Download Phys Chp 10

10.2
SUMMARY
Refraction in Lenses
• Converging lenses bring parallel rays together after they are refracted.
• Diverging lenses cause parallel rays to move apart after they are refracted.
• Rays are refracted at the surfaces of lenses according to Snell’s law.
Section 10.1 Questions
Understanding Concepts
1. Figure 3 shows light rays in air approaching three glass lenses.
Draw diagrams of the same shape, but larger in size, in your
notebook and draw the approximate direction of each light ray as
it travels into the glass and back into the air. (Hint: Draw the
appropriate normals wherever a ray strikes a surface.)
2. Draw a large double concave lens, then draw three parallel rays
on one side of the lens. Apply the method you used in question 1
to show that light rays emerging from the lens diverge. Extend
the rays straight back until they meet, or almost meet. Label this
point appropriately.
(a)
(b)
(c)
Figure 3
For question 1
Applying Inquiry Skills
3. Pretend that the fingers of your wide-open hand, when spread
out flat on a piece of paper, represent diverging light rays that
originate at a single virtual focal point. Devise a way to measure
the distance from the focal point to the tip of your middle finger.
Making Connections
4. The lenses in your science classroom are likely double convex and
double concave lenses. These are not suited for eyeglasses. Why?
Which lenses in Figure 2 are best suited for eyeglasses? Why?
10.2
Images Formed by Lenses
In all lenses, the geometric centre is called the optical centre (O), as shown in
Figure 1. A vertical line drawn through the optical centre is called the optical
axis (OA) of the lens. A horizontal line drawn through the optical centre is
optical centre: (O) the geometric centre
of all lenses
optical axis: (OA) a vertical line through
the optical centre
F
PA
principal
focus
f
focal length
optical axis (OA)
focal plane
Figure 1
Lenses and the Eye 357
principal axis: (PA) a horizontal line
drawn through the optical centre
principal focus: (F ) the point on the principal axis through which a group of rays parallel to the principal axis is refracted
focal length: (f ) the distance between the
principal focus and the optical centre, measured along the principal axis
focal plane: the plane, perpendicular to
the principal axis, on which all focal points lie
focal plane
called the principal axis (PA). If a lens is thin, a group of rays parallel to the
principal axis is refracted through a point on the principal axis called the principal focus (F ). The focal length ( f ) is the distance between the principal focus
and the optical centre, measured along the principal axis.
A beam of parallel rays that is not aligned with the principal axis converges
at a focal point that is not on the principal axis (Figure 1). All focal points,
including the principal focus, lie on the focal plane, perpendicular to the principal axis. When a converging lens refracts light from a distant object (Figure 2),
the rays arriving at the lens are nearly parallel; thus, a real image is formed at a
distance close to one focal length from the lens.
Since light can enter a lens from either side, there are two principal foci. The
focal length is the same on both sides of the lens, even if the curvature on each
side is different. The notation F is always given to the primary principal focus, the
point at which the rays converge or from which they appear to diverge; the secondary principal focus is usually expressed as F.
f
PA
light
dista rays from
nt ob
ject
Characteristics of an Image
F
l
Once an image has been observed or located in a ray diagram, we can use four
characteristics to describe it relative to the object: magnification, attitude, location, and type of the image.
Magnification of the Image
Figure 2
The height of an object is written as ho; the height of the image is written as hi.
To compare their heights, the magnification (M) of the image is found by calculating the ratio of the image height to the object height:
h
M = i
ho
Note that magnification has no units because it is a ratio of heights.
Attitude of the Image
The attitude of the image refers to its orientation relative to the object. For
example, when an image forms on film in a camera, the image is inverted relative
to the object that was photographed. An image is either upright or inverted, relative to the object.
Location of the Image
The distance between the subject of a photograph and the lens of a camera is the
object distance, designated by do. An image of the object is formed on the film
inside the camera. The distance between the image on the film and the lens is the
image distance, di. The image is located either on the object side of the lens, or on
the opposite side of the lens. When discussing object and image distances, we refer
to the location as “between F and the lens,” “between F and 2F,” or “beyond 2F.”
Type of the Image
image point: point at which light from an
object point converges
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An image can be either real or virtual. A real image can be placed onto a screen;
a virtual image cannot. Recall that light diverges from a real object point
(Figure 3(a)). An optical device can converge light from an object point to a
point called the image point. A diverging beam from this image point must
enter your eye in order for you to see the image point. Such an image point is
called a real image point. By using a screen to scatter the light from the real
image point, the image is visible from many angles.
An optical device can also change the direction of a diverging beam from an
object point so that the rays appear to diverge from behind the object point
10.2
(a)
diverging beam
object point
converging beam
diverging beam
real image point
2F
F'
PA
From here, the
real image point
can be viewed
in space.
