Download reflection and refraction of light

Name: ___________________
PHYSICS 119 – SPRING 2008
EXPERIMENT 4: REFLECTION AND REFRACTION OF LIGHT
S.N.: ____________________
SECTION:
Experiment 4
PARTNER:
DATE:
REFLECTION AND REFRACTION OF LIGHT
Objectives
•
•
•
•
To introduce the concept of light rays and ray tracing.
Find the location of images using ray tracing and parallax.
Determine experimentally the law of reflection.
Test the law of refraction (Snell’s Law).
A. Method of Ray Tracing
1. Sketch a person standing under a light bulb and holding a piece of paper. Draw the path the light takes when she
looks at the bulb directly, and the path it takes when she looks at the paper. Write down all the things you
assumed about light that are relevant to your drawing.
2.
To visualize rays of light, lay a piece of paper on the whiteboard and place a black screw and knob in the middle of
the paper so the shaft is vertical. With your eyes nearly at the level of the paper, close one eye and look along the
paper at the screw. Notice that you can’t determine exactly how far away the screw is (try to touch the screw with
the tip of your finger; you’ll probably miss), but you can easily determine the direction your finger would need to
travel from your eye to reach the screw. Draw a line between the screw and your eye. Repeat several times
viewing from angles that are about 30° apart. Describe where these lines intersect. (This question may seem easy
but it may help you later.)
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The lines you have drawn are along the direction light travels from the screw to your eye and are,
therefore, called light rays. They are often a convenient tool for studying the properties of mirrors and
lenses. Compare the drawing you have made in question 1 with Figure 5-1, which shows light rays
from a point object. Notice the rays go out in all directions, consistent with the assumption that the
object emits or reflects light in all directions. In this figure, only nine of the infinite numbers of rays have
been drawn. Notice that the farther the object is from the eye, the more nearly parallel are the rays that
enter the eye.
Fig. 1 This ray diagram shows light rays diverging from a point object. Only a small fraction
of them enter the pupil of the eye since the pupil is quite small, of the order of a few
millimeters in diameter. A similar diagram would apply to an extended object since each
point of the object can be considered to be a point source of light rays.
B. Method of Parallax
If you have ever looked at an automobile speedometer from the passenger's seat, you have seen an example
of parallax. The speedometer needle and the scale don't match up for you as they do for the driver. That is
because the needle is a small distance in front of the scale so that the line passing through your eye and the
needle intersects the scale at a different point than the line passing through the driver's eye and the needle. In
this part of the experiment, you will learn how to use this “parallax” effect to determine the location of an object
or image.
Place a sheet of paper on the table. Close one eye and lean down in your chair so that your open eye is at
table level. Have your partner make a tiny wad of scratch paper and drop it onto the paper sheet. Hold the
finger of one hand above the sheet and then move your hand until you think your finger is directly above the
paper wad. Move your finger straight down to the table and see if your finger was actually directly above the
wad.
Try this again, with your partner dropping the paper wad at
different locations. Do not lift your head above the tabletop.
After three tries, exchange roles with your partner for an
additional three tries.
3.
Explain why, when your finger misses the paper wad,
your finger is behind or in front of the wad, but not to the
left or right of the wad.
Fig. 4.2 Top view
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4.
Suppose that, in trying to locate the wad of paper using one eye, you placed your finger behind the wad, as shown
in the top-view diagram. Predict whether your finger would appear to the left of, to the right of, or in line with the
wad:
a) if you moved your head to the left.
b) if you moved your head to the right.
5.
Check your predictions. Resolve any inconsistencies.
Suppose that you placed your finger in front of the wad of paper rather than behind it.
6. Predict whether your finger would appear to the left of, to the right of, or in line with the wad:
a) if you moved your head to the left.
b) if you moved your head to the right.
7.
Check your predictions. Resolve any inconsistencies.
8.
Describe a method, incorporating your results from questions 3, 4, and 6 by which you can successfully locate the
wad of paper using only one open eye at table level..
9.
Test your method. [Keep your fingertip above the wad of paper until you are sure that it is directly above the wad.]
How close did your finger come to the wad of paper? If it did not work, revise your method and try again. Both
partners should try the adopted method successfully (finger hits paper)
The method that you have described and tested in question 2 for locating the piece of paper is called ray
tracing and the method in question 9 is called the method of parallax. Ray tracing uses the fact that light
travels in straight lines except when crossing surfaces, and parallax is using geometry to locate a point given
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two rays emerging from it. The point is located where the rays intersect. Parallax is often used to compare the
locations of two points. If rays from both points are the same, regardless of the position of the eye of the
observer, the points must coincide. In the test above, the two objects are the piece of paper and your finger.
