transfer at long distances. Many companies (Lucent) have laid fiber over long Distances to provide internet service.
5 I (academic) network of fiber connections 10 gigabit/sec connections
6 Lecture 4, ACT : Critical Angle An optical fiber is surrounded by another dielectric. In case I this is water, with an inde of refraction of 1.33, while in case II this is air with an inde of refraction of Compare the critical angles for total internal reflection in these two cases Case I Case II θ c θ c water n =1.33 glass n =1.5 water n =1.33 air n =1.00 glass n =1.5 air n =1.00 a) θ ci >θ cii b) θ ci =θ cii c) θ ci <θ cii
7 Lecture 4, ACT : Critical Angle An optical fiber is surrounded by another dielectric. In case I this is water, with an inde of refraction of 1.33, while in case II this is air with an inde of refraction of Compare the critical angles for total internal reflection in these two cases a) θ ci >θ cii b) θ ci =θ cii c) θ ci <θ cii n n 1 n 1 >n Case I Case II θ c θ c water n =1.33 glass n =1.5 water n =1.33 air n =1.00 glass n =1.5 air n =1.00 Since n 1 >n TIR will occur for θ > critical angle. Snell s law says sinθ c =n /n 1. If n =1.0, then θ c is as small as it can be. So θ ci >θ cii.
8 Dispersion: n = n(ω) The inde of refraction depends on frequency, due to the presence of resonant transition lines. For eample, ultraviolet absorption bands in glass cause a rising inde of refraction in the visible, i.e., n(higher ω) > n(lower ω): n red = 1.5 n blue = 1.53 Inde of refraction 1.54 white light frequency ultraviolet absorption bands prism Split into Colors
9 Rainbows
10 Rainbows
11 y UP ο E fast slow E s λ/4 E f RCP
12 Polariation Consider our EM plane wave. The E field is polaried in the Y-direction. We say this is linearly Polaried light. E B y = = E E c 0 0 sin( sin( k k ωt) ωt) Most light sources are not polaried in a particular direction. This is called unpolaried light or radiation. polaroid (sunglasses) Long molecules absorb E- field parallel to molecule. (transmission ais)
13 LP Intensity Reduction UP (unpolaried light) I=I 0 I 1 =?? E 1 θ nˆ LP θ E LP I =?? This set of two linear polariers produces LP light. What is the final intensity? First LP transmits 1/ of the unpolaried light: I 1 = 1/ I 0 Second LP projects out the E-field component parallel to the : E = E 1 cosθ I E I = I1 cos θ Law of Malus
14 Preflight 3 y 3) An EM wave is passed through a linear polarier. Which component of the E-field is absorbed? The component of the E-field which is absorbed is. a) perpendicular to the transmission ais b) parallel to the transmission ais c) both components are absorbed
15 Preflight 3 y An EM wave polaried along the y-ais, is incident on two orthogonal polariers. 5) What percentage of the intensity gets through both polariers? a) 50% b) 5% c) 0% 6) Is it possible to increase this percentage by inserting another polarier between the original two? Eplain.
16 What you said y ZERO: the component that is passed by the first is blocked by the second. BUT, adding an intermediate polarier will restore some light!!! Right: Yes, a polarier at an intermediate angle can change the direction of polariation so that some light is able to get through the last filter. Wrong: No. One polarier will get rid of all of the field that is perpendicular to the ais and the other will get rid of all that is parallel. No other filter will add more signal in any direction.
17 Lecture 3, ACT 1 Light of intensity I 0, polaried along the direction is incident on a set of linear polariers as shown. 1A Assuming θ =, what is I, the intensity at the eit of the polariers, in terms of I 0? (a) (b) (c) y E 0 I=I I = I 0 I = I I 0 = 0 4 θ I =? 1B y What is the relation between I and I 30, the final intensities in the situation above when the angle θ = and 30, respectively? I < I = I I 30 > I30 (a) (b) (c) I 30 E 0 I=I 0 y E 0 I=I 0 30 I =? I 30 =?
18 Lecture 3, ACT 1 Light of intensity I 0, polaried along the direction is incident on a set of linear polariers as shown. 1A Assuming θ =, what is I, the intensity at the eit of the polariers, in terms of I 0? 1 1 I = I 0 I = I I 0 = 0 4 (a) (b) (c) y E 0 I=I 0 θ ˆn 1 E 1 ˆn E I =? We proceed through each polarier in turn. The intensity after the first polarier is: ( ) I1 = I0cos 0 = The electric field after the first polarier is LP at θ 1 =. I 0 The intensity after the second polarier is: cos 1 1 ( 90 I I = I ) = I 1 = I 4 0
19 Lecture 3, ACT 1 Light of intensity I 0, polaried along the direction is incident on a set of linear polariers as shown. 1A 1B Assuming θ =, what is the I=I relation between the I, the intensity 0 at the eit of the polariers, in terms of I 0? 1 1 (a) I = I 0 (b) I = I 0 (c) I 4 = 0 What is the relation between I and I 30, the final intensities in the situation above when the angle θ = and 30, respectively? (a) I < (b) I30 (c) I 30 I = I > I 30 In general, the first polarier reduces the intensity by cos θ, while the second polarier reduces it by an additional factor of cos (90 θ). Thus, the final output intensity is given by: ( θ) ( θ) ( θ) ( θ) ( θ) I = I cos cos 90 = I cos sin sin out 0 0 y E 0 This has a maimum when θ =. θ ˆn 1 E 1 ˆn I =? E
θ =θ i r n sinθ = n sinθ
θ i = θ r n = 1 sinθ1 n2 sin θ 2 Index of Refraction Speed of light, c, in vacuum is 3x10 8 m/s Speed of light, v, in different medium can be v < c. index of refraction, n = c/v. frequency, f, does not
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