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Refraction
Figure 7-13
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
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Critical Angle
Figure 7-14
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
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Total Internal Reflection (TIR)
Figure 7-15
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
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Figure 7-16
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
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Multimode Step-Index
Figure 7-17
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
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Multimode Graded-Index
Figure 7-18
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
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Single Mode
Figure 7-19
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
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Fiber Construction
Figure 7-20
WCB/McGraw-Hill The McGraw-Hill Companies, Inc., 1998
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How a light ray reacts when it meets the interface of twotransmissive materials that have different indexes of refraction
can be explained with Snells law . Snells law simply states
n1 sin 1 = n2 sin 2
Where n1 = refractive index of material 1(unit less)
n2 = refractive index of material 2(unit less)
1 = angle of incidence (degrees)2 = angle of refraction (degrees
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( n 1 n2 )
Refracted Ray
1 (angle of incidence)
( n 1 = n 2 ) Unrefracted
Ray Normal
Medium n1
Medium n2
2
2
Incident ray
( n 1 n2 )
Figure 2.5 Refractive model for Snells law
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Normal
Figure 2.6 Critical AngleRefraction
n2 less dense
1= C (minimum)
2
n1 more dense
Incident ray
Refracted ray (more to less dense)
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When 2 = 90 which result to sin 2 = 1, then 1 = C ,Therefore,
Cri tical Angle :
C =
1
21sinn
n
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Total Internal Reflection
2 conditions necessary for TIR to occur:
1. The refractive index of the first medium is greater thanthe refractive index of the second one (n1 > n2)
2. The angle of incidence, 1, is greater than or equal to thecritical angle, c : 1> c
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n 2 less dense
1 C
n 1 more dense
Incident ray Total InternalReflection ofIncident Ray
The Total Internal Reflection
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Example 1
let medium 1 be glass and medium 2 be ethyl alcohol. For an angle ofincidence of 30 , determine the angle of refraction .
47.335514.0sin
sin5514.030sin36.15.1
sinsin
1
2
2
21
2
1
n
n
n1 (glass) = 1.5n2 (ethyl alcohol) = 1.36
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Determine the critical angle for the fiber describe in Example 1
C =
1
21sinn
n
5.1
36.1sin 1
05.65
C =
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Determine the numerical aperture for the fiber describe in Example 1
633.0
36.15.1 22
2
2
2
1 nn NA
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