Download Jacaranda Physics 1 2E Chapter 2

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Chapter
Refracting
light
2
Remember
Before beginning this chapter, you should be able to:
• associate changes in direction of light rays with
refraction
• describe some everyday uses of lenses.
Key
ideas
After completing this chapter, you should be able to:
• describe the bending of light as it passes from one
medium into another
• use the ray model to describe the refraction of light
• mathematically model refraction using Snell’s Law
• use the ray model of light to describe and explain
total internal reflection and mirages
• describe the path of light through an optical fibre
• discuss the limitations of the ray model in
describing and explaining refraction
• demonstrate how a particle model and a wave
model can be used to describe and explain the
reflection and refraction of light
• use the ray model to locate and describe images
formed by convex and concave lenses.
Figure 2.1
T
he path of a light ray bends when
travelling from one material into
another, such as through water. A person’s
legs appear shorter.
Inve
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ations
stig
Investigation 2.1
Seeing is believing
Refraction is the bending of
light as it passes from one
medium into another.
Experience shows that when you are
spearing for fish in the shallows you
must aim the spear below where the
fish appears to be in the water. At the
beach or in a pool, people standing in
the shallows appear to have shorter
legs. Our perception is distorted, but
the reason is not apparent.
When we set up a special situation,
such as in figure 2.2, where a straight
rod is placed in a beaker of liquids that
do not mix, the idea of change of
direction is apparent. This change in
direction of the light is called refraction.
Figure 2.2
An example of
refraction
BENDING OF LIGHT
The ray model can help explain our observations of light. If a fish seems
closer to the surface of the water, the ray of light from the fish must have
bent. To our eye, the ray seems to be coming from another direction (see
figure 2.3). Given that light can travel both
ways along a light path, the fish will see
the spear thrower further towards the
vertical than he actually is.
air
water
Fish is here.
Fish appears
to be here.
Figure 2.3
The rays from the fish bend when they enter the air. To the
eye, the rays appear to come from a point closer to the surface.
The normal is a line that is
perpendicular to a surface or a
boundary between two surfaces.
The angle of incidence is the
angle between the incident ray
and the normal.
The angle of refraction is the
angle between the refracted ray
and the normal.
The ray model not only gives us a way of describing our observations of
the bending of light, but also of taking measurements. The angle that a
ray of light makes with the normal, angle of incidence and angle of
refraction can be measured and investigated.
Snell’s Law
In 1621, the Dutch physicist Willebrand Snell investigated the refraction
of light and found that the ratio of the sines of the angles of incidence
and refraction was constant for all angles of incidence.
The diagram on page 25 shows how an incident ray is affected when it
meets the boundary between air and water. The normal is a line at right
angles to the boundary, and all angles are measured from the normal.
Some of the light from the incident ray is reflected back into air. The rest
is transmitted into the water. The following ratio is a constant for all
angles for light travelling from air to water:
sin θ
--------------------i- = constant.
sin θ r
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angle of
incidence
normal
angle of
reflection
incident ray
boundary
reflected ray
i
i
air
water
r
angle of
refraction
refracted ray
sin θ
The ratio -----------------------i is constant for all angles for light
sin θ r
travelling from air to water.
Figure 2.4
Relative refractive index is a
measure of how much light
bends when it travels from any
one substance into any other
substance.
The absolute refractive index
of a substance is the relative
refractive index for light
travelling from a vacuum into
the substance. It is commonly
referred to as the refractive
index.
Snell repeated his experiments with different substances and found
that the ratio was still constant, but it had a different value. This suggested that different substances bend light by different amounts.
(Remember that some light is always reflected.)
In fact, there is a different ratio for each pair of substances (for
example air and glass, air and water). A different ratio is obtained for
light travelling from water into glass. The value of the ratio is called the
relative refractive index because it depends on the properties of two different substances.
The bending of light always involves light travelling from one substance to another. It is not possible to find the effect of a particular
substance on the deflection of light without adopting one substance as a
reference standard. Once you have a standard, every substance can be
compared with it. A natural standard is a vacuum — the absence of any
substance. The absolute refractive index of a vacuum is given the value of
one. From this, the absolute refractive index of all other substances can
be determined. Some examples are given in table 2.1. (The word ‘absolute’ is commonly omitted and the term ‘refractive index’ usually refers
to the absolute refractive index.)
Table 2.1
Values for absolute refractive index
MATERIAL
VALUE
Vacuum
1.000 0
Air at 20°C and normal atmospheric pressure
1.000 28
Water
1.33
Perspex
1.49
Quartz
1.46
Crown glass
1.52
Flint glass
1.65
Carbon disulfide
1.63
Diamond
2.42
CHAPTER 2 REFRACTING LIGHT
25
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s
link
Refraction applet
The refractive index is given the symbol n because it is a pure number
without any units. This enables a more useful restatement of Snell’s Law,
for example:
nair sin θair = nwater sin θwater.
More generally this would be expressed as follows:
n1 sin θ1 = n2 sin θ2.
This formula is more useful because the two values for a substance —
its refractive index and the angle the light ray makes — are together on
the same side and are combined in the same way on both sides of the
equation.
normal
Figure 2.5 A
graphical depiction of Snell’s
Law for any two substances.
Note that the light ray has no
arrow, because the relationship
is true for the ray travelling in
either direction.
SAMPLE
PROBLEM
2.1
Solution:
medium 1
refractive index n1
1
boundary
medium 2
refractive index n2
2
n 1 sin 1 = n 2 sin 2
A ray of light strikes a glass block of refractive index 1.45 at an angle of
incidence of 30°. What is the angle of refraction?
nair = 1.0; θair = 30°; nglass = 1.45; θglass = ?
1.0 × sin 30° = 1.45 × sin θglass (substitute values into Snell’s Law)
sin 30 °
sin θglass = ----------------(divide both sides by 1.45, the refractive
1.45
index of glass)
= 0.3448
(calculate value of expression)
(use inverse sine to find angle whose sine
⇒
θglass = 20.17°
is 0.3448)
(round off to two significant figures)
⇒
θglass = 20°.
