Refraction is the bending of the path of a light wave as
it passes across the boundary separating two media. Refraction is caused by
the change in speed experienced by a wave when it changes medium. Lesson 1, focused on the topics of "What causes refraction?" and
"Which direction does light refract?" In that lesson, we learned that
light can either refract towards the normal (when
slowing down while crossing the boundary) or away from
the normal (when speeding up while crossing the boundary). The focus of Lesson 2 is
upon the question of "By how much does light refract when it crosses a
boundary?" In the first part of Lesson 2, we learned that a comparison of the angle of refraction to the angle
of incidence provides a good measure of the refractive ability of any given
boundary. The more that light refracts, the bigger the difference between these
two angles. In this part of Lesson 2, we will learn about a mathematical
equation relating these two angles and the indices of refraction of the two
materials on each side of the boundary.
To begin, consider a hemi-cylindrical dish filled with water.
Suppose that a laser beam is directed towards the flat side of the dish at the
exact center of the dish. The angle of
incidence can be measured at the point of incidence. This ray will refract,
bending towards the normal (since the light is passing from a medium in which
it travels fast into one in which it travels slow - FST). Once the light ray enters the water, it travels in a straight line
until it reaches the second boundary. At the second boundary, the light ray is
approaching along the normal to the curved surface (this stems from the
geometry of circles). The ray does not refract upon exiting since the angle of
incidence is 0-degrees (recall the If I Were An Archer Fish page). The
ray of laser light therefore exits at the same angle as the refracted ray of
light made at the first boundary. These two angles can be measured and
recorded. The angle of incidence of the laser beam can be changed to 5-degrees
and new measurements can be made and recorded. This process can be repeated
until a complete data set of accurate values has been collected. The data below
show a representative set of data for such an experiment.
Angle of Incidence (degrees) |
Angle of Refraction (degrees) |
0.00 |
0.00 |
5.00 |
3.8 |
10.0 |
7.5 |
15.0 |
11.2 |
20.0 |
14.9 |
25.0 |
18.5 |
30.0 |
22.1 |
35.0 |
25.5 |
40.0 |
28.9 |
45.0 |
32.1 |
50.0 |
35.2 |
55.0 |
38.0 |
60.0 |
40.6 |
65.0 |
43.0 |
70.0 |
45.0 |
75.0 |
46.6 |
80.0 |
47.8 |
85.0 |
48.5 |
An inspection of the data above reveals that there is no
clear linear relationship between the angle of incidence and the angle of
refraction. For example, a doubling of the angle of incidence from 40 degrees
to 80 degrees does not result in a doubling of the angle of refraction. Thus, a
plot of this data would not yield a straight line. If however, the sine of the
angle of incidence and the sine of the angle of refraction were plotted, the
plot would be a straight line, indicating a linear relationship between the
sines of the important angles. If two quantities form a straight line on a
graph, then a mathematical relationship can be written in y = m*x + b form. A
plot of the sine of the angle of incidence vs. the sine of the angle of
refraction is shown below.
The equation relating the angles of incidence (Θi) and the angle of refraction (Θr) for light passing from air into water
is given as
Observe that the constant of proportionality in this equation
is 1.33 - the index of refraction value of water. Perhaps
it's just a coincidence. But if the semi-cylindrical dish full of water was
replaced by a semi-cylindrical disk of Plexiglas, the constant of
proportionality would be 1.51 - the index of refraction value of Plexiglas. This is not just a coincidence. The same pattern would result for
light traveling from air into any material. Experimentally, it is found that
for a ray of light traveling from air into some material, the following
equation can be written.
where nmaterial = index of refraction of the
material
This study of the refraction of light as it
crosses from one material into a second material yields a general relationship
between the sines of the angle of incidence and the angle of refraction. This
general relationship is expressed by the following equation:
where Θi ("theta i") = angle of incidence
Θr ("theta r") = angle of refraction
ni = index of
refraction of the incident medium
nr = index of
refraction of the refractive medium
This relationship between the angles of incidence and
refraction and the indices of refraction of the two media is known as Snell's Law. Snell's law applies to the refraction of light in any situation,
regardless of what the two media are.
As with any equation in physics, the Snell's Law equation is
valued for its predictive ability. If any three of the four variables in the
equation are known, the fourth variable can be predicted if appropriate
problem-solving skills are employed. This is illustrated in the two examples
below.
In the following two examples, use
Snell's law, the sine button on your calculator, a protractor, and the index
of refraction values to complete the following diagrams. Measure , calculate , and draw in
the refracted ray with the calculated angle of refraction.
|
In each of these two example problems, the angle of
refraction is the variable to be determined. The indices of refraction (ni and nr)
are given and the angle of incidence can be measured. With three of the four
variables known, substitution into Snell's law followed by algebraic
manipulation will lead to the answer.
Solution to Example A
First, use a protractor to measure the angle of incidence. An
appropriate measurement would be some angle close to 45-degrees.
Second, list all known values and the unknown value for which
you wish to solve:
Given: ni = 1.00nr =
1.33Θi =
45 degrees |
Find: Θr = ??? |
Third, list the relevant equation:
Fourth, substitute known values into the equation and
algebraically manipulate the equation in order to solve for the unknown
variable - Θr.
1.00 * sine (45 degrees) = 1.33 * sine (Θr)
0.7071 = 1.33 * sine (Θr)
0.532 = sine (Θr)
sine-1 (0.532) = sine-1 ( sine Θr)
32.1 degrees = Θr
Proper algebra yields to the answer of 32.1 degrees for the
angle of refraction. The diagram showing the refracted ray can be viewed by
clicking the View Diagram button below.
The solution to Example A is given as an example. Try Example B on your
own and click on the See Answer button to check your answer.
Answer: 34.7 degrees
Measure the angle of incidence - the angle between the normal
and incident ray. It is approximately 60 degrees.
List known Values:
ni =1.00
nr =1.52
Theta i = 60 degrees
List Unknown: Find theta r
Substitute into Snell's law equation and perform the necessary
algebraic operations to solve:
1.00 • sine(60 degrees) = 1.52 • sine(theta r)
0.8660 = 1.52 • sine(theta r)
0.570 = sine(theta r)
34.7 degrees =
theta r
Now draw the refracted ray at an angle of 34.7 degrees from the
normal - see diagram below.
Snell's Law provides the quantitative means of answering the
question of "By how much does the light ray refract?" The task of
answering this question involves using indices of refraction and the angle of
incidence values in order to determine the angle of refraction. This
problem-solving process is discussed in more detail on the remaining pages of
Lesson 2.
Laser light is shown passing into and out of a
hemi-cylindrical dish filled with water. The light enters the water (at the
curved side of the dish) along the normal line; no bending occurs upon entry.
The light continues through the water along a straight line until it reaches
the boundary with air (at the flat side of the dish). The angle of incidence in
the water is approximately 39°. At this angle, the light refracts out of the
water into the surrounding air bending away from the normal. The angle of
refraction in the air is approximately 57°. These values for the angle of
incidence and refraction are consistent with Snell's Law.