F
2F'
(b)
rays traced backwards
virtual image point
diverging beam
from lens
object point
PA
2F'
F'
F
2F
diverging beam
from object point
Virtual image
point viewed
from here.
(Figure 3(b)). Such an image point is called a virtual image point. The virtual
image point is located by extending the rays backward until they intersect. Since
the light itself does not intersect at the virtual image point, a virtual image
cannot be formed or captured on a screen. Whenever extending rays backward
to locate virtual image points, use dashed lines on your ray diagrams.
Figure 3
(a) Real image point
(b) Virtual image point
virtual image point: point from which
rays from an object point appear to diverge
Practice
Understanding Concepts
1. Create a chart, table, or detailed diagram to summarize the symbols
and meanings of these properties or variables of lenses: optical
centre, optical axis, principal axis, principal focus, focal length, object
distance, object height, image distance, image height, magnification.
2. The word “MALT” can be used as a memory aid for the four characteristics of images. Print the word vertically. Add the complete word
that corresponds to each letter. In each case, list the possible choices
used to describe an image.
Images Formed by Converging Lenses
An object gives off light rays in all directions, but, for the purpose of locating its
image, we are only interested in those rays that pass through the lens (Figure 4).
1
3
2
2F'
F
2F
F'
Figure 4
Three rays are particularly convenient for
locating image points, since they either pass
through the lens reference positions F and F’
or are parallel to the principal axis. Once
image points are located, we can predict the
four characteristics of the image.
Lenses and the Eye 359
PA
(a)
(b)
The following are the three rays and their rules that can be used:
Rules for Rays in a Converging Lens
PA
PA
lateral displacement
1. A light ray travelling parallel to the principal axis refracts through the
principal focus (F ).
2. A light ray that passes through the secondary principal focus (F)
refracts parallel to the principal axis.
3. A light ray that passes through the optical centre goes straight
through, without refracting.
Note: Only two rays are needed to locate an image point. The third ray
can be used as a check of accuracy.
Figure 5
It may seem strange that the ray that passes through the optical centre in
Figure 4 is not refracted, since most rays passing through the optical centre are
laterally displaced. However, in thin lenses, the lateral displacement of the ray is
so small that we can assume that the ray is not refracted (Figure 5).
In both diagrams in Figure 6, a construction line has been drawn through
the optical centre perpendicular to the principal axis. The actual path of the light
ray is indicated by a solid line.
An image can form either in front of a lens or behind it, and measurements
are made either above or below the principal axis. Therefore, we need a sign convention to distinguish between real and virtual images and to interpret magnification calculations.
(a)
PA
Sign Convention
OA
actual path of light ray
(b)
1. Object and image distances are measured from the optical centre of
the lens.
2. Object distances are positive if they are on the side of the lens from
which light is coming; otherwise they are negative.
3. Image distances are positive if they are on the opposite side of the
lens from which light is coming; if they are on the same side, the
image distance is negative. (Image distance is positive for real
images, negative for virtual images.)
4. Object heights and image heights are positive when measured upward
and negative when measured downward from the principal axis.
PHY11_U4_F10.2.4a
PA
OA
ray diagram
Figure 6
For simplicity, when drawing ray diagrams in
lenses, we can represent all the refraction of
light as occurring just once along the optical axis
instead of twice at the curved surfaces of the
lens. This results in the same image location.
Using this convention, a converging lens has a real principal focus and a positive focal length. A diverging lens has a virtual principal focus and a negative focal
length. The orientation of the image is predicted using the sign convention.
Magnification is positive for an upright image and negative for an inverted image.
Sample Problem 1
A 1.5-cm-high object is 8.0 cm from a converging lens of focal length 2.5 cm.
(a) Draw a ray diagram to locate the image of the object and state its attitude,
location, and type.
(b) Measure the image height from your diagram, and calculate the magnification.
Solution
(a) Figure 7 shows an arrow resting on the principal axis. Three incident rays
and three refracted rays are drawn, according to the rules for converging
lenses. The resulting image, seen in the ray diagram, is real and inverted.
The image is located on the opposite side of the lens between F and 2F.
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10.2
object
F
2F
image
F'
(b) ho = 1.5 cm
hi = 0.70 cm
(negative because it was measured below the principal axis)
PA
Figure 7
h
M = i
ho
0.70 cm
= 1.5 cm
M = 0.47
The magnification of the image is 0.47. Note that a negative magnification is consistent with both the sign convention and the ray diagram.
The position of an object relative to a lens affects how the image is formed.
There are five cases for a converging lens (Figure 8).
(a) object beyond 2F'
(d) object at F'
object
2F
F
2F'
F'
object
image
F
F'
The image is:
between F and 2F, real, inverted, smaller than the object.