SAMPLE
PROBLEM
2.2
Solution:
A ray of light enters a plastic block at an angle of incidence of 40°. The
angle of refraction is 30°. What is the refractive index of the plastic?
nair = 1.0; θair = 40°; nplastic = ?; θplastic = 30°
1.0 × sin 40° = nplastic × sin 30° (substitute values into Snell’s Law)
sin 40 °
nplastic = ----------------sin 30 °
⇒
= 1.286
nplastic = 1.3.
⇒
(rearrange formula to get the unknown
by itself)
(calculate value of expression)
(round off to two significant figures)
AS A MATTER OF FACT
L
ight can be bent by a strong gravitational field, such as that near
the Sun. The gravitational field can act like a convex lens. Light
from a distant star that is behind and blocked by the Sun bends
around the Sun so that astronomers on Earth see an image of the star
to the side of the Sun.
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Speed of light in glass
During the seventeenth century, it was agreed that light changed speed
when it travelled from air into glass, but how? One explanation or model
of light proposed that light travelled faster in glass. Another model of
light proposed that light travelled slower in glass. Scientists at the time
did not have the technology to measure the speed of light through materials such as water or glass.
It was only in the nineteenth century that Augustin-Jean Fresnel and
Jean Bernard Léon Foucault were able to measure the speed of light in
water as being less than the speed in air. The second explanation above
was accepted and the other rejected.
This means that the refractive index now has a physical meaning:
speed of light in a vacuum
absolute refractive index of water = ----------------------------------------------------------------------------------------------speed of light in water
where
speed of light in a vacuum
= 3.0 × 108 m s−1.
SAMPLE
PROBLEM
2.3
The refractive index of glass is 1.5. How fast does light travel in glass?
8
3.0 × 10
1.5 = ------------------------------------------------------------(speed of light in glass)
Solution:
8
3.0 × 10
⇒ speed of light in glass = ----------------------1.5
(rearrange formula to get the
unknown by itself)
= 2.0 × 108 m s−1.
Apparent depth
At the beginning of the chapter, the experience of the spear thrower was
described and explained. A similar phenomenon occurs when a spear
thrower is directly above a fish — the fish appears to be closer to the surface than it actually is. This observation is known as apparent depth.
Swimming pools provide another example of apparent depth — they
look shallower than they actually are. The refraction of light combined
with our two-eyed vision makes the pool appear shallower.
The relationship is illustrated in figure 2.6 and can be expressed as
follows:
real depth
-------------------------------------------------------- = refractive index.
apparent depth
air
water
apparent
depth
real
depth
Figure 2.6
The phenomenon of apparent depth
CHAPTER 2 REFRACTING LIGHT
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2.4
A swimming pool is 2.00 m deep. How deep does it appear as you stand
on the edge of the pool?
Solution:
real depth = 2.00 m; apparent depth = ?; refractive index of water = 1.33
Inve
SAMPLE
PROBLEM
ations
stig
Investigation 2.2
Using apparent depth
to determine the
refractive index
2.00
---------------------------------------- = 1.33
apparent depth
2.00
⇒ apparent depth = ----------1.33
(substitute values into relationship)
(rearrange formula to get the unknown by
itself)
= 1.50 m.
TOTAL INTERNAL REFLECTION
AND CRITICAL ANGLE
Light can play some strange tricks. Many of these involve refraction away
from the normal and the effect on light of a large increase in the angle
of incidence.
Figure 2.7 There are no mirrors in a fish tank but strange reflections
can be seen. It appears that light is being reflected off the side of the fish tank
and the water surface.
The critical angle is the angle of
incidence for which the angle of
refraction is 90°. The critical
angle exists only when light
passes from one substance into
a second substance with a lower
refractive index.
Total internal reflection is the
total reflection of light from a
boundary between two
substances. It occurs when the
angle of incidence is greater
than the critical angle.
28
It has already been mentioned that some light is reflected off a transparent surface, while the rest is transmitted into the next medium. This
applies whether the refracted ray is bent towards or away from the
normal. However, a special situation applies when the refracted ray is
bent away from the normal. This is illustrated in figure 2.8. As the angle
of incidence increases, the angle of refraction also increases. Eventually
the refracted ray becomes parallel to the surface and the angle of refraction reaches a maximum value of 90° (see figure 2.8(b)). The corresponding angle of incidence is called the critical angle. If the angle of
incidence is increased beyond the critical angle, all the light is reflected
back into the water, with the angles being the same. This phenomenon is
called total internal reflection (see figure 2.8(c)).
WAVE-LIKE PROPERTIES OF LIGHT
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(a) Before critical angle
(b) At critical angle
(c) After critical angle
(total internal reflection)
r
air
water
c c
i i
Figure 2.8
Three stages of refraction leading to total internal
reflection
The critical angle can be calculated using Snell’s Law (see the following
sample problem).
SAMPLE
PROBLEM
2.5
Solution:
What is the critical angle for water given that the refractive index of water
is 1.3?
nair = 1.0; θair = 90°; nwater = 1.3; θwater = ?
1.0 × sin 90° = 1.3 × sin θwater (substitute data into Snell’s Law)
sin 90 °
(rearrange formula to get the unknown
⇒ sin θwater = ----------------1.3
by itself)
= 0.7692
⇒
(determine sine values and calculate
expression)
θwater = 50.28°
(use inverse sine to find angle)
θwater = 50°.
(round off to two significant figures)
Total internal reflection is a relatively common atmospheric phenomenon (as in mirages) and it has technological uses (for example, in
optical fibres).
Mirages
There are several types of mirage that can be seen when certain atmospheric conditions enable total internal reflection to occur. These mirages
appear because the refractive index
of air decreases with temperature.
A common type of mirage occurs
in the desert or above a road on a
sunny day. As displayed in figure
2.10, at ground level the air is hot
(A) with a refractive index close to
1 (B). As height increases, the temperature of the air decreases (C) and
its refractive index increases (D).
Figure 2.9
Mirages such as
this are common on hot, sunny days.