Light Rays Striking Surfaces
When a light ray strikes the boundary between two different materials, it can be transmitted and/or reflected.
Usually both happen to some extent. For both reflection and transmission, the direction of travel of the light will
be changed. The transmitted ray is said to be refracted.
C. Reflected Light
Place a 7.6 cm x 15.2 cm flat mirror in its mount. Place a black screw and knob about 10 cm in front of the
mirror with the shaft vertical. In question 2, it was easy to locate the source of the light rays (the screw),
because you could put your finger on it. However, if you now look into the mirror, you will see light that appears
to come from a second screw behind the mirror. This apparent source of light is called an image, and it has a
definite location just like the object but, in this case, you cannot put your finger on it. In fact, no light actually
passes through this image so it is called a virtual image. However, you can use the methods of parts A and B,
ray tracing and parallax, to find where the image is.
Ray Tracing
Fold a sheet of paper in half and place it on the whiteboard under the screw, making sure the fold butts up to the
mirror. Mark the location of the screw. With your eye at tabletop level, look along the paper at the image of the
screw. Draw a line between the center of the image of the screw that you see in the mirror and your eye.
Repeat several times viewing from angles that are about 30° apart for a total of about 4 measurements.
Remove the screw. Unfold the paper and mark the location of the back of the mirror (the reflecting surface).
Use a ruler to extend the rays back and find the points where they intersect. If you are very careful, the lines will
nearly intersect in a single point.
10. Measure the perpendicular distance between the screw and the back of the mirror (the reflecting surface) and the
distance between the image and the back of the mirror and record them.
mirror to screw distance
mm
mirror to image distance
mm
11. Explain why you had to close one eye for the tests you just made.
12. Describe where the light rays that enter your eye when looking at the image of the screw in the mirror, originate. If
not from the image, why do they appear to come from the image (see Figure 5-1)?
The distance from the object (the screw in this case) to the mirror is called the object distance. The distance
from the image of the screw to the mirror is called the image distance. Like other measurements of distance,
they may be either positive or negative. By convention, the object is placed to the left of the mirror and the object
distance is chosen to be positive. The sign convention we chose is that real images and objects have positive
distances and virtual images have negative distances. For now, all objects will be real and the object distance
will be positive. The image formed by a flat mirror is virtual because the light rays responsible for the image do
not actually pass through the image; they only do so when extended geometrically. This is obvious when you
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recognize that none of the light from the screw passes through the mirror even though the image of the screw is
behind the mirror.
13. Record the object and image distances you just measured. Be sure to use the correct sign. Compare the two
distances.
object distance =
mm
image distance =
mm
Parallax
Fold two sheets of paper in half and butt one against the front of the mirror and the other against the back.
Center the screw in front of the mirror about 10 cm from the mirror. Mark the location of the screw. Place a
second screw behind the mirror. Close one eye and position the other eye at the level of the top of the mirror so
that you see both the image of the first screw and the second screw. The object is now to place the second
screw in the position of the image. Move your head from side to side and at the same time move the second
screw until the image and the second screw move together. This process is called eliminating parallax
between the screw and the image.
14. Mark the position of the screw. Measure the image distance and the object distance. Compare the two distances.
object distance =
mm
image distance =
mm
15. Use what you have just learned to determine the location of the image of the point object in the picture below.
Label the image and, with a ruler, carefully draw two dashed lines that go from the image to two different points in
the pupil of the eye. Draw two solid lines representing the rays that go from the object to the mirror, are reflected
off the mirror and enter your eye. Label them 1 and 2.
Fig. 4.3
Fig. 4.3
Figure 4-3
TA Initials
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You may have heard that the angle of incidence equals
the angle of reflection. The ray that goes from the object
to the mirror is called the incident ray and the angle of
incidence !i is the angle between this ray and a line
perpendicular to the mirror (see Fig. 5.4). Notice that the
line that is perpendicular to the surface of the mirror at the
point where the ray hits is called the normal. The angle
between the reflected ray and the normal is the angle of
reflection !r.
16. Carefully measure the angle of incidence and angle of
reflection for the two rays. Show the values on Fig.
4.3 in the manner indicated in Fig. 4.4. Does !i = !r?
Explain.
