Principles of Technology
CH 13 LIGHT AND GEOMETRIC OPTICS
3
Name________
KEY OBJECTIVES
At the conclusion of this chapter you will be able to:
• Explain how light refracts as it passes from one medium to another.
• Define the term absolute index of refraction, and solve problems using this concept.
• State Snell’s law in terms of absolute indices of refraction, and solve problems using this equation.
• Define the terms critical angle and total internal reflection, and relate them to Snell’s law.
• Define the term dispersion.
Lenses, prisms, and fiber-optic bundles are applications of refraction of light.
13.6 REFRACTION
When monochromatic light travels between two media, there is a change in the speed of the light wave. If the
light enters at an oblique angle, it will change direction as it passes into the second medium.
A result of the relationship expressed in this equation (Snell’s law) is that the slower the speed, the smaller the
angle.
1. Which statement is false?
a. When monochromatic light travels between two media, there is a change in the speed of the
light wave.
b. If the light enters at an oblique angle, it will change direction as it passes into the second
medium.
c. A result of the relationship expressed in Snell’s law is that the slower the speed of the
wave, the smaller the angle change from the normal.
d. The speed of light is constant regardless of the media light travels through.
PROBLEM
Monochromatic light passes between two media. If the angle in medium 1 is
45° (θ1) and the angle in medium 2 is 30° (θ2), calculate the ratio of the light speeds
between media 1 and 2 (v1/v2).
(sin θ1 /sin θ2 = v1/v2)
SOLUTION
2.
Monochromatic light passes between two media. If the angle in medium 1 is
80° (θ1) and the angle in medium 2 is 19° (θ2), calculate the ratio of the light speeds
between media 1 and 2 (v1/v2).
(sin θ1 /sin θ2 = v1/v2)
a. 1
b. 2
c. 3
d. 4
Absolute index of Refraction
To simplify refraction problems, a quantity known as the absolute index of refraction is
defined as follows: n = c/v
The equation states that the absolute index of refraction of a medium (n) is the ratio of
the speed of light in a vacuum (c) to the speed of light in a medium (v). The absolute
index of refraction is always greater than or equal to I. The larger the index of refraction,
the slower the speed of light in a medium.
3. Which statement is false?
a. Opaque objects have an index of refraction greater than 1, transparent objects have an
index of refraction of less than 1.
b. The absolute index of refraction of a medium (n) is the ratio of the speed of light in a
vacuum (c) to the speed of light in a medium (v).
c. The absolute index of refraction is always greater than or equal to 1.
d. The larger the index of refraction, the slower the speed of light in a medium.
PROBLEM
The speed of light in a medium is 2.4 x 108 meters per second (v). What is the absolute
index of refraction of the medium?
(c = 3.0 x 108 m/s)
n = c/v
SOLUTION
4.
What is the absolute index of refraction of a medium if the speed of light in the medium is 9.9 x
107 meters per second (v).
(c = 3.0 x 108 m/s)
n = c/v
a. 1
b. 2
c. 3
d. 4
The table below lists the indices of refraction for some common materials.
Snell’s Law
In terms of the index of refraction, the relationship
becomes
We write this as follows:
This relationship is an alternate form of Snell’s law.
From the equations above, we can also write an equation to
express the relationship between the speed of light and index of refraction in two
mediums:
n2/n1 = v1/v2
PROBLEM
A monochromatic light ray is incident on a surface boundary between air and corn oil at
an angle of 60.0 (θ1) to the normal. Calculate the refracted angle (θ2) of the ray in the
corn oil.
SOLUTION
PROBLEM
A ray of monochromatic light in air is incident on a transparent block of material whose
absolute index of refraction is 1.50 (nblock), as shown in the diagram below.
If the angle of incidence is 300 (θair), trace the path of the light ray through block and
back into the air.
SOLUTION
We begin the solution by applying Snell’s law in order to calculate the angle of
refraction:
We now extend the ray into the block at 19.5°. Using geometry (alternate interior angles),
we find that the ray reaches the second surface at an angle of 19.5°.
If we applied Snell’s law again, we would conclude that the ray would emerge in air at
the original angle of 30°, as illustrated below.
5.