CHAPTER 2 REFRACTING LIGHT
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refractive index
temperature
high
A
D
temperature
increasing
1.000 28
(normal air)
refractive
index
increasing
B
C
low
1.0 (vacuum)
ground
level
Figure 2.10
height
tree
level
Temperature and refractive index profiles for the mirage
phenomenon
Rays of light from a car, for example, go in all directions. The air above
the ground can be considered as layers of air. The closer to the ground,
the higher the temperature and the lower the refractive index. As a ray
moves into hotter air, it bends away from the normal. After successive
deflections, the angle of incidence exceeds the critical angle for air at
that temperature and the ray is totally internally reflected. As the ray
emerges, it follows a similar path, refracting towards the normal as it
enters cooler air. An image of the car can be seen below street level (see
figures 2.9 and 2.11). The mirage is upside down because light from the
car has been totally internally reflected by the hot air close to the road
surface.
Figure 2.11
Web
The
mirage of the car appears upside
down due to total internal
reflection in the hot air close to
the ground.
s
link
Mirages and more
warm air
hot air
road
Another mirage that depends on layers of air at different temperatures
is known as the ‘Fata Morgana’ in which vertical streaks, like towers or walls,
appear. This occurs where there is a temperature inversion — very cold at
ground level and warmer above — and very stable weather conditions.
The phenomenon is named after Morgan le Fay (Fata Morgana in
Italian) who was a fairy and half-sister to King Arthur of the Celtic
legend. She used mirages to show her powers and, in the Italian version
of the legend, lived in a crystal palace under the sea. The mirage is often
seen in the Strait of Messina and over Arctic ice. As shown in figure 2.12,
the light rays from a distant point are each refracted by the different
layers of air, arriving at different angles to the eye. The effect is that the
point source (P) becomes a vertically extended source, like a tower or
wall (see figure 2.13).
rough
sea
P
ice
Figure 2.12
Ray paths for the Fata Morgana
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WAVE-LIKE PROPERTIES OF LIGHT
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Figure 2.13
An example of the Fata Morgana. The conditions that
encourage the Fata Morgana are particularly common in the polar regions over ice.
Optical fibres
Web
An optical fibre is a thin tube of
transparent material that allows
light to pass through without
being refracted into the air or
another external medium.
Another example of total internal reflection is in the important technological application of optical fibres. Optical fibres have become a feature
of modern life. A thin, flexible cable containing an optical fibre can be
placed inside a person’s body to transmit pictures of the condition of
organs and arteries, without the need for invasive surgery. The same can
be done in industry when there is a problem with complex machinery.
Optical fibres are also the basis of the important telecommunications
industry. They allow high quality transmission of many channels of information in a small cable over very long distances and with negligible
signal loss (see figure 2.14).
s
link
Fibre optics
Figure 2.14
A bundle of optical fibres. Each fibre in the bundle
carries its signal along its length. If the individual fibres remain in the same
arrangement, the bundle will emit an image of the original object.
An optical fibre is like a pipe with a light being shone in one end and
coming out of the other. An optical fibre is made of glass which is about
10 micrometres (10 × 10−6 m) thick. Light travels along it as glass is
transparent, but the fibre needs to be able to turn and bend around
corners. The optical fibre is designed so that any ray meeting the outer
surface of the glass fibre is totally internally reflected back into the glass.
As shown in figure 2.15, the light ray meets the edge of the fibre at an
CHAPTER 2 REFRACTING LIGHT
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angle of incidence greater than the critical angle and is reflected back
into the fibre. In this way, nearly all of the light that enters the fibre
emerges at the other end.
optical fibre
light ray
Figure 2.15
A light ray travels along an optical fibre through total
internal reflection.
If the glass fibre is exposed to the air, the critical angle for light travelling from glass to air is 42°, which is quite small. Any angle of incidence
greater than this angle will produce total internal reflection. If the fibre
is very narrow, this angle is easily achieved.
However, in both medical and telecommunication uses, fibres are
joined in bundles with edges touching. The touching would enable light
rays to pass from fibre to fibre, confusing the signal. To overcome this, a
plastic coating is put around the glass to separate the glass fibres. The
total internal reflection occurs between the glass and the plastic. The
critical angle for light travelling from glass to plastic is 82°. This value
presents a problem because light meeting the edge of the glass at any
angle less than 82° will pass out of the fibre (see figure 2.16).
This has implications for the design of the optical fibre and the beam
of light that enters the fibre. The fibre needs to be very narrow and the
light entering the fibre has to be a thin beam with all the rays parallel.
light rays
82°
optical fibre
Figure 2.16
Light rays entering the fibre at too sharp an angle are
refracted out of the fibre.
LIMITATIONS OF THE RAY
MODEL
So far in this chapter we have used the ray model to describe how light is
refracted. Ray diagrams illustrate Snell’s Law and have allowed us to visualise a range of optical phenomena such as mirages and to develop
technologies such as optical fibres. However, the ray model, which views
light as a pencil thin beam, does not offer an explanation of why light
refracts. More sophisticated models are needed to provide an explanation for refraction, and in doing so they suggest further experiments
to investigate the properties of light more deeply, and to develop new
technologies.
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WAVE-LIKE PROPERTIES OF LIGHT
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Two very different models of light were developed in the seventeenth century  one by Sir
Isaac Newton (1642–1727) in England and the
other by Christiaan Huygens (1627–1695) in
Holland.
Newton’s model was described as a ‘particle
model’. In his model, light consists of a stream of
tiny, mass-less particles he called corpuscles. The
particles stream from a light source like water
from a sprinkler.
Huygens proposed a wave model of light, where
light travels in a similar way to sound and water
waves. Light leaves a source in the same way that
water ripples move out from a dropped stone. The
disturbance of the water surface travels outwards
from the source.
Figure 2.17
Huygens proposed that light
travelled outwards from a source like circular ripples on a
pond.
HOW DO THE TWO MODELS EXPLAIN
THE PROPERTIES OF LIGHT?
How light travels
Figure
Newton’s particle model Once ejected from a 2.18
Reflection of light
Every point in the
wavefront is a
Source
source of a small
S
wavelet. The new
wavefront is the
envelope of all the
wavelets.
Web
light source the particles continue in a straight
line until they hit a surface.
Huygens’s wave model Huygens proposed a
basic principle: ‘Every point in the wavefront is a
source of a small wavelet. The new wavefront is
the envelope of all the wavelets.’
s
link
Huygens’s principle
applet
Newton’s particle model As particles approach a surface they are repelled by
a force at the surface that slows down and reverses the normal component of
the particle’s velocity, but does not change its tangential component. The particle is then reflected from the surface at an angle equal to its angle of
approach. The same process happens when a billiard ball hits the cushion.
