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Geometric Optics
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his lab will help you understand images formed by lenses. In this lab you will work with
converging lenses, which are often used to create real images on a screen. In order for
the image to be in focus, the distances from the object to the lens and from the lens to the
screen must be set properly.
A converging lens is designed so that light rays that come into the lens parallel to each other converge to a
single point. The lens is thicker at the center than at the periphery. The principal axis of a lens is a line
drawn through the center of the lens perpendicular to the face of the lens. A focus is a point on the principal
axis through which all rays parallel to the principal axis pass after refraction by the lens. The focal length of
a lens, F, is the distance from the center of the lens to the principal focus. For a thin lens, an object placed at
distance do from the center of the lens will produce and image that is in focus at distance di from the center
of lens. There is a simple relation between the focal length, the object distance, and the image distance:
For this equation, do and F are taken to be positive. A positive value of di indicates that the image is on
the opposite side of the lens from the object, and negative di means the image is on the same side as the
object. For positive di , the image is real; for negative di , the image is virtual.
A virtual image is a perceptual phenomenon. Since no light rays actually intersect where a virtual image
occurs, it cannot be projected on a screen like a real image. Rather the human eye/brain traces diverging
rays back to an imagined focus. If you look through the lens, the image will appear to be sitting somewhere
behind the lens, and (this is a key realization) at a possibly different distance behind the lens than the real
physical object that is producing it!
Warm-Up
Ray Tracing and Principal Rays:
The diagram below shows several rays emanating from the eraser on a pencil. Draw the continuation
of each ray after it passes through the lens. Use a straightedge to make your drawings accurate.
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A
B
C
F
F
D
E
The rays A, C, and E on the diagram on the previous page are called principal rays, and
are useful in determining the location of an image. A and E are special because they each
pass through a focus on one side of the lens and are parallel to the axis on the other side.
C is special because it goes through the center of the lens and does not change direction at
all. In some cases, one or more of these rays may not actually pass through the lens
because they are blocked by some other object; however, they may still be used in
determining the image location. The principal rays are only a few of the infinitely
many that can be drawn from one point on the object.
1.
On the diagram on the previous page, use two of the three principal rays from the tip
of the pencil to determine the location of the image of the tip of the pencil. If possible,
use a different color for this second set of rays.
2. The diagram below shows a small light placed near a convex lens, closer to the lens
than the focal length. Draw all three principal rays from the bulb filament and
determine the location of the image. Clearly label the image location.
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3.
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Now observe what happens with your own eyes when you place an object closer to a
glass lens than the focal length and look at it through the lens. What does the image
look like? Is the image larger, or smaller than the object? Is it upright, or inverted?
For this exercise, it is best to use a light box with a plate on it as your object.
4. Where is the image located? Is it behind the object or in front of it? [Note: It is easy
to pay more attention to the difference in the sizes of the object and image than to their
locations. We are asking here for the location of the object.]
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Image formation and focus
“Screen” means viewing screen, not computer monitor.
A. A lens, a light source and a screen are arranged as shown below. A sharp, inverted image
of the filament appears on the screen when it is at the location shown.
Bulb
Lens
Screen
5. Predict how each of the following changes would affect the image on the screen. Support your answers
with ray diagrams.
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The screen is moved closer to or farther from the lens.
Bulb
Screen
Lens
Farther Screen
Closer
Test your predictions.
Magnifier:
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Place the screen on the optical bench 40 to 50 cm from the source. Be sure that the
screen is straight up and down. Put the lens on the bench between the source and the
screen and very close to the screen. Adjust the lens and screen so they are the same
height as the source. Slowly move the lens away from the screen until the image is
sharply focused on the screen.
6. Record the positions of the elements in the sketch. Determine do and di. Using the thin
lens equation, calculate the focal length.
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7. Study the magnification of the lens. Measure the object height h by holding your ruler
up to the object at the source. The object height h is the distance from the bottom of the
arrow to the top of the arrow, not the distance from the table to the top of the arrow. Then
measure the image height h’ which is the height of the image at the screen. Notice that the
image is inverted, so h’ is negative. Calculate the magnification of the lens, which is:
m = h’/h =
8. Measure di / do. How does it compare to m?
9. If the screen were removed, would you still be able to see an image of the bulb? Does it
matter where your eye is located?
Experimental investigation: optical instruments
One-lens instruments
Projection lanterns and simple cameras are one-lens optical instruments. The projection lantern makes use
of the fact than an object placed just beyond the principal focus gives an enlarged real image. The simple
camera is merely a light-tight box with a convex lens at one end. The camera is not made to give an
enlarged image, but to focus the real image of a distant object on a film.
Construct each of the following:
The projection lantern
Select a convex lens of at least 100 mm focal length and mount it on the optics bench aligned in
front of your light source (ensure the “crossed-arrow target” is mounted over the opening of the
light source). Adjust the location of the lens until a sharply focused real image is projected on
the wall 2 or 3 m away.
10. Compute the magnification using two methods.
11. What are the characteristics of the projection lantern image? How is an erect image
produced in a slide projector? At close distances is the image more distinct? Explain your
observation.
The camera
For this optical instrument, your light source will not be necessary and you should set it aside.
Select a relatively short focus lens (<100 mm) and mount it on the optical bench at least 250 cm
from an object you wish to image. Place the screen on the optics bench and move it until the
most distinct image forms on the screen.