Fig. 4.4
D. Refracted Light
As you have seen, the direction of light propagation changes abruptly when light encounters a reflective surface.
If the angle of incidence is not zero, the direction also changes abruptly when light passes through a boundary
between two different media such as between air and glass or glass and water. This results from the difference
in the speed of light in the two media. As stated earlier, the change in direction, or “bending”, of light as it passes
from one material into another is called refraction.
Here are some definitions. The normal to a surface is a line that is perpendicular to that surface. The angle of
incidence, !i, is the angle between an incoming ray, or incident ray, and the normal. The angle of refraction,
!r, is the angle between the ray leaving the surface, the refracted ray, and the normal. To avoid confusion with
the use of !r for both the angle of refraction and the angle of reflection, your text uses !1 for the angle the
incident ray makes with the normal in the first medium and !2 for the angle the refracted ray makes with the
normal in the second medium. If a light ray travels from a material such as air where its speed is relatively large
into a material such as glass, acrylic or water where its speed is less, the light ray will “bend” toward the normal.
Fig. 4.5a
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Fig. 4.5b
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Notice the difference between the two diagrams in figure 5-5. In each case, the light travels from one
medium into a second one in which the speed of light is different. In one case, the ray bends toward
the normal and in the other it bends away from the normal.
17. In this diagram, draw the normal for each case and label the angles of incidence and refraction !1 and !2. In
which diagram does the ray bend toward the normal and in which does it bend away from the normal? For each
case, write “fast” or “slow” on the diagram to indicate the medium in which light travels faster or slower.
18. Figure 5.6 shows rays incident on a semicircular piece of material in which light waves travel more slowly. (In
this lab, the medium is clear acrylic plastic.) Continue each of the rays through this medium and out the other side.
Each ray should bend appropriately. [Hints: The center of the semicircle is marked. It useful to know the location
of the center when drawing a normal to the curved surface of the semicircle. Draw normals to the surfaces where
the rays intersect the surface. Each ray passes through TWO surfaces. The angle of incidence is always the angle
of the ray before it strikes the surface, and the angle of refraction is always the angle of the ray as it goes away
from the surface.]
Fig. 4.6
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TA Initials
Fig. 4.7 Optics bench and components
The objective in setting up the equipment is to produce a set of parallel, narrow beams of light aligned with the
optics bench axis. The components are easily mounted using the magnetic strips. The bench has a raised edge
against which the component carriers, light source and ray table mount are placed in order to assure horizontal
alignment. Vertical alignment depends on careful placing of the components on the component carriers. First
position the light source on one end of the optics bench and the ray table with its mount on the other. The ray
table is two-sided and (unlike the picture) should be placed with the “grid” side up. With the grid side up, it is
simpler to center the cylindrical lens.
Turn on the light source and use the light source position adjustment to center the lamp over the bench.
Carefully center the parallel ray lens on a component carrier and place it on the bench about 6 cm from the light
source. Move it along the bench until there is a broad, parallel beam crossing the ray table. Put the slit plate
near the ray table and adjust the slit position until you have multiple, parallel beams crossing the ray table with a
central beam centered on the table (from 0˚ to 0˚). You may need to make small adjustments to the slits, light
source position and cylindrical lens to accomplish good alignment. When this is achieved, place a short piece of
double sticky tape parallel to the center line about 5mm from the line. This is intended to hold the cylindrical lens
in place. Next place the cylindrical lens on the ray table and carefully align the flat surface of it with the scale
between the 90˚marks. Use the scale to center the lens as precisely as you can. Leave out the slit mask for the
time being.
19. Rotate the lens table to produce the situations you drew in Figure 4.6. Do the rays of light bend as you
predicted? Reconcile any differences by reconsidering your drawing, or by looking for problems in your
experimental setup.
Now add the slit mask (single slit) to the bench as shown to select a single narrow beam. With the cylindrical
lens properly aligned, the narrow light beam will pass straight through the 0˚ mark on one side and emerge at
the 0˚ mark on the other side. Without disturbing the lens, rotate the ray table and observe the path of the beam
of light. If you have aligned the components carefully, the beam will continue to strike the middle of the flat side
of the lens. You will notice that part of the incoming ray passes through the lens and part is reflected. You can
use double-sided tape to tape the lens to the ray table but, if you take sufficient care in not disturbing the lens
this should not be necessary.
20. Explain why it is important to have the center of the flat part of the cylindrical lens at the center of the ray table.
