MasteringPhysics
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PHY208FALL2008
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Week3HW
Due at 11:59pm on Sunday, September 21, 2008
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The next exercise is about reflection and refraction of light
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.
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
of a material is defined as the ratio of the speed of light
in vacuum to the speed
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 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
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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 (
interface at an angle
Part C.1
) to air (
Find an expression for the ratio of the sines of
Let the index of refraction of water be
of to the sine of .
and that of air be
, is correct?
and
. Use Snell's law to find an expression for the ratio of the sine
Express your answer in terms of some or all of the variables
ANSWER:
). If the incident ray strikes the water-air
, which of the following relations regarding the angle of refraction,
,
, and
.
=
Now, note that for the water-air interface
. Therefore,
.
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 (
interface at an angle
Part D.1
) to glass (
Find an expression for the ratio of the sines of
Let the index of refraction of water be
sine of to the sine of .
and that of glass be
Express your answer in terms of some or all of the variables
ANSWER:
). If the incident ray strikes the water-glass
, which of the following relations regarding the angle of refraction
is correct?
and
. Use Snell's law to find an expression for the ratio of the
,
, and
.
=
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.
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Part E
Consider a ray of light that propagates from air (
) to any one of the materials listed below. Assuming 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.
Part E.2
Find an expression for the sine of the angle of refraction
Let the index of refraction of the unknown material be
refraction, .
. Use Snell's law to find an expression for the sine of the angle of
Express your answer in terms of some or all of the variables
ANSWER:
and
.
=
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
Hint F.1
, if the incidence angle is increased, the angle of refraction
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,
greater than the angle of incidence,
. Therefore, as
is increased, at some point
90 and the refracted ray will travel along the interface. The angle of incidence for which
angle
. For any angle of incidence greater than
, is always
will reach its maximum value of
is called the critical
, 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?
Part 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,
material with index of refraction
Express your answer in terms of
to a material with index of refraction
,
, and
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, for a ray of light that travels from a
.
.
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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.
The next two problems are applications of snell's law of refraction
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 23.16
Description: The glass core of an optical fiber has an index of refraction 1.60. The index of refraction of the cladding is 1.48.
(a) What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?
The glass core of an optical fiber has an index of refraction 1.60. The index of refraction of the cladding is 1.48.
Part A
What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?
ANSWER:
Underwater Optics
Description: The net refraction of light passing through several materials is calculated and then used to find the apparent
distance of objects underwater.
Your eye is designed to work in air. Surrounding it with water impairs its ability to form images. Consequently, scuba divers
wear masks to allow them to form images properly underwater. However, this does affect the perception of distance, as you will
calculate.
Consider a flat piece of plastic (index of refraction ) with water (index of refraction
) on one side and air (index of
refraction ) on the other. If light is to move from the water into the air, it will be refracted twice: once at the water/plastic
interface and once at the plastic/air interface.
Part A
If the light strikes the plastic (from the water) at an angle
Hint A.1
, at what angle
does it emerge from the plastic (into the air)?
Angles inside the plastic
There are two important angles within the plastic: the angle immediately after the first refraction (the water/plastic interface)
and the angle immediately before the second refraction (the plastic/air interface). To find out how they relate, draw a picture
with the path the light follows in the plastic and the normals to both surfaces. Once you have labeled both angles, keep in
mind that the surfaces are parallel, and thus their normals are parallel lines. An important theorem from geometry will give
you the relationship between the angles.
Hint A.2 Important theorem from geometry
If two parallel lines are cut by a transversal (a third line not parallel to the first two), then alternate interior angles are
congruent.
Part A.3
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Part A.3 Find the angle in the plastic
Using Snell's law, find
, the angle inside the plastic (i.e., the angle to the normal immediately after the first refraction at
the water/plastic interface).
Hint A.3.a Snell's law
Snell's law states that
, where
medium from which light is incident,
is the angle of incidence,
is the angle of refraction, and
is the index of refraction for the
is the index of refraction for the medium into
which light emerges.
Express your answer in terms of
,
, and
. Remember that the inverse sine of a number
should be entered as
asin(x) in the answer box.
ANSWER:
=
Now, when you calculate
, don't forget that
Express your answer in terms of
,
,
.
, and
. Remember that the inverse sine of a number
should be entered
as asin(x) in the answer box.
ANSWER:
=
Notice that
does not appear in this equation.
