st
1
scenario
MEASUREMENT OF WATER
REFRACTIVE INDEX
GRIGORIOS P. ZYGOURAS
PHYSICIST
MELISSIA LYCEUM , ATTICA , HELLAS
PURPOSE OF EXERCISE
•
To understand that light changes direction passing through different
transparent materials (optical media)
•
To calculate the refractive index of a liquid by using diffraction grating
•
To calculate the refractive index of a liquid by processing two photos,
using SalsaJ software
•
To confirm Snell's law of refraction.
THEORETICAL APPROACH
•
When light passes from one optical medium to another changes
direction
•
This phenomenon is called refraction. Measuring the deviation of the
light ray , we calculate the medium’ s refractive index
•
These measurements are made by editing taken photos , which show
the different path of light when it passes from air into water
•
Two different setups are used to take the appropriate images:
1. with diffraction grating
2. without diffraction grating and changing the angle of incident
ray at the boundary between the two media
•
Two different refraction index calculations are made using :
1. Snell's law and
2. The theory that fringe patterns of the same order have the
same integer wavelength multiple distance from the central fringe
pattern either light is transmitted in the air or in another medium (water).
REQUIRED EQUIPMENT
•
•
•
•
•
•
•
Diffraction grating (200 - 500 lines per mm)
Bubble Laser, or Laser pointer
Transparent bowl
Adhesive tape
Scissors
Translucent paper
Pegs as a means of support, in case of Laser pointer use
• NOTE: The experiment performance is better using Bubble
Laser device than Laser pointer, because the images are
sharper.
REQUIRED EQUIPMENT
TAKING PHOTOS
1.
2.
3.
For best results use :
A dark room to get better definition
Long exposures without flash
A tripod to avoid shaky pictures emerge.
CALCULATION OF THE REFRACTIVE INTEX
WITHOUT DIFFRACTION GRATING
•
Using two images of the light ray :
One falling vertically to the boundary of air - water and one
forming a small angle with it. In the second case the light
path changes as it goes from air to water.
•
We convert the two pictures in one using the software SalsaJ - as
described below - and then we measure the deviation of the two lateral
light rays from vertical.
•
We calculate the refractive index of water from the relations derived on
the basis of the theoretical approach to the phenomenon.
1ST MEASUREMENT METHOD WITH
COMPOSITION OF TWO PHOTOS
(Without diffraction grating)
•
We pour water in the bowl.
•
We stick the translucent paper on one side of the bowl.
•
We keep the source of light (Laser beam) at the opposite side
perpendicular to the surface. One half the beam has to pass through
the water and the other half through the air.
•
Because the light ray is perpendicular to the boundary of the two
media there is no difference between them.
•
We take a photo.
PHOTOS OF EXPERIMENTAL SETUP
THE DIFFERENCE IS DINSTINCTIVE IN THE DARK
PHOTO IN WHICH THE LIGHT RAY IS
PERPENDICULAR TO THE BOUNDARY
OF THE TWO MEDIA
THERE IS NO DEVIATION OF LIGHT RAY PASSING THROUGH THE AIR
AND THE WATER
• We turn slightly the source of the light so that the light ray
strikes with a very small angle of incidence (to be consistent
with the approach that sinφ=tanφ)
• We can now observe the divergence of the light ray when it
passes through the water from the footprint of the translucent
paper
• We take a photo again
IMAGE IN WHICH THE LIGHT RAY STRIKES
WITH A VERY SMALL ANGLE OF INCIDENCE
WE OBSERVE THAT THE LIGHT RAY PASSING THROUGH THE WATER HAS
LESS DEVIATION THAN THAT PASSING THROUGH THE AIR
PHOTOS PROCESSING
•
•
Open the images with software SalsaJ.
We convert images in 8 bit mode to be editable , following the next
steps in the menu Image > format > 8 bit, as shown in the picture
below.
•
•
Then from the same menu select image > adjust > Brightness/contrast.
We suggest the option auto.
As we can see on the photo below.
UNIFICATION OF THE TWO PHOTOS
To measure the deviation of the light rays in the water and in the air, we
unify the two images obtained following the steps that are shown in
the picture below. In the menu click, manipulation> computer images>
add
PUTTING THE TWO PREVIOUS IMAGES
TOGETHER
SCHEMATIC PRESENTATION OF LIGHT RAYS
PASSING THROUGH AIR AND WATER
Διάγραμμα
διάδοσης του
φωτός
στον αέρα
Diagram propagation
of light
in air
φ1
Central
beam
ΔΧα
1st beam
Distance between light source and translucent paper
Διάγραμμα
Diagram
propagation
διάδοσης του
of light
φωτός
in water
στο νερό
Φ2
Απόσταση
Distance between
πηγής light
φωτός
source
και διαφανούς
and translucent
χαρτιού
paper
Central
Κεντρική
δέσμη
beam
ΔΧν
1st beam
d
MATHEMATICAL EQUATIOS USED
FOR OUR CALCULATION
•
From the geometry of the figure we see that tan 1 
and tan 2 
•
•
•
 a
d
for the path of light in air,
v
for the path of light in water.
d
We know that for small φ angle values we have : sin   tan 
From Snell's Law na sin 1  n sin 2 where ηα= refractive index of air and
nv = refractive index of water.
We know that the refractive index of air has a value equal to 1.
So calculating the prices of tanφ and using the equation sin   tan 
we can
calculate the refractive index nv of water from Snell's law :
 a
n sin 1 sin 1 tan 1

