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Unformatted text preview: Assignments 4/23/11 2:36 PM Student View Summary View Diagnostics View Print View withEdit Assignment
Answers Settings per Student MP_Assignment#7 [ Print ] Due: 8:07am on Monday, April 11, 2011
Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy Geometry and Reflections
Description: A case of reflection from multiple surfaces, basically a geometry review.
Learning Goal: To learn and practice the geometry skills necessary for complex reflection setups.
The law of reflection has the very simple form
,
where is the angle between the normal and the incident ray and is the angle between the normal and the reflected ray. Although the law itself is easy to use, many realistic situations involve
successive reflections from multiple surfaces. The law of reflection does not become any more complicated in
such cases, but the geometry of the rays does become complicated. Consider the case of light shining onto
a mirror, which is attached to another mirror at some angle , as shown in the figure . In this problem, we
will find the angle at which light leaves the
arrangement of two mirrors. Part A
If the light strikes the first mirror at an angle
Express your answer in terms of
ANSWER: , what is the reflected angle ? . = http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 1 of 20 Assignments 4/23/11 2:36 PM Part B
Now, find the angle
terms of (shown in the new figure ) in . You can easily find in terms of , then just substitute your expression from Part A. Hint B.1
and Relationship between and are complementary angles. Express your answer in degrees in terms of . Notice that the degrees symbol is already listed for you, so just use the number "23" to indicate 23 degrees.
ANSWER: = Part C
Now, find the angle
of and Hint C.1 shown in the figure in terms . Angles in a triangle The sum of the angles in a triangle is . http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 2 of 20 Assignments 4/23/11 2:36 PM Express your answer in degrees in terms of
ANSWER: and . = Part D
Find the angle
and shown in the figure in terms of . You will need to assume that , as it appears in the picture. Hint D.1
and
Hint D.2 Relationship between and are complementary angles.
How to find You can find in terms of in terms of
by using the law of reflection. Express your answer in degrees in terms of
ANSWER: and . = Virtually any reflection problem, no matter how intimidating it may seem, can be broken down into
simple parts by considering each individual reflection carefully. Is Light Reflected or Refracted?
Description: Mostly conceptual questions on index of refraction and Snell's law. Last few parts deal with
total internal reflection. http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 3 of 20 Assignments 4/23/11 2:36 PM When light propagates through two adjacent materials that have different optical properties, some interesting
phenomena occur at the interface separating the two materials. For example, consider a ray of light that
travels from air into the water of a lake. As the ray strikes the air  water interface (the surface of the lake), it
is partly reflected back into the air and partly refracted or transmitted into the water. This explains why on the
surface of a lake sometimes you see the reflection of the surrounding landscape and other times the
underwater vegetation.
These effects on light propagation occur because light travels at different speeds depending on the medium.
The index of refraction of a material, denoted by , gives an indication of the speed of light in the material. It
is defined as the ratio of the speed of light in vacuum to the speed in the material, or .
Part A
When light propagates from a material with a given index of refraction into a material with a smaller index
of refraction, the speed of the light
Hint A.1 Index of refraction The index of refraction
speed of a material is defined as the ratio of the speed of light
in vacuum to the in that particular material, or
. Since it is the ratio of two positive quantities that have the same units, the index of refraction is a pure
(positive) number. Note that the speed of light in a certain material is inversely proportional to the
index of refraction of that material.
ANSWER: increases. Part B
What is the minimum value that the index of refraction can have ?
Hint B.1 How to approach the problem Remember that the speed of light in a certain material is inversely proportional to the index of refraction of that material. Thus, the minimum value of the index of refraction is calculated for the
medium where the speed of light is maximum. That occurs in vacuum where
.
ANSWER: between 0 and 1
The index of refraction of a material is always a positive number greater than 1 that tells us how fast
http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 4 of 20 Assignments 4/23/11 2:36 PM the light travels in the material. The greater the index of refraction of a material, the more slowly light
travels in the material.
An example of reflection and refraction of light is shown in the figure. An incident ray of light traveling in the
upper material strikes the interface with the lower
material. The reflected ray travels back in the upper
material, while the refracted ray passes into the
lower material. Experimental studies have shown
that the incident, reflected, and refracted rays and
the normal to the interface all lie in the same plane.
Moreover, the angle that the reflected ray makes
with the normal to the interface, called the angle of
reflection, is always equal to the angle of incidence.
