Problem 5.
Mirage
Mirage
Create a mirage like a road or
desert mirage in the laboratory and
study its parameters.
Introduction
• What is mirage?
• How is it formed?
• Which are parameters of a mirage?
Experiment
Effect:
heated
surface
observed
object
Formation of mirage
n1 sin 1  n2 sin 2
Snell’s law:
n1 – index of refraction of first medium
n2 – index of refraction of second
medium
φ1 – ray angle in first medium
φ2 – ray angle in second medium
n1
φ1
n2
φ2
n1 > n2 → ray is refracting away
from vertical
Formation of mirage
- applying Snell’s law on sequence of parallel
planes of infinitesimal thickness whose index of
refraction is changing monotonously:
n0 sin 0  n1 sin 1  ...  n sin 
y
x
Formation of mirage
In one moment (sooner or later), if
sequence of ni isn’t convergent, will be:
n0 sin 0  n1 sin 1  ...  nn sin n  nn1 1
Total reflection occurs and ray is
returning.
Mirage parameters
• index of refraction dependence of temperature
• temperature dependence of shift in direction
parallel with temperature gradient
n1
temperature
n2
n3
…
nN
index of
refraction
Experiment
horizontal line
vessel with
flourescein
termometer
Experiment
• temperature distribution was measured with
thermometer put in the water
• rays are photographed and deviation from
horizontal line is measured
Theoretical model
It is sufficient to derive ray equation.
• assumptions:
• linear dependence of temperature
of depth i.e.:
dT
dy
 kT
• linear dependence of index of
refraction of temperature i.e.:
dn
 kn
dT
Theoretical model
We will use following coordinate system and
following equations:
x
φ
y
-dy
dx
dx
  tan
dy
Snell’s law in differential form:
n sin( )  (n  dn) sin(  d )
…….
Theoretical model
Ray equation:
1 
y  y0  p   re

2
 
n0 sin 0
p
kT kn
1
q
sin 0
r  q  q 1
2
dT
kT 
dy
dn
kn 
dT
x

p


1 
 e  q
r  

x
p
y0 – initial depth
n0 – initial index of
refraction
φ0 – initial ray angle
Results
Temperature gradient 01
370
360
Temperature [K]
350
340
330
320
310
300
290
0
5
10
15
depth [cm]
20
25
30
Results
Mirage gradient 02
Mirage gradient 03
340
340
330
Temperature [K]
Temperature [K]
330
320
310
300
310
300
290
0
5
10
15
20
25
30
depth [cm]
290
0
5
10
15
20
25
depth [cm]
Mirage gradient 04
kT1=-0.78Kmm-1
330
325
kT3=-0.40Kmm-1
320
Temperature [K]
320
315
kT2=-0.32Kmm-1
310
305
kT4=-0.27Kmm-1
300
295
0
5
10
15
depth [cm]
20
25
30
30
Results
Mirage gradient
370
360
gradient 1
Temperatura [K]
350
gradient 3
340
gradient 2
330
gradient 4
320
310
300
290
0
5
10
15
dubina [cm]
20
25
30
Results
24
measurements
ray equation fit
22
depth [mm]
20
18
16
14
12
0
100
200
deviation [mm]
300
400
500
Results
Mirage 3
26
32
24
30
22
28
depth [mm]
depth [mm]
Mirage 2
20
18
26
24
16
22
14
20
12
18
0
20
40
60
deviation [cm]
0
200
400
deviation [mm]
Mirage 4
kn1=-4.86*10-4K-1
32
30
kn3=-4.58*10-4K-1
depth [mm]
28
26
kn2=-4.68*10-4K-1
24
22
20
kn4=-4.75*10-4K-1
18
16
0
10
20
30
deviation [cm]
40
50
60
600
Results
For coefficient of dependence of index of refraction
of water on temperature we got (in interval ≈ 318K –
358K):
dn
4
1
 k n  (4.72  0.06) 10 K
dT
According to
literature:
dn
4
1
 4.5 10 K
dT
Conclusion
• we created mirage in a laboratory
• we constructed theoretical model whose results are
good compared to literature
• mirage parameters are dependence of temperature
on depth and dependence of index of refraction on
temperature
• for water:
dn
4
1
 (4.72  0.06) 10 K
dT
• for practical reasons we photographed and
measured ray in water but we have also created
mirage in air