College Physics II Labs
Lab Activities
Refraction and Total Internal Reflection
1.
Complete laser mini-labs:
a.
Reflection:
Experiment 5 Metrologic ML211 Laser Pointer Education Kit
b.
Refraction:
Experiment 6 Metrologic ML211 Laser Pointer Education Kit
c.
Total internal
reflection:
Demo: Observing Internal reflections in a test tube, p. 6
Metrologic ML211 Laser Pointer Education Kit
Experiment
Total Internal Reflection
2.
Solve Problem #1 (below). Then explain in words how the solution to Problem #2 would
differ.
3.
Solve Problem #3 or #4 (below). You may wish to refer to the formulation (eq.22.2) of
Snell's Law on p.691 (Wilson & Buffa, 4th Ed.):
=
Problems:
1.
Blue light of wavelength 470 nm strikes the side of an equilateral prism.(each angle is
60û). The angle of incidence is 53.8û. The index of refraction for blue light is 1.636.
a.
Calculate the refracted angle relative to a line perpendicular to the first air-glass interface.
b.
If the prism has three 60û angles, calculate the incident angle at the second glass-air
interface.
c.
[Ans:
Calculate the refracted angle relative to a line perpendicular to this surface.
a. 29.6û
b. 30.4û
c. 56û]
2.
Repeat problem #1, but imagine that the prism is immersed in water. Assume that the
index of refraction of water for blue light of this wave length is 1.330.
3.
Sound passes from limestone, where it moves at a speed of 4000 m/s, into another
unknown material. The angle of incidence at the interface is 24û and the angle of refraction in
the unknown material is 38û.
a.
Calculate the speed of sound in the unknown material.
b.
Make a qualitative sketch of the situation and identify relevant quantities.
[Ans: a. 6050 m/s]
4.
Sound traveling in limestone enters the top surface of a cubical salt dome at an angle of
incidence of 35û. After passing through the salt dome, the sound leaves the right side and
reenters the limestone. Calculate its final direction in the limestone and show in a figure the
directions of the ray as it enters, passes through, and leaves the salt dome.
(vlimestone = 4000 m/s and vsalt = 6000m/s)
{Ans: 20û below horizontal}
College Physics II Lab
Total Internal Reflection
Measurement of the Index of Refraction
Background:
In any homogeneous material, light travels in straight lines. However, when light encounters a
boundary, a change in material, some of the light reflects back and some of the light is
transmitted forward into the new material. The transmitted light does not travel in the same
direction as the original light. Instead it is bent at the boundary and travels in a slightly different
direction. When light is bent in this way, the phenomenon is called
refraction.
The refraction of light at the boundary between two materials is
described quantitatively by Snell's Law. Consider the situation depicted
in Figure 1. A ray of light is incident on the surface between two
different materials. The long dashed line represents a line perpendicular
to the surface. The angle  measures the angle between the incident ray
and the line perpendicular to the surface. This angle is the angle of
incidence. The angle  measures the angle between the refracted ray
and the line perpendicular to the surface. This angle is the angle of
refraction. Snell's Law states:
ni*sin() = nr*sin()
The quantity ni is the index of refraction for the material in which the light was incident. The
quantity nr is the index of refraction for the material in which the light was refracted.
The phenomenon of total internal reflection can occur when the light travels from a material with
higher index of refraction to a material with a lower index of refraction. When ni > nr, the
refracted ray is bent away from the line perpendicular to the surface. If the angle of incidence is
large enough, the angle of refraction will be 90û, and the light is refracted parallel to the surface.
The angle of incidence for which this occurs is called the critical angle. If the angle of incidence
is any greater than the critical angle, it is not possible for there to be a refracted ray. [Applying
Snell's Law for this case would lead one to predict that sin() > 1, which contradicts what is
known of the sine function ( -1 ² sin() ² 1).] When there is no refracted ray, all of the energy of
the light goes into the reflected ray.
Through investigation of total internal reflection one can determine the index of refraction of
various substances. By measuring the critical angle for a situation in which light goes from a
material of higher index of refraction to a material of lower index of refraction, and by knowing
the index of refraction of one of the materials, you can determine the index of refraction of the
other material. The critical angle crit is the largest angle for which there is not evidence of total
internal reflection. The angle of refraction would be 90û in that case. Therefore, sin() = 1,
because  = 90û. Air has an index of refraction equal to 1. If the second material, into which the
light was traveling, was air, then nr = 1. If this information is substituted into Snell's Law:
ni*sin() = nr*sin()
ni*sin(crit) = (1)*(1)
ni =
The phenomenon of total internal reflection is important in numerous technologies. Total internal
reflection explains how light can be shined into one end of a glass fiber and transmitted great
distances with little loss of energy. Such fiber optics are used in electronic communication to
replace copper wires and in modern diagnostic and surgical medical devices, which minimize the
amount of trauma to the body.
Materials:
semi-circular plastic dish
protractor or polar graph paper
laser pen pointer
non-dairy creamer
water
aqueous cane sugar solution (i.e. sugar-water)
Procedure:
1.
Set-up:
Place a sheet of polar graph paper on the table. Place the semi-circular plastic dish
at the origin of the polar graph paper and turn it until the line perpendicular to the flat
wall of the dish is directed along the 0û line. Add liquid to the dish and a light sprinkle of
non-dairy creaner as needed. A very small amount of non-dairy creamer added to a liquid
will make the laser beam visible in the liquid, just as sprinkling chalk dust in the air will
make the laser beam visible in air. You may have to experiment to get the right amount of
creamer.
2.
Exploration of total internal reflection:
Shine the laser into the dish through the curved wall of the dish. Aim the beam so
that it hits the midpoint of the flat wall of the dish. Vary the angle of incidence of the
laser beam relative to the flat wall of the dish by moving the laser pointer around the
curve of the dish. Always shine the laser perpendicular to curved wall and so that the
beam strikes the midpoint of the flat wall. As you vary the angle of incidence of the laser
beam relative to the flat wall you will be able to observe the phenomenon of total internal
reflection.
3.
Measurement of the index of refraction of a liquid:
Move the laser pointer until you have established a condition of total internal
reflection. Then move the pointer so that the angle of incidence of the laser light on the
flat wall of the dish becomes smaller. Continue this process carefully until a total internal
reflection is no longer observed. Measure the angle of incidence at this time. This angle is
the critical angle. Determine the index of refraction for the liquid using the relationship:
ni =
[Verify that when total internal reflection occurs, the angle of reflection equals the
angle of incidence.]
4.
Compare your value of nwater with the best known value nwater = 1.33. What is
the % difference?
5.
Repeat step 3 for an aqueous cane sugar solution. What value do you infer for
nsugar-water? Based on your value of nsugar-water, what would you predict to be the
concentration of sugar (by weight) for this solution?
SKIP
6.
Using Dr. Landis' refractometer, this index of refraction for this solution has been
determined to be _____, thus indicating a concentration of ____ % sugar by weight. What
is the % difference between your value of nsugar-water and that obtained by Dr. Landis?
What factors might account for any difference you may find? [Hint: Does  make a
difference to n?