(Both of these angles are measured between the
light ray and the normal to the interface separating
the two materials.) This is known as the law of
reflection.
The direction of propagation of the refracted ray,
instead, is given by the angle that the refracted ray
makes with the normal to the interface, which is
called the angle of refraction . The angle of refraction
depends on the angle of incidence and the indices of refraction of the two materials. In particular, if we let
be the index of refraction of the upper material and
the index of refraction of the lower material, then
the angle of incidence, , and the angle of refraction, , satisfy the relation
.
This is the law of refraction, also known as Snell's law.
Part C
Now consider a ray of light that propagates from water (
the water air interface at an angle ) to air ( ). If the incident ray strikes , which of the following relations regarding the angle of refraction, , is correct?
Hint C.1 Find an expression for the ratio of the sines of Let the index of refraction of water be
the ratio of the sine of
to the sine of and that of air be
. and
. Use Snell's law to find an expression for Express your answer in terms of some or all of the variables , , and . ANSWER:
= Now, note that for the water air interface . Therefore, . ANSWER: http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 5 of 20 Assignments 4/23/11 2:36 PM ANSWER: When light propagates from a certain material to another one that has a smaller index of refraction,
that is,
, the speed of propagation of the light rays increases and the angle of refraction is
always greater than the angle of incidence. This means that the rays are always bent away from the
normal to the interface separating the two media.
Part D
Consider a ray of light that propagates from water (
strikes the water glass interface at an angle
refraction
Hint D.1 ) to glass ( ). If the incident ray , which of the following relations regarding the angle of is correct?
Find an expression for the ratio of the sines of Let the index of refraction of water be
and that of glass be
for the ratio of the sine of
to the sine of . and
. Use Snell's law to find an expression Express your answer in terms of some or all of the variables , , and . ANSWER:
= Now, note that for the water glass interface . Therefore, . ANSWER: When light propagates from a certain material to another one that has a greater index of refraction,
that is,
, the speed of propagation of the light rays decreases and the angle of refraction is
always smaller than the angle of incidence. This means that the rays are always bent toward the
normal to the interface separating the two media.
Part E
Consider a ray of light that propagates from air ( http://session.masteringphysics.com/myct/assignments?courseID=1035557 ) to any one of the materials listed below. Assuming Page 6 of 20 Assignments 4/23/11 2:36 PM that the ray strikes the interface with any of the listed materials always at the same angle , in which material will the direction of propagation of the ray change the most due to refraction ?
Hint E.1 How to approach the problem The direction of propagation of the ray of light will change the most when the difference between the
angle of refraction and the angle of incidence is maximum. Since we are studying a situation where light
propagates from air to a material that has a greater index of refraction, we can make use of the results
obtained in Part D. We know that, in this case, the angle of refraction is always smaller than the angle
of incidence. Thus, the difference between the angle of refraction and the angle of incidence is
maximum when the angle of refraction is smallest.
Hint E.2 Find an expression for the sine of the angle of refraction Let the index of refraction of the unknown material be
sine of the angle of refraction, . . Use Snell's law to find an expression for the Express your answer in terms of some or all of the variables and . ANSWER:
= Your result shows that the sine of the angle of refraction is inversely proportional to the index of
refraction of the unknown material. Therefore, the angle of refraction is minimum in the material
that has the greatest index of refraction.
ANSWER: ice (
water ( )
) turpentine (
glass (
diamond ( )
)
) The greater the change in index of refraction, the greater the change in the direction of propagation
of light. To avoid or minimize undesired bending of the light rays, light should travel through materials
with matching indices of refraction.
Is light always both reflected and refracted at the interface separating two different materials? To answer
this question, let's consider the case of light propagating from a certain material to another material with a
smaller index of refraction (i.e.,
).
Part F
In the case of , if the incidence angle is increased, the angle of refraction http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 7 of 20 Assignments 4/23/11 2:36 PM Hint F.1 How to approach the question Recall that, according to Snell's law, the sine of the angle of refraction is directly proportional to the sine
of the angle of incidence. Thus, as the angle of incidence is increased, the angle of refraction changes
accordingly. Moreover, since the angle of refraction is greater than the angle of incidence, as you found
in Part C, the angle of refraction can reach its maximum value sooner than the angle of incidence.
