Reference
Introduction to Electrodynamics
By D. J. Griffith
Reflection and Transmission at Normal incidence
Reflection and Transmission at Normal incidence
0 1
0 11
ˆ( , ) exp
1 ˆ( , ) exp
I I
I I
E x t E i k x t j
B x t E i k x t kv
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Suppose yz plane forms the boundary between two linear media. Aplane wave of frequency ω traveling in The x direction (from left) and polarizedalong y direction, approaches the interface from left (see figure)
In medium 1 following reflected wave travels back
0 1
0 11
ˆ( , ) exp
1 ˆ( , ) exp
R R
R R
E x t E i k x t j
B x t E i k x t kv
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Incident wave
Transmitted wave
0 2
0 22
ˆ( , ) exp
1 ˆ( , ) exp
T T
T T
E x t E i k x t j
B x t E i k x t kv
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In medium 2 we get a transmitted wave:
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:
0 0 0
0 0 01 1 1 2 2
1 1 1 1 1
R I T
I R T
E E E
E E Ev v v
(1)
00
00
xBxB
xDxD
00
00////
////
xHxH
xExE
(i)
(ii)
(iii)
(iv)
The reflected wave is in phase if v2>v1 or n1>n2
and out of phase if v2<v1.or n1<n2
Using (1) and (2)
0 0 0
1 1 1 2 1 2
2 2 2 1 2 1
0 0 0 0
1r
2
2 1 20 0 0 0
2 1 2 1
,
where
1 2and
1 1
vIf 1(nonmagnetic media) then =
v
2thus we have, and
I R T
R I T I
R I T I
or E E E
v n
v n
E E E E
v v vE E E E
v v v v
(2)
2 1 20 0 0 0
2 1 2 1
1 2 10 0 0 0
2 1 2 1
The real amplitudes are related by
2and
cin terms of refractive index n=
v
2and
R I T I
R I T I
v v vE E E E
v v v v
n n nE E E E
n n n n
Reflected wave is 180o out of phase when reflected froma denser medium. This fact was encountered by you during
Last semester optics course. Now you have a proof!!!
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
20
1
2I vE
2 2
0 1 2
0 1 2
RR
I I
EI n nR
I E n n
Where as the ratio of transmitted intensity to incident intensity is 2 2
02 2 2 1 1 22
1 1 0 1 1 2 1 2
2 4
( )TT
I I
EI v n n n nT
I v E n n n n n
NOTE: R+T=1 => conservation of energy
Use ε α (n)2
Reference: Chapter-21, “OPTICS” by Ajoy Ghatak
Oblique Incidence
Oblique Incidence-1
Prove that: (1) Angle of Incidence = Angle of Reflection(2) Snell’s Law
Oblique Incidence-2
Derive the expressions for reflection coefficient and transmission coefficient
Z
X
Y
Interface of two medium
k1 k3
k2
Plane of incidence
Note: k1,k2,k3 and E1,E2,E3 lies in X-Z Plane. Understand 3-D picture but work With 2-D Fig for calculations
E1
As x=0
k1z=k2z=k3z
Zero
Interface
Incident wave
Refracted wave
Reflected wave
Using fig.
Identity: If
A e(iax)+Be(ibx)=Ce(icx)
Then a=b=c (Ref: Griffith)
Using identity
Snell’s Law
-------------(X)
-------------(Y)
Simplify equation (X) and (y), substitute the value of E20 from (X) to (Y)
BC—(iii)
BC—(i)
x
zy
As Θ1= Θ3
Divide equation (X) by E10 and substitute the value of E30/E10,
Case II: If E is perpendicular to the plane of incidence (do it yourself)
All these four equations are known as Fresnel’s equation
Snell’s Law
ε = n2
Case II: E is perpendicular to plane of incidence.
……(YY)
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)]
900
Polarized light
(b)
(A)
(A)
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┴ .
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