Ref: J. D. Jackson: Classical Electrodynamics; A. Sommerfeld: Electrodynamics.

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1 2 Fresnel Equations Contents 2. Laws of reflection and refraction 2.2 Electric field parallel to the plane of incidence 2.3 Electric field perpendicular to the plane of incidence Keywords: Snell s law, Polarisation, Brewster s law Ref: J. D. Jackson: Classical Electrodynamics; A. Sommerfeld: Electrodynamics. 2. Laws of reflection and refraction In this section we derive laws of reflection and Snell s law from Maxwell s equations and boundary conditions. These are kinematical properties and are followed directly from the wave nature of light and the boundary conditions. The dynamical properties such as intensities of reflected and refracted light, phase changes and polarisations are dictated by the specific nature of electromagnetic fields and their boundary conditions. Consider a plane electromagnetic wave (in medium, with permittivity and permeability ǫ and µ respectively) incident obliquely on a medium 2 (with permittivity and permeability ǫ 2 and µ 2 respectively) at an incidence angle i to the in-

2 2 2 Fresnel Equations terface normal (Fig. 2.). The angles of reflection and refraction with respect to the interface normal (ˆn) are r and r respectively. The interface is given by the plane z = 0. Let the incident plane wave be represented by exp(ik I r iωt) (B I = µ ǫ k k I ). The frequency ω will force the charges(electrons) in the medium 2 to oscillate with the same frequency and in turn the oscillating charges will radiate with the same frequency. So we have one reflected ray, E R exp(ik R r iωt) (B R = µ ǫ k k R E R ), and one refracted (or transmitted) ray E T exp(ik T r iωt) (B T = µ 2 ǫ 2 k k T E T ) oscillating with the same frequency as the incident one. Wave vectors have magnitudes k I = k R = k = ω µ ǫ and k T = k = ω µ 2 ǫ 2. Applying boundary condition at z = 0, [ exp(ik I r iωt)+e R exp(ik R r iω R t)] z=0 = E T exp(ik T r iω T t) z=0 (2.) µ, ki i r =i kr µ 2, 2 r kt Fig. 2.: Electromagnetic waves from one medium to the other

3 2.2 Electric field parallel to the plane of incidence 3 The condition should be satisfied for all points x,y on the plane z = 0 and also for all time. This is a stringent condition and would demand, k I r z=0 = k R r z=0 = k T r z=0. (2.2) The above constraint gives (for all r on z = 0 plane) (k I r) x = (k R r) x = (k T r) x, (2.3) (k I r) y = (k R r) y = (k T r) y. (2.4) Without loss of generality if the incident wave vector k I is taken in the x z plane the first expression of the second condition (2.4) is zero and in turn it would restrict k R and k T in the x z plane. So that makes all the rays in the same plane. The remaining condition (2.3) now reads, ksini = ksinr = k sinr. (2.5) The first equality in (2.5) sets i = r, that is the angle of incidence is equal to the angle of reflection. The second equality is nothing but the Snell s law, sini sinr = k k = µ2 ǫ 2 µ ǫ = v v 2 = n 2 n. (2.6) In the following we will consider polarisation in the incidence plane and perpendicular to the incidence plane separately. The linear combination of the two can produce any desired polarisation.

4 4 2 Fresnel Equations π/2 i π/2 i E R µ, i n^ r =i µ 2, 2 r π/ 2 r E T Fig. 2.2: Electric field parallel to the incident plane 2.2 Electric field parallel to the plane of incidence For the electric field in the plane of incidence (or polarisation parallel to the incidence plane, Fig. 2.2, the magnetic field are perpendicular to the plane of the paper and are coming out of the paper shown by encircling a dot i.e. the arrow head), we have the following boundary conditions (see the last chapter),

5 2.2 Electric field parallel to the plane of incidence 5 ǫ ˆn ( +E R ) = ǫ 2ˆn E T, (2.7) ˆn (B I +B R ) = ˆn B T, (2.8) ˆn ( +E R ) = ˆn E T, (2.9) µ ˆn (B I +B R ) = µ 2ˆn B T. (2.0) Now since B is always perpendicular to the plane of incidence and hence ˆn, the second condition (eqn. (2.6))is trivially satisfied (each term being zero). Simplifying the rest of the conditions using r = i, ǫ ( +E R )sini = ǫ 2 E T sinr, (2.) ( E R )cosi = E T cosr, (2.2) µ (B I +B R ) = µ 2 B T. (2.3) Now using B I = k I /ω, B R = k R E R /ω and B T = k T E T /ω with Snell s law one finds the first and the third conditions above are identical. Problem : Prove the previous statement. So now from the first two independent conditions one finds, E R = ǫ 2sinrcosi ǫ sinicosr ǫ 2 sinrcosi+ǫ sinicosr, (2.4) E T 2ǫ sinicosi = ǫ 2 sinrcosi+ǫ sinicosr. (2.5)

6 6 2 Fresnel Equations Problem 2: Show that there is no reflected ray for tani = n 2 /n, use approximation µ µ 2. The angle, θ W = tan (n 2 /n ) is known as Brewster angle. So for this angle of incidence there is no inplane polarisation component in the reflected light and hence the reflected light is completely plane polarised with the polarisation perpendicular to the plane of incidence. Typical values of reflected and refracted amplitudes are shown in Fig (2.3) for the air-glass interface E E T I Angle of incidence (in radians) For air glass interface (n /n =.5) 2 E R Fig. 2.3: Reflected and transmitted amplitudes as a function of angle of incidence when polarisation of the electric field is in the plane of incidence. 2.3 Electric field perpendicular to the plane of incidence Now we consider the case where the electricfield is perpendicular to the plane of incidence (Fig. 2.4). Since the electric in this case the electric field is normal to the plane of

7 2.3 Electric field perpendicular to the plane of incidence 7 π/2 i B I π/2 i B R µ, i n^ r =i µ 2, 2 r π/2 r B T Fig. 2.4: Electric field perpendicular to the incident plane incidence the first condition, (2.5), vanishes identically, and the rest of the conditions are written as, (B I +B R )sini = B T sinr (2.6) ( +E R ) = E T (2.7) µ (B I B R )cosi = µ 2 B T cosr. (2.8) Among the above three the first two conditions are identical. Problem 3: Like the previous case show that the first two conditions are the same.

8 8 2 Fresnel Equations The last condition, in terms of electric field, is written as, ( E R )cosi = Using the independent conditions we have, ǫ2 µ ǫ µ 2 E T cosr. (2.9) E R = E T = ǫ µ 2 cosi ǫ 2 µ cosr ǫ µ 2 cosi+ ǫ 2 µ cosr, (2.20) 2 ǫ µ 2 cosi ǫ µ 2 cosi+ ǫ 2 µ cosr. (2.2) Fig. 2.5 shows reflected and refracted amplitudes for air-glass interface when the polarisation of the electric field is perpendicular to the plane of incidence E E T I Angle of incidence (in radians) For air glass interface (n 2/n =.5) E R Fig. 2.5: Reflected and transmitted amplitudes as a function of angle of incidence when polarisation of the electric field is perpendicular to the plane of incidence.

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