ANSWER: increases up to a maximum value of 90 degrees. Since the light is propagating into a material with a smaller index of refraction, the angle of refraction,
, is always greater than the angle of incidence, . Therefore, as
is increased, at some point
will reach its maximum value of 90 and the refracted ray will travel along the interface. The angle of
incidence for which is called the critical angle . For any angle of incidence greater than , no refraction occurs. The ray no longer passes into the second material. Instead, it is completely
reflected back into the original material. This phenomenon is called total internal reflection and occurs
only when light encounters an interface with a second material with a smaller index of refraction than
the original material.
Part G
What is the critical angle for light propagating from a material with index of refraction of 1.50 to a material with index of refraction of 1.00 ?
Hint G.1 Find an expression for the sine of the angle of incidence Use Snell's law to find a general expression for the sine of the angle of incidence,
that travels from a material with index of refraction
Express your answer in terms of , , and , for a ray of light to a material with index of refraction . . ANSWER:
= Now find when , = 1.50 , and = 1.00 . Express your answer in radians.
ANSWER:
= In conclusion, light is always both reflected and refracted, except in the special situation when the
conditions for total internal reflection occur. In that case, there is no refracted ray and the incident ray
is completely reflected.
http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 8 of 20 Assignments 4/23/11 2:36 PM Tactics Box 23.2 Ray Tracing for a Converging Lens
Description: Knight Tactics Box 23.2 Ray Tracing for a Converging Lens is illustrated. (vector applet)
Learning Goal: To practice Tactics Box 23.2 Ray Tracing for a Converging Lens.
The procedure known as ray tracing is a pictorial method for understanding image formation when lenses or
mirrors are used. It consists of locating the image by the use of just three "special rays." The following tactics
box explains this procedure for the case of a converging lens.
TACTICS BOX 23.2 Ray tracing for a converging lens Draw an optical axis. Use graph paper or a ruler. Establish an appropriate scale.
Center the lens on the axis. Mark and label the focal points at distance on either side.
Represent the object with an upright arrow at distance . It is usually best to place the base of the
arrow on the axis and to draw the arrow about half the radius of the lens.
Draw the three "special rays" from the tip of the arrow. Use a straightedge.
A ray parallel to the axis (Ray 1) refracts through the far focal point.
A ray that enters the lens along a line through the near focal point (Ray 2) emerges parallel to the axis.
A ray through the center of the lens (Ray 3) does not bend.
Extend the rays until they converge. This is the image point. Draw the rest of the image in the image
plane. If the base of the object is on the axis, then the base of the image will also be on the axis.
Measure the image distance . Also, if needed, measure the image height relative to the object height.
Follow the steps above to solve the following problem: An object is 9.0
focal length of 2.8 from a converging lens with a . Use ray tracing to determine the location of the image. Part A
The diagram below shows the situation described in the problem. The focal length of the lens is labeled ; the scale on the optical axis is in centimeters.
Draw the three special rays, Ray 1, Ray 2, and Ray 3, as described in the tactics box above, and label
each ray accordingly. Do not draw the refracted rays.
Draw the rays from the tip of the object to the lens.
ANSWER: View Part B http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 9 of 20 Assignments 4/23/11 2:36 PM Now, draw the refracted segments of the three special rays considered previously. Use the labels
, and , for the refracted segment of Ray 1, the refracted segment of Ray 2, and the refracted segment of Ray 3, respectively.
Make sure to extend the refracted rays until they all converge. If your rays do not all converge at
the same point, you may need to be more precise in your drawing.
ANSWER: View Part C
Based on the ray diagram drawn above, at what distance
ANSWER: from the lens does the image form? at approximately 2
at approximately 4
at approximately 6 Chromatic Aberration
Description: Calculation of the effect of dispersion on focusing by lenses.
There are two major types of aberration (defective image formation) in lenses: spherical and chromatic .
Spherical aberration comes from the fact that a spherical lens does not focus parallel rays at a single point.
The closer the rays are to the axis of the lens, the better job the lens does of focusing them at one point, but
it is never perfect.
Chromatic aberration comes from the fact that different wavelengths of light travel at different speeds through
the material of the lens; that is, they have different indices of refraction, a property known as dispersion . This
means that a lens, in effect, has different focal lengths for different wavelengths of light.
Consider a lens made to the following specifications: focal length for red light
, focal length
for blue light . Part A
Consider an object tall placed a distance reflects both red and blue light, find the ratio http://session.masteringphysics.com/myct/assignments?courseID=1035557 from the lens. Assuming that the object of the height of the red image to the height of the blue Page 10 of 20 Assignments 4/23/11 2:36 PM image.