n  


 d  
sin 2
sin 2 tan 2  
d
Where Δχα και Δχν are the distances in number of pixels to be calculated from the
image processing
PROCESS OF FINDING VALUES
After you get the result of adding the two images, we can find the
deviation of each beam from the central. We use the linear option from
the menu (see picture below). During this process, we press the shift
key down to keep the line horizontal.
We draw a line as shown below, and then following the steps in the menu
analysis> chart distribution, we get the graph from which we calculate
the deviation of the light rays. The deviation is measured in number of pixels.
We use the results of the measurements for our calculation
MEASURING THE DISTANCES BETWEEN LIGHT RAYS
Moving the cursor on the taken graph, we find the positions of light rays
(See picture below).
The deviation (distance between them) is measured in number of pixels.
We use the results of the measurement for our calculation
Price of χ
Cursor
REFRACTIVE INDEX & MEASUREMENT
VALUES
Coordinates of the beams
AIR
CENTRAL χ
Distances of the side beams from
the central
WATER
1140
972
AIR
Δχα = χ - χ1α = 900
WITH ANGLE
χ1
240
WATER
Δχν = χ - χ1ν = 686
286
According to the relationship that we have in our theoretical calculations
and our measurements the refractive index of water has a value = 1.312.
As we can see the price is too close to real which is 1.333
2nd MEASUREMENT METHOD USING DIFFRACTION
GRATING
•
•
•
•
•
•
•
We pour water in the bowl up to a level
We stick the translucent paper on one side of the bowl
We stick the diffraction grating on the opposite side so that it covers
both visual media.
We put the source of light (Laser beam) in front of the diffraction
grating perpendicular to the surface. We take care of half the beam
passes through the water and the other half from the air.
Due to the vertical incidence of the beam , the central fringe pattern in
water and in the air are on the same line. The rest, however, have
different deviations.
We take a photo.
We use the first and second fringe pattern for our calculations
FINDING THE REFRACTIVE INTEX
WITH DIFFRACTION GRATING
•
In this method we use a photo of the beam,
when it strikes vertically to the boundary of air - water.
•
In this case the light follows a different path in water than in air.
•
We convert the image using the software SalsaJ, as described
below, and then measure the deviation of the two lateral beams
from vertical.
•
From the relations derived on the basis of the theoretical approach
to the phenomenon, we calculate the refractive index of water.
PHOTOS USING THE DIFFRACTION GRATING
COURSE OF LIGHT RAYS PASSING THROUGH AIR AND WATER
1st fringe
pettern
Diagram propagation of light in air
φ1
Central
fringe pattern
ΔΧα
1st fringe
pattern
Distance between light source and translucent paper
1st fringe
pattern
Diagram propagation of light in water
Φ2
Central
fringe pattern
ΔΧν
1st fringe
pattern
Distance between light source and translucent paper
d
MATHEMATICAL EQUATIONS
USED FOR CALCULATION
•
From the geometry of figure we have
and
•
•
•
tan 2 
v
d
tan 1 
 a
d
in the air,
in the water.
We know that for small φ angle values we have sin   tan 
Snell's Law gives na sin 1  n sin 2
where ηα= refractive index of air and
nv = refractive index of water
We know that the refractive index of air has a value equal to 1. So calculating
the prices of tanφ and using the equation sin   tan  we can calculate the
refractive index nv of water from the Snell's law :
 a
n sin 1 sin 1 tan 1