Figure 2.19
Newton’s particle model of reflection
i i´
mirror
Huygens’s wave model As each part of the wavefront arrives at the surface,
it produces a reflected wavelet. The new wavelets overlap to produce the next
wavefront, which is travelling away from the surface at an angle equal to its
angle of approach.
D
B
Figure 2.20
The wave model of reflection. C and D are parallel,
incoming rays. AB is the wavefront. When A hits the mirror a circular wavelet
is produced. By the time B has reached the mirror at E, the reflected wavelet
has travelled out to F. The line EF is the reflected wavefront.
E
C
F
mirror
A
(continued )
CHAPTER 2 REFRACTING LIGHT
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Refraction of light
Newton’s particle model In approaching a denser medium, the particles
experience an attractive force which increases the normal component of the
particle’s velocity, but does not affect the tangential component. This has the
effect of changing the direction of the particles, bending them towards the
normal where they are now travelling faster in the denser medium. Snell’s
Law can be explained by this model.
Inve
i
Figure 2.21
The particle
model of refraction. The particles are
pulled towards the denser medium,
resulting in a change in direction.
ations
stig
air
water
r
Investigation 2.3
Refraction of particles
Huygens’s wave model When the wavefront meets a heavier medium the
wavelets do not travel as fast as before. This causes the wavefront to change
direction. In this case the wavefront bends towards the normal when it
enters a medium where the wave is slowed down. Snell’s Law can be
explained by this model.
B
C
E
A
F
The wave model of refraction.
C and D are parallel, incoming rays. AB is the
wavefront. When A hits the surface a circular wavelet
of slower speed and so smaller radius is produced. By
air
the time B has reached the surface at E, the refracted
water
wavelet has only gone as far as F. The line EF is the
refracted wavefront, heading in a direction bent towards
the normal compared to the incoming wavefront, AB.
A point of difference
Now, with these two explanations of refraction, there is a clear distinction
between the two models. When light bends towards the normal as it enters
water (a denser medium), the particle model says it is because light travels
faster in water (the denser medium), whereas the wave model says it is
because the light is travelling slower.
In the seventeenth century they did not have the technology to measure
the speed of light in water. However, the particle model became the
accepted explanation, partly because of Newton’s status, and partly because
Huygens’s principle suggested that light should bend around corners like
sound, and there was no evidence of this at the time. (Newton himself
actually thought that the particles in his model needed to have some wavelike characteristics to explain some of his other observations of light and
colour.)
New evidence emerges
In 1802, Thomas Young (1773–1829) showed that in fact light could bend
around an edge. This led to the rejection of the particle model, as it had no
mechanism to explain how particles could bend around a corner, and consequently the wave model re-emerged. In 1856, Foucault measured the speed
of light in water and showed that it was indeed slower than that in air,
clinching the argument for the wave model.
34
WAVE-LIKE PROPERTIES OF LIGHT
Inve
Figure 2.22
D
ations
stig
Investigation 2.4
Refraction of waves
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FORMING IMAGES WITH
LENSES
Inve
A convex lens is a lens that is
thicker in the middle than at the
edges.
A concave lens is a lens that is
thinner in the middle than at
the edges.
Converging lenses are those
that bend incident parallel rays
towards a focus. That is, they
converge light.
Diverging lenses are those that
bend incident parallel rays away
from each other. That is, they
diverge light.
ations
stig
The refraction of light at a boundary can be put to use if the boundaries
are curved. There are two possibilities for curved boundaries — curving
inwards or curving outwards. A convex lens has its faces curving outwards. A lens that curves inwards is a concave lens. The simple ray tracing
in figure 2.23 illustrates what each lens does to the light rays.
As rays enter the glass they are bent towards the normal. When they
reach the air on the other side of the lens, they are bent away from the
normal. In the case of the convex lens, the emerging rays come together
or ‘converge’ at a point called the focus (F). For the concave lens, the
rays move apart or ‘diverge’, so that they appear to come from a point,
also called the focus, on the other side of the lens. For these reasons,
convex lenses are sometimes called converging lenses, and concave lenses
can be called diverging lenses.
(a) Convex lens
Investigation 2.5
The convex lens as a
magnifying glass
(b) Concave lens
F
F
Figure 2.23
Refraction of rays through (a) a
convex and (b) a concave lens
focal
plane
F
Figure 2.24
principal
axis
Rays
converge at a point on the focal
plane, which passes through the
focus point (F).
In fact, the focus is more than just a point. It is a plane — a focal plane
through the focus point. In figure 2.24, for example, parallel light rays
from a distant object coming in at an angle to the lens are still brought to
a focus, not at the focus but elsewhere in the focal plane. This is of use in
designing telescopes.
Convex lenses — locating
images using ray tracing
There are similarities in the nature of images produced by a convex lens
and those produced by a concave mirror. The features of a convex lens
are illustrated in figure 2.25. The convex lens has two symmetrical curved
surfaces, which means that it has two focus points.
principal
axis
convex lens
pole
focus
Figure 2.25
focus
The convex lens has two focus points.
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The ray model can now be used to locate an image (see figures 2.26
and 2.27).
s
link
Convex lens applet
object
3
convex lens
image
1
Inve
4
F
ations
stig
2
Investigation 2.6
Describing images
produced by a convex lens
convex
lens
2
object
image
F
F
1
3
Figure 2.27 The object
is inside the focus. This makes the rays
diverge after passing through the
lens. ‘Backtracking’ these rays reveals
that they appear to be coming from a
point behind the object. The image is
located at this point.
eMo
F
ling
del
Figure 2.26
The location of the image is determined according to
the point where the three rays cross. All the rays that pass through the lens pass
through the image.
The ray diagram in figure 2.26 demonstrates the following:
• Ray 1 leaves the head of the object parallel to the principal axis, is
refracted by the lens and passes through the focus on the other side of
the lens.
• Ray 2 passes through the focus on the same side as the object, is
refracted by the lens and emerges travelling parallel to the principal
axis.
