Reference Introduction to Electrodynamics By D. J. Griffith

1. Reflection and Transmission at Normal incidence Reference Introduction to Electrodynamics By D. J. Griffith

2. Reflection and Transmission at Normal incidence Suppose yz plane forms the boundary between two linear media. A plane wave of frequency ω traveling in The x direction (from left) and polarized along y direction, approaches the interface from left (see figure) Transmitted wave Incident wave In medium 1 following reflected wave travels back -sign in BR is because Poynting vector must aim in the direction of propagation

3. (i) In medium 2 we get a transmitted wave: (ii) At x=o the combined fields to the left EI+ER and BI+BR, must join the fields to the right ET and BT in accordance to the boundary condition. Since there are no components perpendicular to the surface so boundary conditions (i) and (ii) are trivial. However last two [(iii) & (iv)] yields: (1) (iii) (iv)

4. (2) Using (1) and (2) The reflected wave is in phase if v2>v1 or n1>n2 and out of phase if v2<v1.or n1<n2

5. Reflected wave is 180o out of phase when reflected from a denser medium. This fact was encountered by you during Last semester optics course. Now you have a proof!!!

6. Reflection coefficient (R) and Transmission coefficient (T) • Intensity (average power per unit area is given by): • If μ1= μ2 = μ0, i.e μr=1 , then the ratio of the reflected intensity to the incident intensity is Where as the ratio of transmitted intensity to incident intensity is Use εα(n)2 NOTE: R+T=1 => conservation of energy

7. Oblique Incidence Reference: Chapter-21, “OPTICS” by Ajoy Ghatak

8. Oblique Incidence-1 • Prove that: • Angle of Incidence = Angle of Reflection • Snell’s Law

9. Y X k2 Z k1 k3 Plane of incidence Interface of two medium Note: k1,k2,k3 and E1,E2,E3 lies in X-Z Plane. Oblique Incidence-2 E1 Derive the expressions for reflection coefficient and transmission coefficient Understand 3-D picture but work With 2-D Fig for calculations

10. Identity: If A e(iax)+Be(ibx)=Ce(icx) Then a=b=c (Ref: Griffith) Zero Refracted wave Interface Reflected wave Incident wave As x=0 Using identity k1z=k2z=k3z Using fig.

11. Snell’s Law

12. x z y BC—(iii) As Θ1= Θ3 -------------(X) BC—(i) -------------(Y) Simplify equation (X) and (y), substitute the value of E20 from (X) to (Y)

13. Divide equation (X) by E10 and substitute the value of E30/E10,

14. ε = n2 Snell’s Law Case II: If E is perpendicular to the plane of incidence (do it yourself) All these four equations are known as Fresnel’s equation

15. ……(YY) Case II: E is perpendicular to plane of incidence. Since the Y-axis is tangential or parallel to the interface, the y-component of E must be continuous across the interface. E10 + E30 = E20 ……….(XX) Using (XX) and (YY), we can get [Boundary Condition (iv)]

17. Polarized light 900

18. (A) (b) (A)

19. More to do….. • (C) Phase Change in reflection( Role of Brewster’s angle). Ref: Page-21.4 & 21.6 , “Ghatak” • (D) Total Internal Reflection • Ref: Page-21.5 & 21.6 , “Ghatak” If θ2=900 and θ1= θC , determine r//, t// and r┴ & t┴ .

20. Numerical to do… (1) For an Air glass Interface (n1=1.0 and n2=1.5) (a) determine r// and t// for normal incidence. (b) determine Brewster’s angle. (c) if incidence angle is 300 determine r// and t// for oblique incidence. Example: 21.5, 21.6, 21.7 Page-21.11, “Ghatak” 3rd edition. Problem: 21.4 Page21.17 “Ghatak” 3rd edition