Hint A.1 Find the height of the red image Find the height
of the red image. This is a standard thin lens problem. Use the focal length
given in the problem introduction.
Hint A.1.1 Find the image distance
To find the height of the image, you must first find the image distance. Use the thin lens equation:
, where is the focal length, is the object distance, and is the image distance. What is the image distance ?
Express your answer in centimeters, to four significant figures.
ANSWER:
= Now use the relation to find the height of the red image. Express your answer numerically in centimeters, to three significant figures.
ANSWER:
= Hint A.2 Find the height of the blue image Find the height of the blue image. This is a standard thin lens problem, which may be solved with the
same techniques that you used to find the height of the red image. Use the focal length
given
in the problem introduction.
Express your answer numerically in centimeters, to three significant figures.
ANSWER:
= Express your answer numerically to three significant figures.
ANSWER: = Part B http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 11 of 20 Assignments 4/23/11 2:36 PM In light of your answer to Part A, what would you expect to see if a circular piece of white paper with
radius
were placed
from the lens with its center on the axis of the lens?
Hint B.1 What is white, really? The white paper is reflecting light of all colors including red and blue (the lower and upper frequencies
our eyes can detect).
Hint B.2 What is happening at the center of the image ? The light coming from the center of the paper through the center of the lens does not get refracted. It
merely travels through the lens along a straight line.
Hint B.3 Find which color image of the edge is larger Considering the information found in Part A, which image of the edge of the paper is larger, the image
due to red light or the image due to blue light?
ANSWER: Red
Blue ANSWER: The image would have blue edges.
The image would have a dark blue spot in the center.
The image would have red edges.
The image would have a dark red spot in the center. ± A Laser and a Lens
Description: ± Includes Math Remediation. A laser is projected through a lens onto a screen. Find the
location of the point on the screen.
A laser is mounted as shown in the figure (distances
with positive focal length
at a distance and are given) above the axis of a converging lens . The laser beam travels parallel to the axis of the lens. A large screen is placed to the right of the lens. The laser beam passes through the lens and makes a dot on the screen at a distance , measured upward from the axis of the lens. Assume that a positive value means that the dot is above the axis, while a negative value means that the dot is below the axis. http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 12 of 20 Assignments 4/23/11 2:36 PM Part A
Find the distance
Hint A.1 . How to approach the problem Make your own diagram and draw the refracted ray. Then use the geometry of similar triangles to find
the position of the dot on the screen.
Hint A.2 Drawing the refracted ray Since the incident ray is parallel to the axis of the lens, the refracted ray passes through the focal point
on the right side of the lens. Hint A.3 Find a pair of similar triangles Once you have drawn your own diagram with the refracted ray shown, similar to the picture shown
here, consider the two similar triangles ACF and EFG that the refracted ray forms with the axis of the
lens, one on each side of the focal point F. The
triangle ACF has sides of length and . What
are the lengths of the corresponding sides of the
triangle EFG? http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 13 of 20 Assignments 4/23/11 2:36 PM ANSWER: and
and
and
and Now use triangle similarity to find . Note that your final answer should refer to the location of the dot on the screen. If the dot on the screen is above the axis of the lens, the corresponding length
should be positive. If the dot on the screen is below the axis of the lens, the corresponding
length should be negative. Express your answer in terms of some or all of the variables , , , and . Note that not all of the variables may be required for the answer.
ANSWER:
= Notice that the sign of the height depends upon the sign of . If the focal point is in front of the screen, the dot will be below the axis of the lens. If the focal point is on the screen, the image will be
at the level of the axis of the lens, and if the focal point is behind the screen, the dot will be above
the axis of the lens.
Now consider a specific case. Let the laser be 50 centimeters to the left of the lens and at height 15
centimeters above the axis of the lens. The lens has focal length 30 centimeters, and the screen is 1 meter
to the right of the lens.
Part B
What is the position of the dot on the screen ? Express your answer in centimeters, to two significant figures.
ANSWER: = Part C http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 14 of 20 Assignments 4/23/11 2:36 PM If the laser is moved farther from the lens, what happens to the dot on the screen ?
Hint C.1 How to approach the problem Consider the expression found in Part A. Does the position of the dot on the screen depend on the
distance of the laser from the lens?
ANSWER: It moves up.
It moves down.
It does not move. Part D
If the laser is moved up, farther above the axis of the lens, what happens to the dot on the screen ?
Assume that the ray still strikes the lens.