• Ray 3 travels towards the centre of the lens. If the angle of the
incoming ray is small and the lens is considered thin, then the ray
would appear to continue on in the same direction.
All three rays pass through the same point. This is where the image of
the head of the object is located. Note that the image could have been
located with any two of the rays, but the other can be used to confirm the
original location of the image.
The features of the image can now be described — location, size,
orientation and nature (see chapter 1, page 12).
Convex lenses are used in a variety of applications. There are different
applications for each location of the object. A range of these applications
is given in table 2.2 (see page 38).
Model of a convex lens
Convex lenses — using a
formula to locate and
describe images
u
Ho
image
object
F
F
f
Hi
v
Figure 2.28
Representation of the elements of
a formula to locate images in a
convex lens
36
The similarities between the ray diagrams for the concave mirror and the
convex lens suggest that the same formula (see below) can be used for
both, but with a new sign convention which reads:
Real image distances are expressed as positive quantities and virtual
image distances are expressed as negative quantities.
1 1 1
--- + --- = --u v
f
height of image ( H i )
v
magnification (M) = ------------------------------------------------------- = --height of image ( H o ) u
WAVE-LIKE PROPERTIES OF LIGHT
Jac Phys 1 2E - 02 Page 37 Tuesday, October 21, 2003 1:43 PM
SAMPLE
PROBLEM
2.6
Solution:
Skill
cks
che
Significant figures
(p. 493)
Skill
cks
che
Describe fully the image of a 4.0 cm object 15 cm in front of a convex
lens with a focal length of 10 cm.
u = 15 cm; v = ?; f = 10 cm; M = ?; Ho = 4.0 cm; Hi = ?
1
1
1
------ + --- = -----(substitute data into the lens formula)
15 v 10
1
1
1
--- = ------ − -----⇒
(rearrange to get the unknown by itself)
v 10 15
1
3
2
= ------ − ------ = ------ (combine fractions with lowest common
30 30 30 denominator)
⇒
v = 30
(invert fractions to get v as numerator)
The image is 30 cm from the lens and on the opposite side of the lens to
the object.
Hi
Magnification = ------Ho
v
= --u
Using spreadsheets
(p. 504)
H
30
-------i = -----(substitute values into magnification formula)
4.0 15
H
-------i = 2.0
4.0
⇒
Hi = 2.0 × 4.0 (rearrange to get the unknown by itself)
= 8.0
(calculate expression)
The image is 8.0 cm high, real, inverted and located 30 cm from the lens.
eMo
⇒
ling
del
Inve
Using a spreadsheet
to locate the image
from a slide projector
ations
stig
Web
Investigation 2.7
Locating an image
for a concave lens
s
link
Concave lenses — locating
images using ray tracing
Concave lenses curve inwards so they are thinner in the middle, as
opposed to the convex lens which is thicker in the middle. This means
that the rays of light are bent away from the principal axis at both the
front and the back surfaces of the lens (see figure 2.23(b), page 35. For
this reason they are also called diverging lenses.
Concave lenses have a principal focus, like object
convex lenses, but it is located on the same
side of the lens as the object and is called
a virtual focus. Light rays parallel to
the principal axis spread out after
passing through the concave lens as if
F
F
they are coming from the virtual focus.
image
concave lens
Figure 2.29
The rays diverge for all
object positions. An upright, diminished, virtual image
is located between the focus and the lens.
Concave lens applet
Concave lenses — using a
formula to locate and
describe images
As with the formula method with the convex mirror, the focal length of a
concave lens can be represented as a negative number because it is a
virtual focus.
CHAPTER 2 REFRACTING LIGHT
37
Jac Phys 1 2E - 02 Page 38 Tuesday, October 21, 2003 1:43 PM
SAMPLE
PROBLEM
2.7
Solution:
Locate and describe the image formed by a 6.0 cm high object 15 cm in
front of a concave lens with a focal length of 10 cm.
Ho = 6.0 cm; Hi = ?; u = 15 cm; v = ?; f = −10 cm
1 1 1
Using --- + --- = --u v
f
1
1
1
------ + --- = ---------⇒
15 v −10
1
1
⇒
= − ------ − -----10 15
(−3 − 2)
= ---------------------30
(substitute data into the lens formula)
(rearrange to get the unknown by itself)
(combine fractions with lowest common
denominator)
–5
= -----(simplify expression)
30
30
⇒
v = -----(invert to get v in the numerator)
−5
= −6.0.
The image is located 6.0 cm in front of the lens. It is virtual and upright.
Hi
v
Magnification = ------- = --Ho u
H
6.0
-------i = ------- = 0.4
(substitute data into magnification
6.0 15
formula)
Hi = 0.4 × 6.0 = 2.4. (rearrange to get the unknown by itself
and calculate expression)
The image is 2.4 cm high.
Because they spread light rays further apart, concave lenses do not
form real images. Therefore, they are usually used in combination with a
convex lens.
Table 2.2
Simple applications of convex lenses
LOCATION OF OBJECT
USES
DESCRIPTION OF IMAGE
Very large distance away from
lens
Objective lens of refracting
telescope
Real, inverted, diminished and
located near the opposite focus
Beyond twice the focal length
from lens
Human eye; camera
Real, inverted, diminished and
located on other side between one
and two focal lengths from lens
At twice the focal length from
lens
Correction lens for terrestrial
telescope
Real, inverted, same size and
located two focal lengths from lens
Between twice the focal length
from lens and the focus
Slide projector; objective lens of
microscope
Real, inverted, magnified and
located on other side of lens
beyond two focal lengths
At the focus
Searchlight; eyepiece of
refracting telescope
No image. The emerging parallel
rays do not meet.
Between focus and lens
Magnifying glass; eyepiece lens
of microscope; spectacles for
long-sightedness
Virtual, upright, magnified and
located on same side of the lens
and further away
38
WAVE-LIKE PROPERTIES OF LIGHT
Jac Phys 1 2E - 02 Page 39 Friday, October 24, 2003 2:08 PM
PHYSICS IN FOCUS
Inve
Flat lenses?
ations
stig
Investigation 2.8
Examining a flat lense
lens works by changing the direction of the light ray at the front
surface and then again at the back surface. The glass in the
middle is there to keep the two surfaces apart. Augustin-Jean Fresnel
devised a way of making a lens without the need for all the glass in
the middle.
