Snell's Law
Refraction of Light
By: Brett Rapponotti and Corrine Yap
Purpose
To determine the relationship between the
angle of incidence and the angle of refraction
of a light wave traveling through a medium.
The media investigated in this experiment are
air to plastic and water.
Equipment Used
other medium,
plastic filled
with water
circle, divided into
four quadrants,
with one quadrant
divided into
separate ten
degree sectors
laser
beam
medium
(plastic)
Procedure
1.
2.
3.
4.
5.
6.
7.
8.
Begin with the solid glass semicircle and align it with the diameter of the circle,
facing the quadrant divided into 10o sectors.
Place the laser at the “0o” position and shine it through the glass. It should follow
the diameter perpendicular to the glass. Mark the light on the opposite end of the
paper circle.
Now place the laser at 10o. Mark the refracted light on the opposite end of the
circle after it has gone through the glass medium.
Repeat step 3, each time adding 10o, until 90o has been reached.
Using the protractor, measure the angles of refraction. This is best done by
connecting each point to the center of the circle using a ruler and then measuring the
angles from the center.
Plot Angle of Refraction vs. Angle of Incidence in LoggerPro.
Draw semichords from the marked points to the perpendicular diameter, creating
right triangles. Do the same for the original angles, and plot Semichord of Refraction
vs. Semicord of Incidence.
Determine the relationship between the angles and the semichords and also plot
Sine of Angle of Refraction vs. Sine of Angle of Incidence.
Data
Graphs
What Now??
The angle of incidence and angle of refraction
do not have a linear relationship, although it
may seem that way. So we had to find a
different relationship. We decided to draw
semichords from the points on the perimeter
of the circle to the radii and measure them.
More Graphs
Thoughts...
How is the semichord related to the angle,
then?
radius
angle
semichord
sin(θ) = semichord/radius
So.....
Even More Graphs!
Mathematical Analyses
Sine of angle of refraction = sin(ϴR)
,
Sine of angle of incidence = sin(ϴI)
sin(ϴR) = ksin(ϴI)
k = ∆sin(ϴR)/∆sin(ϴI)
kglass = 0.6532 (calculated by LoggerPro)
sin(ϴR) = 0.6532sin(ϴI)
Sine of angle of refraction = 0.6532*Sine of
angle of incidence
Hm. Now what could that possibly mean?
Index of Refraction
a dimensionless number that describes the
relationship between the speed of light and
the nature of the medium through which it
travels.
Index of refraction --> n
speed of light --> c
speed of wave --> v
n = c/v
nair = 1
nplastic = 1.495
kplastic (calculated by LoggerPro) = 0.6532
1/0.6532 = 1.531
Taking error into consideration,
n = 1/k
Therefore,
sin(θR) = sin(θI)/n for plastic.
WATER
More on Water
Even More on Water
Mathematical Analysis
Sine of angle of refraction = sin(ϴR)
,
Sine of angle of incidence = sin(ϴI)
sin(ϴR) = ksin(ϴI)
k = ∆sin(ϴR)/∆sin(ϴI)
kwater = 0.7621 (calculated by LoggerPro)
sin(ϴR) = 0.7621sin(ϴI)
Sine of angle of refraction = 0.7621*Sine of
angle of incidence (water)
Therefore,
the same conclusion can be drawn.
nwater = 1.33283
kwater = 0.7621
1/0.7621 = 1.312
Taking error into consideration,
nwater = 1/kwater
So,
sin(θR) = sin(θI)/n for water as well.
But why 1?
We can't just say, oh, let's divide one by the
constant because it works out that way! One
is a significant number. Remember that
nair = 1, which is also nvacuum.
The angle of incidence was measured as the
light wave was traveling through air, while
the angle of refraction was measured once the
light traveled through a different medium.
Therefore, nair = nmedium of incidence and nplastic or
nwater = nmedium of refraction
WHOA!
sin(θR)/nI = sin(θI)/nR
Rearranging,
nRsin(θR) = nIsin(θI)
This is Snell's Law!