No image is formed because the refracted rays are
parallel and never meet.
(b) object at 2F'
(e) object between lens and F'
object
2F'
F
F'
2F
image
image
The image is:
at 2F, real, inverted, the same size as the object.
object
F
F'
(c) object between F' and 2F'
object
2F'
The image is:
behind the object, virtual, erect, larger than the object.
F
2F
F'
image
The image is:
beyond 2F, real, inverted, larger than the object.
Figure 8
Lenses and the Eye 361
Practice
Understanding Concepts
Answers
3. (a) M = –1.0
(b) M = –1.6
(c) M = 2.0
3. A 15-mm-high object is viewed with a converging lens of focal length
32 mm. For each object distance listed below, draw a ray diagram
using all the rules to locate the image of the object. State the characteristics of each image, including the magnification for
(a) do = 64 mm
(b) do = 52 mm
(c) do = 16 mm
Making Connections
4. Below is a list of optical devices that match the arrangements of
lenses and objects in Figure 8. For each of the following cases,
explain where the lens must be placed relative to the object:
(a) a copy camera produces an image that is real and the same size
(b) a hand magnifier produces an image that is virtual and larger
(c) a slide projector produces an image that is real and larger
(d) a 35-mm camera produces an image that is real and smaller
(e) a spotlight produces parallel light; there is no image
(f) a photographic enlarger produces an image that is real and larger
Images Formed by Diverging Lenses
In a diverging lens, parallel rays are refracted so that they radiate outward from
the principal focus (F ) as shown in Figure 9.
principal focus (virtual)
focal plane
diverging lens
F
f
(focal length)
Figure 9
A diverging lens
The rays used to locate the position of the image in a diverging lens are similar to those used with converging lenses (Figure 10). As a result, one set of rules
is used for all lenses. The important difference is that the principal focus in the
converging lens is real, whereas in the diverging lens it is virtual.
1
3
F'
PA
F
Figure 10
Rays refracting through a diverging lens
2
As with converging lenses, we assume with ray diagrams in diverging lenses
that all refraction occurs at the optical axis of the lens. This makes the ray diagram easier to draw.
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10.2
Rules for Rays in a Diverging Lens
1. A light ray travelling parallel to the principal axis refracts in line with
the principal focus (F ).
2. A light ray that is aimed toward the secondary principal focus (F )
refracts parallel to the principal axis.
3. A light ray that passes through the optical centre goes straight
through, without refracting.
Figure 11 illustrates the formation of an image by a diverging lens. For all
positions of the object, the image is virtual, upright, and smaller. The image is
always located between the principal focus and the optical centre.
object
F
image
Figure 11
Image formation in a diverging lens
Sample Problem 2
A 1.5-cm-high object is located 8.0 cm from a diverging lens of focal length
2.5 cm.
(a) Draw a ray diagram to locate the image of the object and state its attitude,
location, and type.
(b) Measure the image height from your ray diagram, and calculate the magnification.
Solution
(a) The refracted rays must be extended straight back to where they meet on
the side of the lens where the object is located, as in Figure 12. The
resulting image is virtual, upright, and located between F and the lens.
object
F
image
PA
F'
Figure 12
Lenses and the Eye 363
(b) ho = 1.5 cm
hi = 0.40 cm (positive because it was measured above the principal axis)
h
M = i
ho
0.40 cm
= 1.5 cm
M = 0.27
The magnification of the image is 0.27.
Practice
Understanding Concepts
5. A 12-mm-high object is viewed using a diverging lens of focal
length 32 mm. Using all the rules, draw a ray diagram to locate
the image of the object for each situation listed below. State the
characteristics of each image, and write a conclusion describing
what happens to the magnification of the image viewed in a
diverging lens as the object distance decreases for
(a) do = 64 mm
(b) do = 32 mm
Answers
5. (a) M = 0.33
(b) M = 0.50
INQUIRY SKILLS
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
Investigation 10.2.1
Images of a Pinhole Camera, Converging Lens,
and Diverging Lens
The principle of the pinhole camera was described early in the 10th century by
the Egyptian scholar Alhazen, who used it to indirectly view a solar eclipse. The
purpose of this Investigation is to study the images formed by a pinhole camera
and compare them with images produced by lenses.
Question
What are the differences among images formed by a pinhole camera, a converging lens, and a diverging lens?
Materials
opaque screen with a pinhole (pinhole camera)
small light source (miniature light bulb)
translucent screen (white paper)
converging lens (f = 20 cm)
diverging lens (f = 20 cm)
metre stick
optical bench
Prediction
(a) Predict the differences among the images formed by the pinhole camera,
the converging lens, and the diverging lens.
