Bryan Berns
ECE 313 – Optics Lab
Lab I – Principles of Optical Waveguiding
Lab Purpose:
Use snell’s law and its derivative formulas to discover the refractive indexes of other materials
given the known values of one material (i.e. air). Also find the critical angle of a certain interface by
finding the angle of incidence which begins to totally internally reflect.
Using find the numerical aperture of a given fiber by using it’s intensity profile, which can be
found by measure the intensity of light at stepped degrees from it’s intensity maximum (normally the
perpendicular).
Data, Calculations, and Graphs:
Finding plexiglass refractive index:
n1 sin 1  n2 sin 2
Trial 1 :
Trial 2 :
n2  1.61
n2  1.57
1sin 25  n2 sin 43 1sin 15  n2 sin 24
n2  1.59
Finding water refractive index:
 n2 

 n1 
c  sin 1 
 n 
63  sin 1  2 
 1.59 

n2  1.42
Using known water refractive index to find index of plexiglass:
n1 sin 1  n2 sin 2
1.33sin 79  n1 sin 62

n1  1.48
Finding oil refractive index:
 n2 

 n1 
c  sin 1 
 n 
80  sin 1  2 
 1.59 

n2  1.57
Using known oil refractive index to find index of plexiglass:
n1 sin 1  n2 sin 2
1.46sin 81  n1 sin 77.5

n1  1.50
Single Mode Intensity Reading
48
60
26
.5
34
35
40
30
-10
7.
5
17
2.
7
10
2
1.
6
1.
55
9
20
-7
0
-1
-4
2
5
8
1.
8
Intensity (mV)
50
11
Degree Offset
HalfPower  46 / 2  23
f ( x)  33 at x  3,5
x
4
2
NA  sin 4  .07
200
17
7
Multimode Mode Intensity Reading
10
5
41
55
68
100
-10
-7
-4
0
-1
-50
2
5
2.
2
2.
5
3.
5
20
50
5
Intensity (mV)
150
8
11
Degree Offset
HalfPower  177 / 2  89
f ( x)  89 at x  1.5,2.5
x
2
2
NA  sin 2  .034
Question:
Fiber optic telecommunication uses light at a wavelength of about 1.55 um. This corresponds to what
optical frequency (in Hz)?
f 
c


3.0  108
 1.93  1014 Hz 
1.55  10 6
What is the wave number of this light?
k
2


m  40536.7 cm-1 
100cm
If two light signals are separated by 100 GHz, what is the corresponding wavelength separation.
 
c
 3  106 nm 
9
100  10
Given that the refractive index of the cladding is 1.45 in single mode operation fiber (NA=.14),
calculate the index of the core.
NA  n1  n2
2
2
.14  n1  1.45
2
2

n1  1.456
Repeat for multimode fiber.
NA  n1  n2
2
2
.20  n1  1.45
2
2

n1  1.463
Conclusion:
Given the index of one material, we can calculate the refractive index of another by measuring the
incident angle and the refracted. One can also estimate the given NA of a fiber by measuring the
intensity out of like emitting at different angles.
Since I’ve already taken an optics course, this lab just reinforced snell’s law and basic principles of
fibers.