W3 Reflection
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School
University of California, Irvine *
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Course
47180
Subject
Mathematics
Date
Feb 20, 2024
Type
docx
Pages
7
Uploaded by AmbassadorFreedomAnt102
Week 3 Lab Report – Reflection
Date: 10/19/23
Section:
47199, Th 2:30-5:20
La
b
Instructor: Alex Stewar
t
Lab Partners: 1.
Kalysa Bui (85561106)
2.
Miles Nguyen
(47028628)
Total Score: /24
3.3.2 Specular Reflection
(8 points)
Objective
: Verify Eq. 3.1 for specular reflection
Draw a diagram. (1 points)
Image 3.3.2 - 1
Compass resolution = ?
Image 3.3.2 - 2
M aligned perpendicular on the OB
Image 3.3.2 - 3
Pointer set at 0° on the compass
1
Image 3.3.2 – 4
Photometer and Mirror positions
on the optical bench. Distance between
the mirror and photometer is measured
on the optical bench ruler.
Meter stick measures laser-to-mirror (L-M)
distance.
Image 3.3.2 - 5
Top view of set up. Angles to be measured are
shown.
Data and Analysis
: (3 points)
Tri
al #
θ
bench
direct compas
s reading
(degree
s)
Distanc
e from
laser to
mirror
(cm)
Distanc
e from
laser to
optical
bench
(cm)
θ
bench
from
distance
s/
using
trigono
metry
(degree
s)
θ
mirror
direct compas
s reading
(degree
s)
θ
mirror
calculat
ed from
θ
bench
obtaine
d from
distance
s
(degree
s)
θ
bench
/2
from
compass
reading
(degrees)
Flux
recorded
from
photomete
r (mV)
Flux minus
background
(mV)
background
= ~2.5 mV
%
Discrepancy
1
50
32.5
22.5
46.2
30
23.1
25
285
282.5
7.9
2
60
34
26
40.11
20
20.05
30
435
432.3
39.7
3
40
25.5
15
54.0
20
27
20
575
573.9
30
Show calculations for the discrepancy for one of the trials, say trial #1. (1 point)
Trial #1
: cos θ = 22.5 cm / 32.5 cm
θ = cos^-1 (0.692)
θ = 46.18693854° = 46.2°
% Discrepancy
:
2
%
discrepancy
=
|
X
1
−
X
2
|
|
X
1
+
X
2
2
|
×
100%
%
discrepancy
=
|
25
−
23.1
|
|
25
+
23.1
2
|
×
100%
% discrepancy = 7.9002079% = 7.9%
You do not need to estimate the errors, but state and explain which of the two methods (compass reading vs. distance measurement) is more accurate. (1 point)
Compass reading is more accurate because there can be user error when measuring the distances between the laser and the optical bench and between the laser and the mirror. It is more accurate to simply measure the angle of the compass reading because the calculation of the angle from the
distances can be off if the distances are measured incorrectly.
Can you conclude from your results whether the mirror is a perfect specular reflector and why? (2 points)
The mirror is not a perfect specular reflector. Specular reflection follows the rule that the angle of incidence is equal to the angle of reflection. Since there were some differences between the angle of incidence and angle of reflection for this mirror as shown by the high percent discrepancy, the mirror does not perfectly reflect the laser. However, the high discrepancy may be due to the measurements of the distances, not due to the imperfections of the mirror, so the mirror may be close to a perfect specular reflector.
3.3.3 Diffuse Reflector
(7 points)
Objective
: Measure how the amount of light reflected by a diffuse reflector varies with the angle of reflection.
*For this section, keep the laser fixed at a given θ
bench
and rotate the white card. It is much easier than trying to change the position of the laser.
θ
bench value should
be between 0 and 90 degrees for simplicity.
3
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Image 3.3.3 – 1
Position of photometer and white card.
Distance is measured on the optical bench
ruler.
