Reflected and transmitted rays
plane of incidence
i
n
r
θ1 θ1
η1
boundary
θ2
t
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η2
Index of refraction
Absolute index of refraction:
ηabs
c
=
v
where c = 2.99 x 108 m/s. Note: v ≤ c for all
transparent materials, so η ≥ 1.
Relative index of refraction:
η2
η=
η1
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Snell’s Law
plane of incidence
i
sin θ1 v1 η2
= =
= η = η12
sin θ 2 v2 η1
n
r
θ1 θ1
η1
boundary
θ2
t
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η2
Transmission
sin θ1 η2
S1
=
= η = , S2η = S1
sin θ 2 η1
S2
 
cosθ1 = C1 = −N ⋅V
 
cosθ 2 = C2 = −N ⋅ T
Square both sides of S1η = S2 :
S η =S
2
2
2
2
1
And use S + C = 1 to get cosine forms :
2
2
(1 − C )η = 1 − C
C − 1)
(
C = 1+
2
2
2
2
C2 =
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2
2
1
2
1
η
C
(
1+
2
2
1
)
−1
η2
N
V
θ1
-N’
θ2 T
V’
Transmission
N
V






 
T = − N ′ + k V ′ + N ′ = kV ′ + (k − 1) N ′

T =1


N ′ = C2 N

 

 C2


− N ′ 
, V′ =
V ′ C1 = − N ′ , V ′ =
C1
C1

 C2 
V
V′ =
C1

S2 1
S2C1
S2 T

 =
=
k=
S1 V ′ S1 C2 C2 S1
C1
(
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)
θ1
-N’
θ2 T
k
V’
V’+N’


T = −N′ + k

S2 T

 =
k=
S1 V ′




 
V ′ + N ′ = kV ′ + (k − 1) N ′
(
C2 =
(
1+
)
S2 1
S2C1
=
S1 C2 C2 S1
C1
N
V
θ1
)
C12 − 1
η
2
 S2C1 C2  ⎛ S2C1 ⎞ 
T=
V +⎜
− 1⎟ C2 N
C2 S1 C1
⎝ C2 S1 ⎠
⎞ 
1  ⎛ 1
= V + ⎜ C1 − C2 ⎟ N
⎝η
⎠
η
(
-N’
θ2 T
k
⎞ 
S2  ⎛ S2
V + ⎜ C1 − C2 ⎟ N
=
S1
⎝ S1
⎠
)
2
⎛
C1 − 1
C1
1 
= V + ⎜ − 1+
2
η
η
⎜⎝ η
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Transmission
⎞ 
⎟N
⎟⎠
V’
V’+N’
Transmitted ray direction, t (cont’d.)
When η1 < η2, t bends toward the normal direction
at the hit point
n
i
r
θ1 θ1
η1
η2
θ2
t
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Transmitted ray direction, t (cont’d.)
When η1 > η2, t bends away from the normal
direction at the hit point
n
i
r
θ1 θ1
η1
θ2
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t
η2
Total internal reflection
There exists a critical angle, θc, when the
transmitted ray direction t is parallel to the
boundary and θ2 = π/2
η1
θ2 = π/2
η2
t
i
θ1 θ1
n
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r
θ1 approaches θc
As θ1 increases, t bends toward the boundary
θ2
t
η1
η1
η2
η2
θ1 θ1
i
r
n
θ1 << θc
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…
θ2
i
θ1 θ1
n
θ1 < θc
t
r
θ1 is greater than or equal to θc
When θ1 ≥ θc, t is parallel to the boundary and it carries no energy (total internal reflection
occurs)
η1
θ2 = π/2
η1
η2
t
η2
i
θ1 θ1
r
n
θ1 = θc
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i
θ1 θ1
n
θ1 > θc
r
Checking for TIR
Recall that the angle of refraction θ2 is
given by:
1
2
1/2
cos θ2 = [1 - 2 (1 - cos θ1)]
η
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TIR occurs when θ1 = θc
When θ1 = θc, the expression:
1
2
1 - 2 (1 - cos θ1)
η
becomes zero.
When θ1 > θc, this expression is negative, so θ2
is imaginary and t is a complex number.
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TIR condition
Thus, TIR occurs when:
1
2
1 - 2 (1 - cos θ1) < 0
η
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Spawning secondary rays
t1
i
r1
c(p) = clocal(p) + kr(p)*cr(pr) + kt(p)*ct(pt)
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