Humans estimate distance based on several different factors, such as shadows and relative positions. The most important
method for estimating distance, triangulation, is performed unconsciously. Triangulation is based on the fact that light from
distant objects strikes each eye at a slightly different angle. Your brain can then use that information to determine the angle
as shown in the figure. In the figure, points L and R represent
your left and right eyes, respectively. The distance between your
eyes is , and the distance to the object, point O, is .
Part B
What is the distance to the object in terms of
Express your answer in terms of
and ?
and .
ANSWER:
=
Part C
If the distance to the object is more than about 0.4
the formula for the distance
, then you can use the small-angle approximation
. What is
to the object, if you make use of this approximation?
Express your answer in terms of
and .
ANSWER:
=
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=
Your eyes determine
by assuming that
and
(in the figure) are equal. This is true, unless the light rays are bent before
they reach your eyes, as they are if you're wearing a scuba mask underwater.
Underwater, the situation changes, as shown in the figure..Your
eyes will calculate an apparent distance using the angle
that
reaches your eyes, instead of the correct geometric angle
is the same
. This
that you calculated in Part A. Note that there are
no important geometric considerations arising from the refraction
except the substitution of
for
, because the refraction takes
place so close to your eyes. If the problem discussed someone
looking out of the porthole of a submarine, the geometry would
become more complicated.
Part D
Now use the expression found in Part C for the distance between your eyes and the object at point O, and find the ratio of the
apparent distance to the real distance,
. Remember that the apparent distance is the distance calculated by your eyes using
the angle
instead of the angle
and
. Since we are dealing with small angles, you may use the approximations
.
Part D.1
Use the small-angle approximations
From Part A, you have the expression
.
Apply both small-angle approximations to this equation to get a simpler expression for
Express your answer in terms of
ANSWER:
Part D.2
,
, and
Find
. Once you've used the small-angle approximations, plug your
. Since you are putting
into the equation instead of
, this
.
Express your answer in terms of ,
,
, and
.
=
Express your answer in terms of
ANSWER:
instead of
into the equation from Part C,
gives the apparent distance
ANSWER:
.
=
Because of the refraction your eyes use
equation for
.
and
.
=
Part E
Given that
Hint E.1
and
, by what percent do objects underwater appear closer than they actually are?
How to approach the problem
To find the fraction by which objects appear closer, simply find the difference between the apparent distance and the true
distance, then divide by the true distance. You can then convert this fraction to a percent to get the final answer. For
example, if an object is 1
away and it appears to be 0.6
away then it is 40% closer (
%).
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Express your answer to two significant figures.
ANSWER:
%
These problems deal with ray optics and lenses.
Problem 23.5
Description: A student has built a l-long pinhole camera for a science fair project. She wants to photograph the Washington
Monument, which is 167 m (550 ft) tall, and to have the image on the film be h high. (a) How far should she stand from the
Washington...
A student has built a 19.0
-long pinhole camera for a science fair project. She wants to photograph the Washington
Monument, which is 167 m (550 ft) tall, and to have the image on the film be 5.10
high.
Part A
How far should she stand from the Washington Monument?
ANSWER:
m
A Laser and a Lens
Description: 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
length
and
are given) above the axis of a converging lens with positive focal
. The laser beam travels parallel to the axis of the lens. A large screen is placed at a distance
The laser beam passes through the lens and makes a dot on the screen at a distance
to the right of the lens.
, 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.
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.
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Part A.3
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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?
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
screen is below the axis of the lens, the corresponding length
Express your answer in terms of some or all of the variables
should be positive. If the dot on the
should be negative.
,
,
, 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
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
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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.
Focusing with the Human Eye
Description: A man looks at three objects at different distances. Explain how he must adjust the lenses in his eyes to bring the
various objects into focus.
Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: a
squirrel sitting on a rock next to him, a tree a few meters away, and a distant mountain. As Joe is taking in the view, he thinks
back to what he learned in his physics class about how the human eye works.