n  


 d  
sin 2
sin 2 tan 2  
d
Where Δχα and Δχν are the distances in number of pixels to be calculated from
the image processing
FIND THE POSITIONS OF THE FRINGE
PATTERN
We use the linear option from the menu (see picture below). We draw a
line along the photo to find the positions of the fringes, as described
below. During this process, we press the shift key down to keep the line
horizontal.
Following the procedure described in the previous case , we get the
graph from which we calculate the distances.
MEASUREMENT OF THE POSITION OF THE FRINGE
PATTERN
•
Moving the cursor on the line we draw , the position of the fringe
pattern appears as in the picture below.
Central fringe
pattern
1st left fringe
pattern
2nd left fringe
pattern
1st right fringe
pattern
2nd right fringe
pattern
•
Price of
χ
1ST MEASUREMENT using SNELL’S LAW
•
We measure the positions of the first two fringe patterns and the
central and then we calculate the distances from the central, when
light is passing through air and water.
•
The values ​of positions in number of pixels are listed in the table :
AIR
•
WATER
Central (χ)
2189
2176
1st Left (χ1)
1490
1641
1st Right (χ2)
2887
2694
2nd Left (χ3)
971
1246
2nd Right (χ4)
3398
3071
Subtracting the values ​of each fringe from the central we find their
distances. We use these values ​ in the final equation of Snell’s
law to calculate the refractive index of water..
Table showing distances of the lateral fringes pattern from the central
AIR
WATER
χ - (χ1) = Δ(χ1α)
699
χ - (χ1) = Δ(χ1ν)
535
χ - (χ3) = Δ(χ3α)
1218
χ - (χ3) = Δ(χ3ν)
930
(χ2) – χ = Δ(χ2α)
698
(χ2) – χ = Δ(χ2ν)
518
(χ4) – χ = Δ(χ4α)
1209
(χ4) – χ = Δ(χ4ν)
895
We use these values ​in the final equation of Snell’s law
to calculate the following refractive index values
η1 = 1,306
η2 = 1,347
η3 = 1,309
η4 = 1,350
With an average value of n = 1.328, which is very
close to actual value n = 1.333
Another group of measurements listed in the table below show the
distances between central and lateral fringe pattern , giving good results
as we see.
AIR
η1 = 1,329
WATER
χ - (χ1) = Δ(χ1α)
718
χ - (χ1) = Δ(χ1ν)
540
η2 = 1,355
χ - (χ3) = Δ(χ3α)
1273
χ - (χ3) = Δ(χ3ν)
952
η3 = 1,337
(χ2) – χ = Δ(χ2α)
709
(χ2) – χ = Δ(χ2ν)
523
η4 = 1,322
(χ4) – χ = Δ(χ4α)
1213
(χ4) – χ = Δ(χ4ν)
917
Average value n = 1.335, which is close to the actual value of the
refractive index of water n = 1.333
These values ​are obtained as above, using the diffraction grating and
Snell’s law
COURSE OF LIGHT RAYS PASSING THROUGH AIR AND WATER
FOR THE 2nd MESUREMENT METHOD
1st fringe
pettern
Diagram propagation of light in air
φ1
Central
fringe pattern
ΔΧα
1st fringe
pattern
Distance between light source and translucent paper
1st fringe
pattern
Diagram propagation of light in water
Φ2
Central
fringe pattern
ΔΧν
1st fringe
pattern
Distance between light source and translucent paper
d
2nd MEASUREMENT METHOD FROM THE POSITION OF
FRINGE PATTERNS
•
We derive to the same results using the equation that calculates the
distance of the fringe patterns from the central in the air or in the
water:    and    , Ν: the same integer number for
the first fringe pattern in the air and in the water.
•
From the geometry of the figure above tan 1 
tan 2 

   tan  2 d
d
. From the above equation   tan 2 d

   tan 1d
d
and
and finally
(1)
and
•
From the definition of the refractive index n 
c
c: light speed in
u
vacuum or in the air and u light speed in another optical medium.
•
From the basic wave speed equation c   f we finally have:
n
•
 f 

 f 
(2)
From equations (1) and (2) n 


Calculating Δχα and Δχν we
can derive to the calculation of the refractive index of water or any
other medium.
•
As mentioned values are similar to those we used before, thus
results are the same.
CALCULATION OF MEASUREMENT ERROR
THEORETICAL
VALUE
WITH
DIFFRACTION
GRATING
1,333
WITHOUT
DIFFRACTION
GRATING
1,333
MEASUREMENT
VALUE
ERROR %
1st
MEASUREMENT
1,328
1st
MEASUREMENT
0,38
2nd
MEASUREMENT
1,335
2nd
MEASUREMENT
0,15
1,312
1,58