Hint D.1 How to approach the problem Recall the expression found in Part A. If you double the value of
ANSWER: , how is affected ? It moves up.
It moves down.
It does not move. ± A Sparkling Diamond I
Description: ± Includes Math Remediation. Apply Snell's law to derive the dispersion of red and blue light
in a diamond. "A Sparkling Diamond II" continues this thread, calculating the critical angle for total internal
reflection.
A beam of white light is incident on the surface of a diamond at an angle . Since the index of refraction depends on the light's wavelength, the different colors
that comprise white light will spread out as they pass
through the diamond. The indices of refraction in
diamond are
for red light and
for blue light. The surrounding air has
. Note that the angles in the figure are not
to scale. http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 15 of 20 Assignments 4/23/11 2:36 PM Part A
Calculate , the speed of red light in the diamond. To four significant figures, . Express your answer in meters per second to four significant digits.
ANSWER:
= Part B
Calculate , the speed of blue light in the diamond. To four significant figures, . Express your answer in meters per second to four significant digits.
ANSWER:
= Part C
Derive a formula for
Hint C.1 , the angle between the red and blue refracted rays in the diamond. Apply Snell's law for blue light Snell's law states that
through material and , where is the angle of the incident ray as it passes is the angle of the refracted ray, which passes through material . These angles are measured from the line normal to the interface between the two materials.
Type an expression for
, where
is the angle of the blue refracted ray.
Express your answer in terms of and . Use . ANSWER:
= http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 16 of 20 Assignments 4/23/11 2:36 PM Hint C.2 Apply Snell's law for red light Write an expression for , where Express your answer in terms of is the angle of the red refracted ray. and . Use . ANSWER:
= Hint C.3 Compare the angles Which refracted angle is larger ?
ANSWER: Since the index of refraction is higher for blue light, the blue ray gets bent more, which means that
(see the figure). Thus,
.
Express the angle in terms of
enter the inverse sine of , , and . Use . Remember that the proper way to in this case is asin(x). ANSWER:
= Part D
Calculate
Hint D.1 numerically for . How to approach the problem Plug the given numerical values into the formula you derived from Part C.
Express your answer in degrees to three significant figures.
ANSWER: = The red and blue light rays are split by almost a third of a degree as they pass through the diamond.
This explains why diamonds are cut to have faceted surfaces—if the rays are spread out enough,
each color will shine out of a different facet on the surface of the diamond, producing a brilliant
http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 17 of 20 Assignments 4/23/11 2:36 PM sparkle. Problem 23.14
Description: An underwater diver sees the sun 50 degree(s) above horizontal. (a) How high is the sun
above the horizon to a fisherman in a boat above the diver ?
An underwater diver sees the sun 50 above horizontal.
Part A
How high is the sun above the horizon to a fisherman in a boat above the diver ?
ANSWER: Problem 23.17
Description: A thin glass rod is submerged in oil. (a) What is the critical angle for light traveling inside the
rod?
A thin glass rod is submerged in oil.
Part A
What is the critical angle for light traveling inside the rod?
ANSWER: = Problem 23.32
Description: (a) Find the focal length of the glass lens in the figure .
Part A
Find the focal length of the glass lens in the figure . ANSWER:
cm
http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 18 of 20 Assignments 4/23/11 2:36 PM cm Problem 23.33
Description: (a) Find the focal length of the meniscus polystyrene plastic lens in the figure.
Part A
Find the focal length of the meniscus polystyrene plastic lens in the figure.
Express your answer using two significant
figures. ANSWER: = Problem 23.2
Description: A 5.0  cm  thick layer of oil is sandwiched between a 1.0  cm  thick sheet of glass and a 2.0 cm  thick sheet of polystyrene plastic. (a) How long (in ns) does it take light incident perpendicular to the
glass to pass through this 8.0  cm  thick sandwich ?
A 5.0  cm  thick layer of oil is sandwiched between a 1.0  cm  thick sheet of glass and a 2.0  cm  thick sheet of
polystyrene plastic.
Part A
How long (in ns) does it take light incident perpendicular to the glass to pass through this 8.0  cm  thick
sandwich ?
ANSWER:
ns http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 19 of 20 Assignments 4/23/11 2:36 PM Score Summary:
Your score on this assignment is 0%.
You received 0 out of a possible total of 11 points. http://session.masteringphysics.com/myct/assignments?courseID=1035557 Page 20 of 20 ...
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This note was uploaded on 12/28/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.
 Fall '08
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