The glass surface of the lens is a series of concentric rings. Each
ring has the slope of the corresponding section of the full lens,
but its base is flat. The slopes of the rings get flatter towards the
centre.
A
Figure 2.30
A side view of a convex Fresnel lens showing how
it is constructed
This design substantially reduces the weight of the lens, so lenses
of this type are used in lighthouses. Their relative thinness means
they are also used where space is at a premium, such as in overhead
projectors, and as a lens to be used with the ground-glass screens in
camera viewfinders.
Flat lenses, or Fresnel lenses as they are called, are now attached to
the rear windows of vans and station wagons to assist the driver when
reversing or parking.
MICROSCOPES: AN
APPLICATION OF LENSES
A convex lens can be used as a simple magnifying glass, but to obtain
greater magnification another lens is needed.
eMo
Compound microscope
ling
del
Using a spreadsheet
to model a
microscope
Skill
cks
che
Using spreadsheets
(p. 506)
In its simplest form, a compound microscope has two convex lenses,
one fatter and with a shorter focal length than the other. The fatter
one is located near the object to be magnified and is called the objective lens. The other lens is close to the eye and is called the eyepiece
(see figure 2.31).
An object placed just outside the focus of the objective lens
produces a real, inverted, magnified image. The eyepiece lens is
adjusted so that this image is formed inside the focus of the eyepiece
lens.
The real, inverted image now becomes the object for the eyepiece
lens. Since it is inside the focus of the eyepiece, it produces a final
magnified image which is virtual and inverted compared with the
original object.
CHAPTER 2 REFRACTING LIGHT
39
Jac Phys 1 2E - 02 Page 40 Tuesday, October 21, 2003 1:43 PM
Inve
objective lens
eyepiece lens
convex lenses
ations
stig
object
F
F
•o
Investigation 2.9
Calculating with a
microscope
•o
•
Fe
Fe
•
I1
I2
Figure 2.31
Ray
diagram for a compound
microscope
HUMAN AND ANIMAL VISION
The human eye is an extraordinary device. It is able to respond to an
enormous range of light brightness. The strongest light that the eye can
safely detect is 10 000 million times as bright as the weakest light it can
detect. It is able to focus on objects from many billions of kilometres away
to objects a few centimetres away. It can also detect colour. The parts of
the human eye are shown in figure 2.32.
ciliary
muscle
iris
(coloured)
aqueous
humour
pupil
cornea
fovea
centralis
vitreous
humour
macula
optic
nerve
blindspot
lens
sclera
Cross-section of a human eye
choroid
retina
Figure 2.32
The retina takes the role of the
screen of the eye. It is covered
with nerve cells that detect the
brightness and colour of the
light falling on it.
The cornea is the curved front
surface of the eye. It refracts
light towards the pupil so that it
can pass through towards the
lens.
40
The purpose of the eye is to produce a sharp, real image on a screen
— the retina. Light passes through many refracting elements in the
human eye on its way to the retina (see table 2.3). It achieves a focus in
two stages. The curved cornea at the front of the eye does two-thirds of
the focusing, due to the large difference in refractive index between the
cornea and the air. The lens does the fine focusing as there is little difference in the refractive index of the lens and that of the liquid on either
side of it.
In the rest of the eye, the differences in refractive index are marginal,
but they are crucial when the eye wishes to produce clear images of far
and near objects.
WAVE-LIKE PROPERTIES OF LIGHT
Jac Phys 1 2E - 02 Page 41 Tuesday, October 21, 2003 1:43 PM
Table 2.3
Refractive index of the parts of the eye
PART OF THE EYE
REFRACTIVE INDEX
Tears
1.33
Cornea
1.37
Aqueous humour
1.33
Lens cover
1.38
Lens centre
1.41
Vitreous humour
1.33
AS A MATTER OF FACT
T
Web
he eye is an amazing instrument. It has an automatic aperture
adjustment called the iris. The act of blinking operates the
cornea’s built-in scratch remover, lens cleaner and lubricator. In dim
light, the eye operates as a supersensitive, black and white television
camera. It allows us to see objects with less than 0.1 per cent, or onethousandth, of the light we need for colour vision.
The optic nerve packages the visual information from the retina so
that the brain receives about 30 discrete ‘frames’ per second in a
similar way to television and cinema.
s
link
Modelling the eye
applet
Accommodation is the
adjustment to the thickness of
the lens in the eye to ensure
that the image on the retina is
sharp. When the thickness of
the lens changes, so does its
focal length.
The near point of your eye is
the closest an object can be to
your eye so that your eye can
produce a sharp image of the
object on its retina.
Accommodation mechanisms
Consider the light from a light globe that passes through a convex lens
onto a screen to produce an image of the globe’s filament. If the lens is
moved to a new position, the screen needs to be moved to obtain a sharp
image. The human eye can produce a sharp image on its ‘screen’ (the
retina) of objects at various distances. But the eye’s screen stays put, so
something else has to change to achieve a sharp image. The only thing
that can change is the focal length of the lens.
Convex lenses form images of very distant objects at their focus. As
objects get closer to the lens, the image is formed behind the focus. For
the human eye with its fixed ‘screen’, this means that the focal length
must shorten to keep a sharp image on the retina. This adjustment is
called accommodation.
How short can the focal length of the lens in an eye become? When
you bring an object closer to your eye from an arm’s length away, there is
a point at which the object becomes fuzzy in appearance. This point is
called your near point. As you age, your near point becomes further away
(see table 2.4). That is why some older people hold the newspaper further away to read it.
Table 2.4
Inve
AGE (YEARS)
ations
stig
Investigation 2.10
Measuring your near
point
Average near points by age
NEAR POINT (CM)
AGE (YEARS)
NEAR POINT (CM)
10
7
40
22
15
8.5
50
40
20
10
60
65
30
15
70
200
CHAPTER 2 REFRACTING LIGHT
41
Web
Jac Phys 1 2E - 02 Page 42 Tuesday, October 21, 2003 1:43 PM
s
link
Animal eyes
To enable the focal length of the eye to change, the lens has to
change its shape. To achieve a shorter focal length, the lens needs to
become fatter. This raises important questions. How does the lens
change shape? Which is the natural rest position of the lens in the eye?
