Download lab - UCSC Physics

GEOMETRIC OPTICS
INTRODUCTION
In this section we study the properties of rays of light that are reflected or refracted at the
interface between transparent materials. We will assume that the reflecting or refracting
objects are much larger than the wavelength of light, so that we can neglect interference
and diffraction effects, which will be studied in a subsequent lab. In the present case, the
light appears to travel in straight lines in a vacuum or in a homogeneous, transparent
medium such as water or glass. However, the light rays may be deflected as they pass
from one material to another; this phenomenon is called refraction. Refraction occurs
because of the unusual propagation properties of light in transparent media.
Let us consider a monochromatic light ray that originates in air, which, to a high
approximation, resembles the vacuum insofar as light propagation is concerned. Suppose
that this light has a frequency f in air (or vacuum). The first and most important rule of
ray optics is as follows:
Rule #1. If a monochromatic light ray with frequency f in air or vacuum passes into a
transparent medium, the frequency of the light in that medium will not change; i.e. the
frequency will also be f in the medium. We say that the frequency is invariant.
Rule #2. The speed of light in a vacuum is exactly 299,792,458 m/s. This quantity is
called c.
Rule #3. The speed of light in a transparent medium (v) is slower than the speed of light
in a vacuum (c). The index of refraction of the material is defined as the ratio of the speed
in vacuum (c) to the speed in the material. Therefore, n = c/v, or v = c/n.
Rule #4. The relationship between the wavelength, the frequency, and the speed of a
wave in any medium is always f ! " v , or f " v / ! .
Rule #5. (Which follows from rules #1, 3, and 4). For a beam of light passing from one
medium with index of refraction n1 into a second medium of index of refraction n2,
f1 " f 2 #
v1
!1
"
v2
!2
#
c / n1
!1
"
c / n2
!2
# n1!1 " n2 ! 2
(1)
Rule #6. For a beam of light passing from one medium with index of refraction n1 into a
second medium of index of refraction n2, the angle of incidence and the angle of
refraction are related by Snell’s law:
n1 sin $1 " n 2 sin $ 2
(2)
1. INDEX OF REFRACTION
In this experiment we will test Snell’s law, given by the equation in rule #6 above. To do
so, we prepare a narrow beam of light using a combination of the slit plate and the slit
mask, giving us a single beam of light. We then shine this beam of light on to the flat side
of a semi-cylindrical transparent object. With the lights dim, one can see the primary
beam, the reflected beam, and the transmitted beam, and measure their angles. From
these measurements one can determine the index of refraction of the material.
Experiment: Measure the index of refraction of the plexiglass semi-cylinder at your
bench. Do this by measuring the angle of the reflected ray and of the refracted ray for
every 10 degrees variation in the incidence angle. Make a plot of sin !1 vs. sin !2, and
determine the index of refraction of the material.
2. TOTAL INTERNAL REFLECTION
From the previous experiment, we saw that if we consider light passing from a medium
of low index of refraction (air) into a medium of high index of refraction, the light is
deflected toward the direction perpendicular to the interface (also known as the normal).
Therefore, for any incident angle between zero and 90 degrees, there will be a refracted
ray, refracted into a smaller angle. However, the converse is not true. If light within a
medium of high index of refraction attempts to pass into a region of low index, there will
be a critical incidence angle for which the exiting light emerges just parallel to the
surface (i.e. at 90 degrees). Using Snell’s law, setting !2 = 90 degrees, we get
sin $1 "
n2
n
sin $ 2 " 2
n1
n1
(3)
For any incident angles greater than this critical angle, there will be no transmitted beam;
instead, the incident beam will be 100% reflected. This phenomenon is known as total
internal reflection and can be vividly demonstrated.
Using the previous configuration, have the light now incident upon the curved surface of
the transparent block. Because the block is shaped as a semi-cylinder, the light enters the
block and heads toward the center of the cylinder, which is located at the exit surface.
Now, however, we can control the incidence angle of the beam as it tries to emerge. So,
as you vary the angle of the block, at what incidence angle results in an outgoing beam at
90 degrees? If you go beyond this angle, does any transmitted light emerge?
The above phenomenon has many practical applications. It makes it possible for light to
travel within optical fibers without losses; it also makes it possible to construct mirrors
with essentially 100% reflectivity, which is useful in instruments like binoculars and
single-lens reflex cameras.