364 Chapter 10
10.2
Procedure
1. Position the light source at the end of the optical bench (Figure 13).
viewing direction 1
opaque screen with
pinhole or a
lens (at midpoint)
object
50 cm
image
(translucent
screen)
35 cm
viewing
direction 2
Figure 13
Setup for Investigation 10.2.1
2. Place the opaque screen 50 cm away from the light source.
3. In a dark part of the room, place the translucent screen about 35 cm from
the opaque screen. Slowly move the translucent screen forward and backward. Make observations from both viewing directions at different positions
along the principal axis. Record what happens to the image.
4. Remove the translucent screen and make observations from both viewing
directions, as shown in Figure 13.
5. Replace the opaque screen with a converging lens.
6. (a) Replace the translucent screen and move it until you get a relatively
sharp image on the screen. It may help to angle the converging lens up
slightly. Slowly move the screen forward and backward along the principal axis. Record what happens to the image.
(b) Cover the upper half of the lens with a piece of paper. Observe and
record the effect on the image. Cover the left half of the lens. Observe
and record the effect on the image.
7. Repeat step 4.
8. Replace the converging lens with a diverging lens.
9. Repeat step 4.
10. Place the converging lens between the opaque screen and the light source.
Observe the resulting image with a screen. Move the screen back and forth
along the principal axis. Change the relative positions of the lens, pinhole,
and screen to obtain the “sharpest” image possible. Record the positions of
the pinhole, lens, and screen.
Analysis
(b) Using the evidence you have collected, answer the Question. Your explanation should include neatly drawn ray diagrams and a discussion of linear
propagation, formation of real and virtual images, and scattering of a real
image by a screen.
(c) Which arrangement of pinhole camera, lens, and screen produced the
sharpest image?
Evaluation
(d) Evaluate your predictions.
(e) Describe the sources of error in the investigation and evaluate their effect on
the results. Suggest one or two improvements to the experimental design.
Lenses and the Eye 365
SUMMARY
Images Formed by Lenses
• An image can be described by four characteristics: magnification, attitude,
location, and type.
• The magnification of an image is the ratio of the image height to the object
height.
• Images are real or virtual; real images can be scattered by a screen and
viewed from many angles.
Section 10.2 Questions
Understanding Concepts
1. When a converging lens is used as an ordinary magnifying glass,
is the image it produces real or virtual? Include a sketch with
your explanation.
2. Copy Figure 14 into your notebook and draw the path of each ray
after it has been refracted.
1
2
3
PA
F‘
F
Figure 14
3. Copy Figure 15 into your notebook and draw the path of each ray
after it has been refracted.
1
3
F‘
PA
F
2
Figure 15
4. Images in a diverging lens are always virtual. When an object is
placed between the focus and optical centre of a converging lens,
the image is also virtual. What are the differences between these
virtual images? Explain by studying the four characteristics of
each image.
5. Use ray diagrams to explain why a half-covered lens still produces a complete image of an object.
6. For each situation described below, draw a ray diagram to locate
the image of the object, then describe the four characteristics of
each image. (In each case, try to use all three rules for drawing
ray diagrams.)
(a) f = 3.0 cm; do = 7.5 cm; ho = 2.0 cm
(b) f = 4.4 cm; do = 2.2 cm; ho = 1.2 cm
(c) f = 2.8 cm; do = 5.0 cm; ho = 2.2 cm
366 Chapter 10
10.3
Applying Inquiry Skills
7. (a) Draw a ray diagram of your own design of a variable-length
pinhole camera made of common household materials, such
as cardboard tubes or shoe boxes.
(b) How would you use your design to determine the relationship between the magnification of the image and the distance between the pinhole and the image?
10.3
Mathematical Relationships
for Thin Lenses
You have studied how a converging lens produces images for light sources placed
at different positions along the principal axis. Now you will study the quantitative relationship between object and image positions.
Investigation 10.3.1
Predicting the Location of Images Produced by a
Converging Lens
The purpose of this investigation is to determine the relationship among the
focal length, the image distance, and the object distance of a converging lens.
INQUIRY SKILLS
Questioning
Hypothesizing
Predicting
Planning
Conducting
Recording
Analyzing
Evaluating
Communicating
Question
What is the relationship among the focal length, the image distance, and the
object distance of a converging lens?
Materials
converging lens (f =10 cm → 25 cm)
small light source (miniature light bulb)
translucent screen (white paper)
optical bench
Prediction
(a) Predict what will happen to the size and distance of an image as an object
is moved closer to a converging lens.
Procedure
Part 1: Real Image
1. Create a table similar to Table 1.
2. In a dark part of the room, hold the lens so that light from a distant object
passes through it and onto the screen, as in Figure 1. Move the screen back
and forth until the image is clearly focused. Measure the focal length, f, of
the lens, the distance between the lens and the screen.
Lenses and the Eye 367