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# lec20 - 20. Lecture 20 20.0.4 Fresnel Equations In the...

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Unformatted text preview: 20. Lecture 20 20.0.4 Fresnel Equations In the previous lecture we discussed that light can experience a 1800 phase shift when reﬂected from an interface. Here we are going to show that this follows from the properties of light as an electromagnetic wave. Typical media that we consider are air, water, glass, etc. which are dielectrics. Namely they are made of tiny dipoles that orient themselves with the electric ﬁeld. In the presence of an electromagnetic wave they oscillate producing a slow down of light and the property of refraction. What is important for us here is that they generate a surface charge and therefore the normal component of the electric ﬁeld can have a jump at the interface. On the other hand the component of the electric ﬁeld parallel to the interface is continuous. Besides, the magnetic ﬁeld is also continuous since the materials we consider are not magnetic and therefore generate no magnetic ﬁelds at the interface. To understand the calculation we draw ﬁg.113 were we consider the case when the electric ﬁeld is parallel to the interface in which case it has to be continuous: Ei + Er = Et (20.1) that is, the electric ﬁeld in medium 1 is the sum of the incident and reﬂected ones and has to be equal to the electric ﬁeld in medium two which is the transmitted one. The magnetic ﬁeld has two components and we have −Bi cos θ + Br cos θ = −Bt cos θ￿ −Bi sin θ − Br sin θ = −Bt sin θ ￿ (20.2) (20.3) ￿ ￿ Notice that the direction of the magnetic ﬁeld is determined by the fact that E × B points in the direction of propagation of the wave. Moreover, the strength of the magnetic ﬁeld is related to the electric ﬁeld by the speed of light: Bi = Ei , c1 Br = Er , c1 Bt = Et . c2 (20.4) Using eq.(20.4) to replace the magnetic ﬁeld in eqns.(20.2) and (20.3) we get Ei Er Et cos θ + cos θ = − cos θ￿ c1 c1 c2 Ei Er Et − sin θ − sin θ = − sin θ￿ c1 c1 c2 − (20.5) (20.6) Taking into account that the speed of light c1,2 in each media are given in terms of the index of refraction by c c1 n2 c 1,2 = ⇒ = (20.7) n 1,2 c2 n1 – 131 – we ﬁnd that eq.(20.6) reduces to n1 sin θ (Ei + Er ) = n2 sin θ￿ Et (20.8) Comparing with eq.(20.1) we see that we need to satisfy n1 sin θ = n2 sin θ￿ (20.9) which is Snell’s law!. Therefore we are left with two equations: Ei + Er = Et (20.10) n2 −Ei cos θ + Er cos θ = − Et cos θ￿ (20.11) n1 namely eqns.(20.1) and (20.2). We have two equations and we need to compute two quantities, Er and Et , it is then just a matter of algebra. Before proceeding we can take the simplest case of normal incidence when θ = θ￿ = 0. This gives Ei + Er = Et n2 −Ei + Er = − Et n1 (20.12) (20.13) from where we derive 2 n1 Ei (20.14) n1 + n2 n1 − n2 Er = Ei (20.15) n1 + n2 which in particular implies that, if n2 > n1 the reﬂected electric ﬁeld is opposite to the incident one, namely a 180o phase shift. Equivalent to shifting the wave by a half wave-length. If the incidence is not normal the same calculation gives for the reﬂected electric ﬁeld n1 cos θ − n2 cos θ￿ Er = Ei (20.16) n1 cos θ − n2 cos θ￿ sin θ￿ cos θ − sin θ cos θ￿ = Ei (20.17) sin θ￿ cos θ − sin θ cos θ￿ sin(θ￿ − θ) = Ei (20.18) sin(θ￿ + θ) Et = where we used Snell’s law and the identity sin(α + β ) = sin α cos β + cos α sin β . We again derive that n1 > n2 ⇒ θ < θ￿ ⇒ Er , Ei same sign ⇒ no phase shift ￿ n1 < n2 ⇒ θ > θ ⇒ Er , Ei opposite sign ⇒ 180o phase shift (20.19) Besides, from Er and Et we can also compute how much of the wave is reﬂected and how much transmitted. That this calculations agree with the experiment leave very little doubt that light is indeed an electromagnetic wave. – 132 – Figure 113: Refraction of an electromagnetic wave. 20.1 Gratings A grating is a special device composed of a large number of slits parallel to each other. It works similarly to the two slits but it is very eﬀective at separating light in its diﬀerent constituent colors because the interference maxima are quite narrow. Typical gratings have hundreds of slits per millimeter. In ﬁg.114 we see that the diﬀerence in path length between two neighboring rays is given by ∆L = d sin θ (20.20) If it is an integer number m of wave-lengths d sin θM = mλ (20.21) then two consecutive paths interfere constructively. Moreover it is easy to see that any path will interfere constructively with any other one and then we will have a maximum. – 133 – It is also clear that if we change the angle slightly then consecutive paths might still interfere constructively but not paths originating from far apart slits because the path diﬀerence would be larger. This makes the peaks very narrow. This phenomenon is very important to decompose light in its component wave-lengths. This way we can study light coming from distant stars and understand their chemical composition. The reason is that each element absorbs particular wave-lengths and therefore leave their characteristic imprint in the spectrum. Sort of a “ﬁngerprinting” for chemical elements. Gratings can also be reﬂective, an everyday example is a CD which has parallel grooves very close together. If you look at light reﬂected from a CD it is easy to see a rainbowlike pattern. Prisms work similarly using that the refraction index depends on the wave-length but they are not eﬃcient because to separate the spectrum they have to be thick and therefore absorb too much light. Nevertheless the actual rainbow in the sky is produced in this way by refraction inside tiny drops of water. Figure 114: Grating. The angles corresponding to maxima are easily computed as d sin θM = nλ . – 134 – Figure 115: A laser has longer coherence length than a light bulb and makes it easier to look at the interference patterns of gratings as seen in the next ﬁgure. 20.2 Diﬀraction Interference can also occur when light goes through only one slit in which case is known as diﬀraction. By shining light on a slit one can see on a screen behind a pattern of light and dark stripes as schematically indicated in ﬁg.117. If we divide the slit into tiny imaginary slits we see that, if a ray coming from the top of the slit interferes destructively with a ray coming from the middle, then for every ray in the top half of the slit there is a ray in the bottom half that cancels it. This gives the condition for a minimum as ￿ ￿ a 1 sin θm = m + λ (20.22) 2 2 where a is the width of the slit and m is an integer. The ﬁrst minimum or dark band appears at an angle λ sin θm = (20.23) a Since the opening is usually signiﬁcantly larger than the wave-length the angle is small and we can use the approximation sin θm ￿ θm giving θm ￿ λ a – 135 – (20.24) Figure 116: Pattern generated by laser light going through a grating. If we put a screen at a distance D the width h of the bright band in the middle is h = 2D tan θm ￿ – 136 – 2 λD a (20.25) Figure 117: Diﬀraction through a slit. – 137 – ...
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## This note was uploaded on 12/07/2011 for the course PHY 219 taught by Professor Na during the Fall '11 term at Purdue.

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