Data and Analysis
: Measure the intensity of the reflected light versus angle for approximately 10 angles between 0° and 90°. Measure the background by closing the laser shutter. (3 points)
Background: 1.74 mV
Plot the results: Plot V −
Bkg vs. θ
white card
(corrected angle). (2 points)
4
θ
white card
[degrees on Compass reading]
θ
white card
[corrected angle]
V (mV)
V −
Bkg
(mV)
180 (180-180=0)
0
5.40
3.66
170 (180-
170=10)
10
6.10
4.36
160 (180-
160=20)
20
4.08
2.34
150(180-
150=30)
30
4.56
2.82
140 (180-
140=40)
40
3.93
2.19
130 (180-
130=50)
50
3.13
1.39
120 (180-
120=60)
60
2.97
1.23
110 (180-
110=70)
70
2.86
1.12
100 (180-
100=80)
80
2.40
0.66
90 (180-90=90)
90
2.14
0.40
Is the white card an ideal diffuse reflector? (1 point)
The white card is not an ideal diffuse reflector. Ideal diffuse reflectors scatter light uniformly in all directions, so if the white card was an ideal diffuse reflector, the photometer would measure the same intensity for all angles of the card. However, the intensity varies as the card’s angle changes, so the card is not an ideal diffuse reflector. This means the card angles the light into the photometer more directly at certain angles, when it should scatter the light in the same way for all angles if it was an ideal diffuse reflector.
How are the results for the reflector different from the mirror? (1 point)
The diffuse reflector scatters light, not quite uniformly, in all directions while the mirror reflects most of the light at a specific angle straight into the photometer, so the intensity measured by the diffuse reflector is much lower at all angles (about 0-4 mV) than it is for the mirror (about 280-
580 mV).
3.3.4 Internal Reflection
(9 points)
Objective
: Find the critical angle for propagation.
Image 3.3.4 - 1
Positions of photometer, fiber optic cable, and
concave lens on optical bench.
What is your background reading in mV? (1 point)
The background reading is -0.01 mV.
Table 3.3.4 (3 points)
Note: Corrected Reading has α
= 0 when the face of the optical fiber is perpendicular to the laser
beam.
5
α
(degrees)
[Compass Reading]
α
(degrees)
[Corrected Reading]
V (mV)
V −
Bkg (mV)
180
0
20.93
20.94
170
10
5.25
5.26
160
20
2.06
2.07
150
30
0.41
0.42
140
40
0.06
0.07
130
50
0.00
0.01
120
60
-0.01
0.00
Make a figure of V −
Bkg vs. angle α
. (1 point)
Measure and write down the single loop signal and the double loop signal, can you conclude that there is no loss of intensity from curving the fiber or at least the difference is not significant? (1 point)
The single loop signal is 20.43 mV and the double loop signal is 17.22 mV. There seems to be a small loss of intensity from the data collected but the loss is negligible.
Calculate the theoretical critical angle. (Equation and numerical result: 1 point)
sin
α
crit
=
√
❑
α
crit
=
arcsin
√
❑
α
crit
=
arcsin
√
❑
α
crit
=
¿
32.58270626
° = 33°
What is the experimentally determined angle from the plot? (1 point) Discrepancy? (1 point)
The experimentally determined angle from the plot is 30°. The discrepancy is 9.5%.
Draw the image patterns observed at α
≈
0
o
and α
≈
20
o
. Explain the observed patterns. (1 point)
6
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The image is a solid red dot for α ≈ 0° because the laser goes straight through to the end of the fiber optic cable with few, if any, reflections against the inside cladding of the cable and produces a solid image. For α ≈ 20°, the laser beam is reflecting more often against the cladding of the cable as it goes down the cable. However, at greater angles than the critical angle, the angle of the beam is large and the angle of incidence at the cladding is large, so the beam propagates into the cladding and is lost from the fiber instead of going down the fiber and creating an image. Since α ≈ 20° is approaching the critical angle of 30°, the angle at which the image disappears, the image is more of a hazy donut than it is a solid circle.
7