3. SINGLE-LENS OPTICAL SYSTEMS
In this experiment we will develop a full understanding of the optics of a single lens;
using this understanding we will then understand many everyday optical systems,
including the camera, the slide projector, the magnifier, and the human eye.
We will first gain a qualitative understanding of how a lens works by using the preceding
apparatus. To do so, prepare a bundle of beams by removing the slit mask, and by
moving the slit plate and the rotary table as far from the light source as possible. Now
shine the light beams on the curved surface of the cylinder lens and observe the pattern of
beams as they re-emerge from the flat side. Please describe what you observe.
The above experiment demonstrates the basic features of a lens: that a bundle of parallel
rays may be focused to a point by a lens. More importantly, the light from a luminous
plane object can be faithfully projected on to an image plane by means of a single lens.
Moreover, if the lens is thin compared to the path lengths involved, it can be represented
by a single number, f, called the focal length. In that case, the object distance s, the image
distance s´, and the focal length are represented by the famous lens equation:
1 1 1
& "
s s% f
(4)
It is important to note that s, s´, and f can individually be positive, negative, or even zero.
The sign conventions follow from the following definitions:
Incoming and Outgoing Rays: If we consider a bundle of rays moving toward the surface
of an optical element (in our case a lens), we say that the rays are incoming rays, and that
they are striking the incoming side of the lens. By contrast, the rays emerging away from
the surface of an optical element such as a lens are said to be outgoing rays, and that they
are emerging from the outgoing side of the lens.
Object point: If a bundle of incoming rays, when traced either forward or backward,
converge to a point, we say that the point is an object point. The distance of the object
point to the lens is called s. If the object point is on the same side of the lens as the
incoming light, then s > 0, and the object point is said to be real. Conversely, if the object
point is on the same side of the lens as the outgoing light, then s < 0, and the object point
is said to be virtual.
Usually an object point is a point on a physical object, such as a flower, that is emitting or
reflecting light. In this case the object point is a real object point, because the light is
necessarily incoming light from the standpoint of a nearby lens.1
1
In a compound optical system it is possible that an object point is on the outgoing side of a lens, in which
case the object point is a virtual object point and s < 0. Such systems are beyond the scope of this lab.
Image point: If a bundle of outgoing rays, when traced either forward or backward,
converge to a point, we say that the point is said to be an image point. The distance of the
image point to the lens is called s´. If the image point is on the same side of the lens as
the outgoing light, then s´ > 0, and the image point is said to be real. If the image point is
on the same side of the lens as the incoming light, then s´ < 0, and the image point is said
to be virtual.
Focal length f: In general, the focal length of a lens is positive (f > 0) if a lens is thicker in
the center than on the edges. Such a lens will cause a parallel beam of incoming rays to
converge at a distance f on the outgoing side. Such a lens will also render the light from a
point source (point object) located at a distance f from a lens into parallel outgoing rays.
If a lens is thinner at the center than at the edges, its focal length is negative (f < 0).
Magnification: Here is one of many places where the complicated sign conventions pay
off. In every possible case, the ratio of the image size to the object size is given by the
quantity M, where M = -s´/s. What’s more, if M is positive, the image will be erect with
respect to the object; conversely, if M is negative, the image will be inverted with respect
to the object.
4. REAL OBJECTS AND REAL IMAGES
The simplest and commonest optical system produces a real image (s´ > 0) from a real
object (s > 0). This is how the slide projector, the camera, and the eye (by itself) all work.
It is very straightforward to set up a demonstration of this single-lens system. A
schematic of a single lens system is shown in Figure 1.
Figure 1. A schematic illustration of the principle of a single lens optical system, with a
focal length f = +26.25 mm. The luminous object is located at s = +35 mm, and the real
image is located at s´ = +105 mm. The magnification of this lens is M = -s´/s = -3. The
three principal rays are shown, as well as several additional rays. For clarity, the lens is
shown edgewise as a thin plane.
On the lab bench is a light source that brightly illuminates a test object. Putting this light
source at one end of the optical rail, and putting the viewing screen at the opposite end,
move the f = +75 millimeter lens back and forth until the object is brought into focus onto
the screen. Record the size of the object and the image, and calculate the observed
magnification M. Also, record the values of s and s´. Did your measurements of s, s´, and
M agree with what you expected? Are there one or more additional locations for the
placement of the lens that give you a sharp focus; and if so, record s, s´, and the
magnification M.