Is the lens relaxed in the short focal length position and needs to be
‘stretched’ to see distant objects? Or is it relaxed in the long focal
length position and needs to be ‘squashed’ to see near objects? Muscles
can act only by contraction, so how are the muscles arranged in the eye
to squash the lens?
In fact, the relaxed human eye has a long focal length located at the
retina to form images of very distant objects. Ligaments are attached to
the outside edge of the lens which are continually pulling outwards; however, around the lens itself is a circular muscle. When this circular muscle
contracts, it produces a smaller circle, making a lens that is smaller in diameter, but fatter front to back.
Do the eyes of animals work in the same way as human eyes? What lens
mechanism would be appropriate for an animal in the wild? In fact, most
animals’ eyes are relaxed in the long focal length position because, if the
animal is roused from sleep by the sound of a distant predator, the eye is
ready to focus on it.
While many animals use an adjustable focal length lens like humans,
some use other strategies. One strategy is to move the lens backwards and
forwards. In most fish and snakes, the lens moves within the eyeball in the
same way that a camera lens is moved to produce a sharp image on film.
Correcting eye defects
Sometimes our eyes do not work properly and corrective lenses need to
be prescribed. Corrective lenses were the first prosthetic devices to be
invented (in about 1300) but, until early in the twentieth century, they
were available only to the wealthy. The two main eye defects are hypermetropia and myopia.
PHYSICS IN FOCUS
The power of the lens
O
ptometrists and opticians describe the focusing ability of a lens
in terms of its power. The power of a simple lens is defined as:
1
Power = --- .
f
The unit of the power of a lens is the dioptre (D). For example, a
concave lens of a focal length of 25 cm has a power given by:
1
1
Power = --- = -------------- = −4.0 D.
f
– 0.25
Myopia, or short-sightedness, means that the person’s vision of nearby
objects is good, but distant objects are unclear. Parallel rays of light from
distant objects are brought to a focus in front of rather than on the
retina. The eyeball is too large for the ‘relaxed focal length’ of the eye
(see figure 2.33). This defect is not usually noticed until the eyeball
approaches its final size in adolescence. It can be overcome by placing a
concave lens in front of the eye that spreads the rays apart slightly before
they enter the eye (see figure 2.34).
42
WAVE-LIKE PROPERTIES OF LIGHT
Jac Phys 1 2E - 02 Page 43 Tuesday, October 21, 2003 1:43 PM
Figure 2.33
Myopic vision causes light rays to focus in front of the
retina.
(a) Spectacle lens
F
Figure 2.34
(b) Contact lens
F
Corrective concave lenses are used in spectacles and
contact lenses.
Figure 2.35 Contact
lenses can be a solution to eye
defects. They alter the curvature
of the front of the eye.
Sometimes myopia can be caused by excessive curvature of the cornea,
the principal focusing agent in the eye. Recent medical developments
use low temperature ultraviolet lasers to ‘shave off’ some tissue from the
front of the cornea, to flatten the curvature.
Hypermetropia, or long-sightedness, is the reverse of myopia. The lens
produces an image of near objects behind instead of on the retina
(see figure 2.36). Near objects are not clear, while distant objects are in
focus. The muscles around the lens cannot contract enough to shorten
the focal length to bring the image onto the retina. The eye needs help
to achieve this, and an extra convex lens can help (see figure 2.37).
Figure 2.36
Hypermetropia causes light rays to focus behind the
retina. This produces a ‘fuzzy’ image.
CHAPTER 2 REFRACTING LIGHT
43
Jac Phys 1 2E - 02 Page 44 Tuesday, October 21, 2003 1:43 PM
(a) Spectacles
(b) Contact lens
Figure 2.37 Convex lenses are used in spectacles and contact lenses
to correct hypermetropia.
As people age, the muscles around the lens slowly weaken and the lens
becomes thicker and less flexible. The loss in flexibility means that the
near point moves further away and the newspaper needs to be held out
to be read. The thickening of the lens, even when the muscles are
relaxed, means that distant objects start becoming fuzzy. This means that
the eye is showing signs of myopia for distant objects and hypermetropia
for near objects, which require different corrective solutions. Hence the
need for bifocal lenses, which have a convex surface at the bottom to
look down to read the newspaper and a concave surface at the top to
look up to drive the car (see figure 2.38).
lens for watching
TV and driving
lens for
reading
Figure 2.38
44
WAVE-LIKE PROPERTIES OF LIGHT
Example of a bifocal lens design
Jac Phys 1 2E - 02 Page 45 Tuesday, October 21, 2003 1:43 PM
• Light bends as it travels from one medium to
another. A measure of a medium’s capacity to
bend light is given by its refractive index.
• If light travels into a medium of a higher refractive index, the light is bent towards the normal.
If light travels into a medium of a lower refractive index, the light is bent away from the
normal. This change in direction is summarised
in Snell’s Law. Snell’s Law can be expressed as
n1 sin θ1 = n2 sin θ2.
• When light travels into a medium of a lower
refractive index, there will be an angle of incidence for which the angle of refraction is 90°.
This angle of incidence is called the critical
angle. For angles of incidence greater than the
critical angle, all the light is reflected back into
the medium. This phenomenon is called total
internal reflection.
• Two improvements on the ray model of light
are the particle model and the wave model.
Both of these models explain various light
phenomena, but they differ in their prediction
of the speed of light in a medium. The wave
model’s prediction was found to be correct.
• When refractive materials are shaped into
convex and concave lenses, both real and virtual images can be formed.
• The ray model explains the formation and properties of images in convex and concave lenses.
• The formation of images in convex and concave lenses can be mathematically modelled
1 1 1
v
with the equations --- + --- = --- and M = --- .
u v
f
u
• The formation and properties of images in
optical devices such as optical fibres, microscopes and spectacles can be modelled mathematically or by ray tracing.
QUESTIONS
Understanding
1. What is the angle of refraction in water
(n = 1.33) for an angle of incidence of 40°? If
the angle of incidence is increased by 10°, by
how much does the angle of refraction increase?
2. A ray of light enters a plastic block at an angle
of incidence of 55° with an angle of refraction
of 33°. What is the refractive index of the plastic?