Light enters the eye at the curved front surface of the cornea,
passes through the lens, and then strikes the retina and fovea on the
back of the eye. The cornea and lens together form a compond lens
system. The large difference between the index of refraction of air
and that of the aqueous humor behind the cornea is responsible for
most of the bending of the light rays that enter the eye, but it is the
lens that allows our eyes to focus. The ciliary muscles surrounding
the lens can be expanded and contracted to change the curvature of
the lens, which in turn changes the effective focal length of the
cornea-lens system. This in turn changes the location of the image
of any object in one's field of view. Images formed on the fovea
appear in focus. Images formed between the lens and the fovea
appear blurry, as do images formed behind the fovea. Therefore, to
focus on some object, you adjust your ciliary muscles until the
image of that object is located on the fovea.
Part A
Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be
__________ the distance between the front and back of his eye.
Hint A.1
How to approach the problem
Draw a picture of the object (the tree), the lens, and the image it produces. Be sure to include the focal point of the lens.
Where must the fovea be in your sketch if this object is in focus? Is the focal point between the lens and the fovea, on the
fovea, or behind the fovea?
ANSWER:
greater than
less than
equal to
Part B
Joe's eyes are focused on the tree, so the squirrel and the mountain appear out of focus. This is because the image of the
squirrel is formed ______ and the image of the mountain is formed _____.
Hint B.1
Image of the squirrel
The squirrel is closer to the lens (the eye) than the tree. As long as Joe's eyes stay fixed on the tree, their focal length does
not change. Using the lens equation, determine whether the image of the squirrel is closer to the lens than the image of the
tree or farther away.
Hint B.2
Image of the mountain
The mountain is farther from the lens (the eye) than the tree. As long as Joe's eyes stay fixed on the tree, their focal length
does not change. Using the lens equation, determine whether the image of the mountain is closer to the lens than the image of
the tree or farther away.
ANSWER:
between the lens and fovea / between the lens and fovea
between the lens and fovea / behind the fovea
behind the fovea / between the lens and fovea
behind the fovea / behind the fovea
Part C
Joe now shifts his focus from the tree to the squirrel. To do this, the ciliary muscles in his eyes must have _____ the curvature
of the lens, resulting in a(n) _______ focal length for the cornea-lens system. Note that curvature is different from radius of
curvature.
ANSWER:
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ANSWER:
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increased / increased
increased / decreased
decreased / increased
decreased / decreased
Part D
Finally, Joe turns his attention to the mountain in the distance but finds that he cannot bring the mountain into focus. This is
because he is nearsighted. But when Joe puts on his glasses, he can see the mountain clearly. To adjust for his
nearsightedness, his glasses must contain _____ lenses.
Hint D.1
Focusing on distant objects
The image of a distant object like the mountain always forms at (or very close to) the focal point of the fovea-lens system.
When Joe is looking at the most distant object he can see clearly, where is the focal point?
Hint D.2
The role of corrective lenses
Nearsightedness and farsightedness are both caused by the fact that the ciliary muscles cannot make the focal length of the
lens arbitrarily large or small. The corrective lenses must make the image of the distant mountains form someplace that his
eyes are naturally able to focus on.
ANSWER:
converging
diverging
A Two-Lens System
Description: Calculate the image size and position for a two-lens system. Then, find the lens that would put the image in the
same position if it replaced the two lenses.
A compound lens system consists of two converging lenses, one at
other at
An object
with focal length
.
centimeter tall is placed at
.
with focal length
, and the
Part A
What is the location of the final image produced by the compound lens system? Give the x coordinate of the image.
Hint A.1
How to handle multiple optics
The image formed by the first lens acts as the object for the second lens.
Part A.2 Find the object distance for the first lens
How far is the object from the first lens?
Express your answer in centimeters, to three significant figures or as a fraction.
ANSWER:
=
Part A.3 Find the image distance from the first lens
Ignoring the second lens, determine where the image is formed just by the first lens. Give its distance from the lens.
Express your answer in centimeters, to three significant figures or as a fraction.
ANSWER:
=
Part A.4 Find the object distance for the second lens
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How far is the image produced by the first lens from the second lens?
Express your answer in centimeters, to three significant figures or as a fraction.
ANSWER:
=
Part A.5 Find the image distance from the second lens
Using the result of the previous hint, determine how far the final image is from the second lens.
Express your answer in centimeters, to three significant figures or as a fraction.
ANSWER:
=
Express your answer in centimeters, to three significant figures or as a fraction.
ANSWER:
=
Part B
How tall is the image?
Hint B.1
How to approach the problem
The total magnification
is the product of the magnifications caused by the two lenses seperately:
. If you
have difficulty finding the individual magnifications, use the other hints.