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12. Starting with your camera focused at “infinity” (viz, something out the window far away),
determine how close you can go to an object without having to refocus the camera on the “film”
(e.g. screen). Start perhaps with a light or back wall of the classroom, then point the camera at
things progressively nearer to you and look at the images that are produced on your screen.
When does it start to get blurry? Sources of light emission will give you the best chance of a
good observation; try fluorescent lights at different distances to you or computer monitors for
example.
A two-lens instrument: the telescope
A telescope (specifically a kind called a refractor) consists of two converging lenses placed so that their
foci nearly coincide as shown in the diagram. The objective lens, the one that is pointed at the object to be
observed (such as a planet, the moon, or a distant ship on the ocean), is a converging lens with a long focal
length. The eyepiece is a converging lens with a comparatively short focal length; this is the one you
actually peek through. The idea is that the image produced by the objective lens becomes the “object” for
the eyepiece lens; the eyepiece lens then produces a second image, and that second image is the one your
eye sees when you look through the telescope.
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Specifically, the objective lens of a simple telescope gathers light from a distant object (at distance d0 ) and
produces a real, inverted image essentially at the position of the focus fO of the objective lens (in reality the
image is slightly farther from the objective lens than fO but the difference may be microscopic – notice the
ray diagram above has exaggerated it to help you see clearly what is going on. For the purposes of
calculation, it helps to simply consider that the real image occurs at the focus). This real image is slightly
inside the principal focus fe of the eyepiece lens, and acts as the object for the eyepiece. The eyepiece in
turn produces an enlarged, virtual image.
Your textbook states that the angular magnification is approximately
M = - fo/fe ,
where fo and fe are the focal lengths of the objective and eyepiece, respectively. In this lab we’re going to
see if this is really a good approximation. Before we start, it’s important to realize this does not mean that
the virtual image produced by the telescope’s eyepiece has greater size than the object you’re looking at!
Telescopes make far-away objects appear bigger, but they do not achieve this by creating images bigger
than the object they’re pointed at. The virtual image produced by the eyepiece is closer to the eye than the
object is and has a greater angular size.
Angular Size To see how this could be possible you need to understand the concept of angular size. Because the object
you’re looking at is far away, it has a small angular size (see illustrations below), even if physically it may
have large dimensions (as does a planet). The telescope acts to create a virtual image which usually has a
much smaller size calculated in absolute terms, but because the virtual image is comparatively close to you,
it has a much larger angular size. (Here’s a good illustration of what angular size means: Try holding a
quarter right up to your eye. Because it subtends a large angle in your field of view, it can block your view
of a much larger object, say the whiteboard of a classroom.)
Figur e 1. The meaning of angular size. You (the viewer) are at the left
point of the tr iangle viewing an object with physical dimensions d. The
object occupies an angle δ in your field of view.
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As you can see from the diagram above, angular size may be found by drawing a triangle with the actual
size of the object, d as the far leg. Draw a midline to divide this into two right triangles, each with legs D
and (1/2)d. Now you can use trigonometry to find what δ/2 is, and the full angle is twice that amount.
Figure 2. A large object far away may have a small angular size. A small object very
nearby may occupy a very large angular size in your field of view.
• Build a telescope using a long-focal length lens as the objective lens and a short focal
length lens as the eyepiece. Set them so their foci overlap only slightly, by perhaps a
few mm. While viewing some distant object, adjust the relative positions of the lenses
until a distinct image is seen.
13. Make a general statement concerning the relative size of the image as seen in the
instrument and the size of the object as seen with the unaided eye.
14. Now it’s time to do some calculation. Your TA will help you with this part. You will
need to apply the lens equation to the objective lens to determine, algebraically, where it
produces a real image. Then you will need to apply the lens equation again for the
eyepiece lens to see, again just algebraically, where that real image (which is the object
that the eyepiece lens “looks at”) is turned into a virtual image. Calculate the physical
size of the virtual image produced by the eyepiece and the distance of the virtual image
from the eyepiece lens. You may make a simplifying assumption to do this: assume that
the foci of the two lenses are at the same exact position.
This will take some time and some algebra. You should plan to ask your TA for guidance
if you get stuck!
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Now let’s examine a specific case. Your TA has drawn a picture on the chalkboard,
which you should measure the height of. Go to the back of the classroom and set
your telescope up on a table or prop it on a chair to hold it steady. Look through and
focus it on the picture.
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Measure the distance d0 in m from your objective lens to the picture. (Floor tiles are
50 cm square.) You already know the focal lengths fo and fe of your lenses and the
size of the picture. From your algebraic expression above, calculate the absolute size y
of the image you see through the scope and distance x from your eyepiece to the
virtual image (i.e., the image you see when you look into that eyepiece.)
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Confirm that the image is where you calculated it to be by doing the following: With
one eye, look through the telescope and make sure the image is in focus. Continuing
to look through the telescope with that eye, open your other eye and look (not through
the telescope) at the hand of your lab partner. Have your partner move until the
image in the telescope and the hand viewed with the other eye are simultaneously in
focus. Measure the distance from the eyepiece to your partner’s hand. How does this
compare to your calculated value of x.
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Now have your partner hold a meter stick so that you can measure the size y of the
image you are seeing. Using this and your measured value of x, together with your
measurements of the object and the distance to it, find the angular magnification of
the telescope.