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With the light incident first on the flat surface, measure the angles of incidence !1and angles of refraction !2 at the flat
surface for every 10° from 10° to 80° and enter the results in the table below. [The refracted beam will become quite broad
at large angles because of the width of the incident beam]. Estimate uncertainties by repeating a couple of measurements
“blind” (not knowing what you got before) and going to the other side of normal.
Angle of
Incidence !1
"!1
Angle of
Refraction !2
"!2
sin !1
" (sin !1)
sin !2
" (sin !2)
0.0˚
10.0°
20.0°
30.0°
40.0°
50.0°
60.0°
70.0°
80.0°
21. Make a graph in which you plot !2 versus !1. Use the grid below. Label your axes and provide appropriate scales.
Note: the graph should not be a straight line. Show uncertainties in both !2 and !1 by using error bars in both
horizontal and vertical directions.
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The Law of Refraction, commonly known as Snell’s Law,
n1 sin! 1 = n2 sin! 2 ,
relates the angle of incidence !1 to the angle of refraction !2. The constants n1 and n2 are the indices of
refraction for the first and second materials. The index of refraction of a material is defined as the speed of light
8
in vacuum (3.00 x 10 m/s) divided by the speed of light in the material. For visible light, the speed of light in a
material is always less than it is in vacuum, so the index of refraction has a value greater than 1. The speed of
light in air is very close to that in vacuum so the index of refraction for air is very nearly one. Its exact value
depends on temperature and pressure but is 1.00029 at 0º C and one atmosphere pressure. For work carried
out to two or three significant figures we therefore use nair = 1.00. The index of refraction of water is 1.33 and
that of most types of glass is 1.50 - 1.60. The speed also depends on the wavelength of the light, as we will see
in the next section. The index of refraction is a property of the material just as density is a property of the
material.
22. Plot sin!2 versus sin!1 in the grid below. Label your axes and provide suitable scales.
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23. Write an equation for sin!2 in terms of sin!1 using Snell’s Law. The resulting equation is linear.
Explain what it means for an equation to be linear and show that the equation for sin!2 is linear. If
sin!2 is plotted vs. sin!1, what is the slope of the resulting straight line in terms of n1 and n2? What
is the intercept?
slope =
intercept =
24. Calculate the index of refraction of acrylic from your data. Estimate the uncertainty in the slope. [Hint: it will
help tremendously if you have plotted error bars on your graph!] Show your work. What are the units of n?
nacrylic =
±
Total Internal Reflection and Dispersion
Rotate the ray table 180° so the light beam enters the curved side of the lens first and travels toward the center
of the flat side.
25. What are the angles of incidence and refraction as the light passes from air to acrylic through the curved surface of
the lens? Do these angles depend on how you rotate the ray table? Explain.
!1 =
!2 =
After the ray passes through the curved surface, it will be incident on the flat surface at some angle. We will use
this angle as the angle of incidence for all calculations in this section. Rotate the ray table and observe the rays
that are incident on, and refracted through, the flat surface. In addition, notice the ray reflected from the flat
surface.
26. What is the maximum possible angle of refraction with this arrangement? Note the approximate angle of
incidence for this angle of refraction and use it for the drawing below. This angle of incidence is called the critical
angle, !C. Draw the diagram of the light ray when the angle of incidence is at its critical value. Indicate the
normal and the critical angle.
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27. Describe what happens to the beam as you rotate the ray table to this critical angle and beyond. You
should notice that at large angles of refraction the beam is no longer white, but is rainbow colored.
Is the critical angle greater for red light or for blue light?
28. Use Snell’s Law to derive an expression for the critical angle ! C in terms of the index of refraction n . Calculate
the critical angle from the value of the index of refraction you found in 24. Show your work.
29. Keeping in mind what you observed in 28, measure the critical angle for red light and blue light and use your
measurements to determine the index of refraction of acrylic for red and blue light. [Note: take care with your
measurements, the critical angles are nearly the same] Comment on the consistency of these values with your
value from 24.
!C(blue) =
±
n(blue) =
±
!C(red) =
±
n(red) =
±
This variation of the index of refraction with wavelength is called dispersion. For visible light, all transparent
materials are like acrylic in that the index of refraction increases for decreasing wavelength. For simple lenses,
this results in a shorter focal length for blue light than for red light. The resulting lens defect is called chromatic
aberration. It shows up most clearly in images of objects with sharp edges -- the edges appear colored.
Combinations of two or more lenses with different indices of refraction can be used to correct for chromatic
aberration.
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