Part B.2
Find the magnification of the first lens
What is the magnification of the first lens? Recall that magnification is defined in two ways:
and
.
Express your answer to three significant figures or as a fraction.
ANSWER:
Part B.3
=
Find the magnification of the second lens
What is the magnification of the second lens? Recall that magnification is defined in two ways:
and
.
Express your answer to three significant figures or as a fraction.
ANSWER:
=
Express your answer in centimeters, to three significant figures or as a fraction.
ANSWER:
=
Part C
Is the final image upright or inverted, relative to the original object at
ANSWER:
?
upright
inverted
Now remove the two lenses at
and
and replace them with a single lens of focal length
at
. We want to choose this new lens so that it produces an image at the same location as before.
Part D
What is the focal length of the new lens at the origin?
Part D.1
Find the object distance for the third lens
How far is the object from the new lens?
Express your answer in centimeters, to three significant figures or as a fraction.
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ANSWER:
Part D.2
=
Find the image distance for the third lens
How far is the image from the new lens?
Express your answer in centimeters, to three significant figures or as a fraction.
ANSWER:
=
Express your answer in centimeters, to three significant figures or as a fraction.
ANSWER:
=
Part E
Is the image formed by
the same size as the image formed by the compound lens system? Does it have the same
orientation?
Part E.1
Find the magnification of the third lens
What is the magnification of the third lens? Compare the result with your answer for Parts B and C.
Express your answer to three significant figures or as a fraction.
ANSWER:
ANSWER:
=
The image is the same size and oriented the same.
The image is the same size and oriented differently.
The image is a different size and oriented the same.
The image is a different size and oriented differently.
A Microscope for Biology
Description: Find the length of a microscope tube needed to view an object in focus, given its distance from the objective and
the objective focal length.
In a two-lens system, the image produced by one lens acts as the object for the next lens. This simple principle finds applications
in many optical instruments, including some of common use such as the microscope and the telescope.
Part A
The microscope available in your biology lab has a converging lens (the eyepiece) with a focal length of 2.50
mounted on
one end of a tube of adjustable length. At the other end is another converging lens (the objective) that has a focal length of 1.00
. When you place the sample to be examined at a distance of 1.30
from the objective, at what length will you need to
adjust the tube of the microscope in order to view the sample in focus with a completely relaxed eye?
Note that to view the sample with a completely relaxed eye, the eyepiece must form its image at infinity.
Hint A.1
How to approach the problem
Since the microscope is a two-lens system, there are two steps to follow to solve the problem. First calculate the location of
the image formed by the objective. Then consider this image to be the object for the eyepiece and find its distance from the
eyepiece. The length of the tube will be the sum of the image distance for the objective and the object distance for the
eyepiece.
Part A.2
Find the image distance for the objective
What is the image distance
for the objective, given that its focal length is 1.00
and the object distance is 1.30
?
Hint A.2.a The thin-lens equation
The image distance
can be found from the thin-lens equation
,
where
is the object distance and
is the focal length of the lens.
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MasteringPhysics
9/25/08 10:15 AM
Express your answer in centimeters.
ANSWER:
Part A.3
=
Find the object distance for the eyepiece
What is the object distance
for the eyepiece, given that it forms an image at infinity and its focal length is 2.50
?
Part A.3.a Find where the object is located
At what point should the object for the eyepiece be located if the eyepiece is to form an image at infinity?
ANSWER:
at the center of curvature of the eyepiece
at the focal point of the eyepiece
at a point between the focal point and the center of curvature of the eyepiece
at infinity
at the focal point of the objective
Express your answer in centimeters.
ANSWER:
=
The length of the tube is then the sum of the image distance for the objective and the object distance for the eyepiece.
Express your answer in centimeters.
ANSWER:
=
Problem 23.68
Description: A slide projector needs to create a h-high image of a h1-tall slide. The screen is l from the slide. (a) What focal
length does the lens need? Assume that it is a thin lens. (b) How far should you place the lens from the slide?
A slide projector needs to create a 98.0
-high image of a 2.20
-tall slide. The screen is 294
from the slide.
Part A
What focal length does the lens need? Assume that it is a thin lens.
ANSWER:
cm
Part B
How far should you place the lens from the slide?
ANSWER:
cm
Summary
0 of 10 items complete (0% avg. score)
0 of 50 points
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