A ray of monochromatic light in air whose absolute index of refraction is 1.00 (nair) is incident
on water whose absolute index of refraction is 1.33 (nwater). If the angle of incidence is 300 (θair),
Calculate the refracted angle (θwater) of the ray in water?
n1θ1 = n2θ2
a. 1
b. 2
c. 3
d. 4
Critical Angle
If monochromatic light passes from medium 1 in which its speed is slower to mediums 2
in which its speed is faster, we can draw a ray diagram as follows:
As angle θ1 is made larger, angle θ2 also becomes larger (n sin θ1 = n sin θ2)
There is one unique angle in medium 1 that will produce an angle of 90° in medium 2, as
illustrated.
This unique angle is called the critical angle (θc). We can derive an equation for the
critical angle using Snell’s law:
n1 sin θc = n2 sin 90°
Then, since sin 90° = 1,
6. Which statement is false?
a. From Snell’s law we can also write an equation to express the relationship between the
speed of light and index of refraction in two mediums.
b. As angle θ1 is made larger, angle θ2 also becomes larger (n sin θ1 = n sin θ2)
c. The critical angle (θc) is one unique angle in medium 1 that will produce an angle of 90° in
medium 2.
d. The angle θ1 is always larger than the angle θ2 and the critical angle (θc)
PROBLEM
Calculate the critical angle (θc) between diamond (n1) and Lucite (n2).
7.
Calculate the critical angle (θc) between Canada Balsam with an index of refraction of 1.53
(n1) and fused quartz with an index of refraction of 1.46 (n2).
sin θc = n2 / n1
o
o
o
o
a) 24
b) 51
c) 73
d) 90
In the event that medium 2 is a vacuum or air (n = 1), the above equation simplifies
to this expression:
8.
Calculate the critical angle (θc) between diamond with an index of refraction of 2.42 (n1)
and air.
sin θc = 1 / n1
o
a) 24
b) 51o
c) 73o
d) 90o
Total Internal Reflection
What is so special about the critical angle? At 90° the maximum
refracted angle possible, light would skim the surface boundary
between the two media. If the critical angle were exceeded, the light
could no longer escape but would be reflected back into the medium.
In this case, the law of reflection would hold as with any other pair of
incident and reflected rays.
We note that the equation sin θc = 1/n given above, implies that larger
values of n yield smaller critical angles. A material such as diamond
(n = 2.42) has a critical angle of only 24°. Because of this small value, much of the light
inside a diamond is totally internally reflected. This total internal reflection is responsible
for much of the diamond’s sparkle.
9. Which statement is false?
a. At 90° the maximum refracted angle possible, light would skim the surface boundary
between the two media.
b. Internal reflection may occur in material such as diamonds, water or glass however, it has
never been observed by scientists.
c. If the critical angle exceeds 90°, the light can no longer escape but would be reflected back
into the medium.
d. The total internal reflection of light is responsible for much of the diamond’s sparkle.
PROBLEM
A monochromatic ray of light in air enters a triangular prism whose absolute index of
refraction is 1.50. The prism is in the shape of an isosceles right tri angle, and the
incident ray is perpendicular to the surface as shown in the diagram.
Complete the path of the light ray as it enters the prism.
SOLUTION
We begin by noting that the angle of incidence is 0°; consequently the angle of
refraction will also be 0°. The ray will enter the prism without bending and will make
an angle of 45° as it strikes the second surface (hypotenuse) of the prism.
We now must ask: Will the critical angle of the block-air combination be exceeded?
Since the angle of the ray is 450, the critical angle is exceeded and total internal
reflection results. The ray continues through the prism and exits into the air as shown:
Dispersion
If white (polychromatic) light is passed through a prism, it is separated into its
component colors as shown in the diagram.
This phenomenon is known as dispersion. When water droplets in the air disperse light,
rainbows may be produced.
Dispersion occurs because the speed of light inside the prism depends on the color of the
light. From the diagram we can see that red light travels fastest because it has the largest
refracted angle (that is, it is bent the least).
10. Which statement is false?
a. If white (polychromatic) light is passed through a prism, it is separated into its component
colors.
b. When water droplets in the air disperse light, rainbows may be produced. This phenomenon
is known as dispersion.
c. The refraction (bending) of light does not occur during dispersion.
d. Dispersion occurs because the speed of light inside the prism depends on the color of the
light. Red light travels fastest because it has the largest refracted angle (that is, it is bent the
least).