Now, by moving the image plane closer to the luminous test object, verify the lens
equation a second time.
How close together can you place the object plane and the image plane and still get a
sharp focus, with the +75 millimeter lens placed midway between them? Is this consistent
with the prelab question?
Now perform the following experiment. Bring the crossed arrow target to a focus on the
imaging screen. What happens if you cover half of the lens with a card? Explain your
observation.
5. DEMONSTRATION: THE EYE (OPTIONAL)
The human eye is the classic model of a single lens optical system. Amazingly, most
persons under the age of 50 or so can involuntarily change the focusing power of the lens
of the eye, so that objects at infinite distance can be brought into sharp focus, or,
alternatively, objects as close as 250 mm can be brought into sharp focus. This
phenomenon is known as accommodation.
There are three common, but correctable, departures from normal vision in young
persons:
1. Myopia (near-sightedness): For a myopic person, the farthest distance that can be
brought into focus is less than infinity. To such a person, stars, or even distant
mountains, appear blurry. This is because the lens, when relaxed, focuses too strongly
for the given distance between the retina and the lens. This can easily be corrected
with eyeglasses that weaken the overall focusing; i.e. with diverging (f < 0) lenses. If
you wear glasses and the lenses are thinner at the center than at the edges, then you
are myopic.
2. Hyperopia (far-sightedness): A far-sighted person has a lens that, when relaxed,
focuses too weakly to even bring objects at infinity into focus. However, with slight
accommodation, objects at infinity can be brought into focus. To such persons, stars
and distant mountains appear in sharp focus, because the lens automatically
accommodates for the distance. Since the lens is relatively weak in relation to the
position of the retina, at full accommodation such a person cannot usually focus on
objects as close as 250 mm. Again, such a condition can be easily corrected with
eyeglasses that strengthen overall focusing; i.e. with converging lenses. If you wear
glasses and the lenses are thicker at the center than at the edges, then you are
hyperopic.
3. Astigmatism: If your eyes are astigmatic, then your lenses do not have the same
focusing power in the horizontal plane as in the vertical plane; i.e. they are not
azimuthally symmetric. This is an effect that cannot be corrected by accommodation,
but it can easily be corrected with eyeglasses.
Finally, over the age of 50 or so, you will gradually lose the power of accommodation.
This condition is called presbyopia. Your lenses will have only a limited range of
focusing, and you are stuck with being able to only focus on things within a certain range
of distances, depending upon whether you have normal, myopic, or hyperopic lenses. To
correct for this condition you need a pair of glasses for every distance at which you need
to focus. In practice it is usually sufficient to have a pair of glasses for distance viewing,
and a pair of glasses for close viewing. These can be combined in a single element known
as a bifocal, or even as a trifocal.
In this demonstration you will find a model of the normal human eye that lacks
accommodation power (as would occur in persons over about the age of 50). Such an eye
sharply focuses objects at great distances on to the retina. For this model, the “retina” can
be placed in three positions: normal, far, and near, representing the effective locations of
the retina for the normal eye, the far-sighted eye, and the near-sighted eye.
Notice that an object is brought into sharp focus on to the “retina” when the retina is
placed at the normal position, but that it is blurry when the retina is placed at the near or
at the far positions. Draw a ray diagram to represent the three cases.
PRE-LABORATORY QUESTIONS
1. Plexiglass (also called lucite, or poly-methyl-methacrylate) has an index of refraction
of n = 1.49. What is the speed of light in plexiglass? At what angle of incidence does
one expect total internal reflection?
2. You have at your disposal a lens with focal length +150 mm. A luminous object is
placed at distances of 100, 150, 200, 300, and 500 mm to the left of the lens,
respectively. In each case, where is the image plane; is the image virtual or real; and
what is the magnification?
3. (Slightly challenging) You have at your disposal a lens with given focal length f. You
may place the lens anywhere between an object plane and an image plane, which are
separated by a distance d. What is the smallest value of d for which you can obtain a
real image of the real object? (Hint: (1) substitute d-s for s% in the lens equation; (2)
solve for d as a function of s; and (3) find the value of s that minimizes d. (4)Use this
information to determine s% and d.