3. A ray of light passes through a rectangular
glass block with a refractive index of 1.55.
The angle of incidence as the ray enters the
block is 65°. Calculate the angle of refraction
at the first face of the block, then calculate
the angle of refraction as the ray emerges on
the other side of the block. Comment on
your answers.
4. Immiscible liquids are liquids that do not mix.
Immiscible liquids will settle on top of each
other, in the order of their density, with the
densest liquid at the bottom. Some immiscible
liquids are also transparent.
(a) Calculate the angles of refraction as a ray
passes down through immiscible layers as
shown in the figure below.
light
ray
25°
air
n = 1.00
acetone
n = 1.357
glycerol
n = 1.4746
n = 1.4601
carbon
tetrachloride
glass beaker
n = 1.53
(b) If a plane mirror was placed at the bottom
of the beaker, calculate the angles of
refraction as the ray reflects back to the
surface. Comment on your answers.
5. Calculate the critical angle for light travelling
through a diamond (n = 2.5) towards the surface.
6. A ray travelling through water (n = 1.33)
approaches the surface at an angle of incidence of 55°. What will happen to the ray?
Support your answer with calculations.
7. (a) Calculate the refractive index of the glass
prism shown in the figure below so that the
light ray meets the faces at the critical angle.
Is this value of the refractive index the minimum
or maximum value for
such a reflection?
(b) Draw two parallel
rays entering the block.
How do they emerge?
8. Calculate
the
refractive
index of the plastic coating
on an optical fibre if the
critical angle for glass to
plastic is 82.0° and the refractive index of glass is 1.500.
45°
CHAPTER 2 REFRACTING LIGHT
45
CHAPTER REVIEW
SUMMARY
Jac Phys 1 2E - 02 Page 46 Tuesday, October 21, 2003 1:43 PM
9. When light passes from air into a glass block,
some light is reflected off the surface as well.
(a) Describe how the wave model of light
explains this observation.
(b) Why can’t the particle model explain this
observation?
10. Use ray tracing to determine the full description of the following objects:
(a) a 4.0 cm high object, 20 cm in front of a
convex lens with a focal length of 15 cm
(b) a 3.0 mm high object, 10 cm in front of a
convex lens with a focal length of 12 cm
(c) a 5.0 cm high object, 200 cm in front of a
convex lens with a focal length of 10 cm.
11. A bull’s eye has a diameter of about 5.0 cm.
The bull needs a clear image of the food it is
eating, so its near point is about 30 cm. Calculate the minimum focal length of the lens in
the bull’s eye.
12. What does ‘accommodation mechanism’ mean?
Give an example.
Application
13. A ray of light enters a parallel-sided glass block
(n = 1.55) at an angle of incidence of 35°, as
shown in the figure below.
(d) Some or all of the light will be reflected
from the adjacent face. Draw the subsequent path of this light, showing angles.
(e) Would your answer to (c) be different if
the initial angle of incidence was larger or
smaller than 35°C? (Hint: Assume the ray
meets the adjacent face at the critical
angle and then work back to the initial
angle of incidence.)
(f) Does the answer to (e) depend on the
refractive index of the material? Set up the
spreadsheet (shown at the base of the page)
to investigate how the initial angle of incidence varies with the refractive index if the
ray emerges from the adjacent face at 90°.
Increase the refractive index in small
steps to find the value for which the angle
of incidence becomes 90°. What is unusual
about this value for the refractive index?
14. Describe what a diver would see when looking
up at a still water surface.
15. A right-angled glass prism (n = 1.55) is placed
under water (n = 1.33), as shown in the figure
below. A ray of light enters the longest side
along the normal. What happens to the ray of
light?
air
CHAPTER REVIEW
35°
5 cm
water
n = 1.55
(a) Calculate the angle of refraction at the top
surface and the angles of incidence and
refraction as the ray emerges at the bottom
surface.
(b) If the initial ray (at the same angle)
entered the block further to the right, the
emerging ray would hit the corner of the
block. If the block is 5.0 cm wide, how far
from the edge should the initial ray enter
the block for this to occur?
(c) If the initial ray is moved even closer to the
edge, the emerging ray will meet the
adjacent side edge. What angle of incidence does the light ray make with the adjacent face? Will the ray pass out of the block
or will it be totally internally reflected?
46
WAVE-LIKE PROPERTIES OF LIGHT
45°
16. A fish looking up at the surface of the water
sees a circle, inside which it sees the ‘air
world’. Outside the circle it sees the reflection
of the ‘water world’. If the fish is 40 cm below
the surface, calculate the radius of the circle
(nwater = 1.33).
17. Light enters an optical fibre 1.0 µ m in diameter,
as shown in the figure on page 47. Some light
goes straight down the centre. Another ray is
angled, leaving the central line and meeting the
outside edge at slightly more than the critical
angle of 82° then reflects back to the central line.
(a) How much further did this
ray travel?
(b) Calculate the speed of light in
the glass and determine the
Jac Phys 1 2E - 02 Page 47 Tuesday, October 21, 2003 1:43 PM
(c) Looking from the back of the slide projector, the slide contains a letter ‘L’. What
shape will appear on the screen?
(d) The slide projector is moved closer to the
screen. The image becomes unclear.
Should the lens system be moved closer to
or further away from the slide?
1 µm
82°
light rays
optical fibre
18. (a) You are carrying out a convex lens investigation at a bench near the classroom window
and you obtain a sharp image of the window
on your screen. A teacher walks past outside
the window. What do you see on the screen?
(b) The trees outside the classroom are
unclear on the screen. What can you do to
bring the trees into focus?
19. Use the ray tracing or the formula method to
determine the magnification of an object
placed under the following two-lens microscope. The object is placed 5.2 mm from an
objective lens of focal length 5.0 mm. The eyepiece lens has a focal length of 40 mm. The
poles of the lenses are 150 mm apart.
20. Light rays are shown passing through boxes in
the figure at the base of the page. Identify the
contents of each box from the options (a)–(g)
given below. Option (b) is a mirror. All others
are solid glass. Note : There are more options
than boxes.
21. A convex lens with a focal length of 5.0 cm is
used as a magnifying glass. Calculate the size
and location of the image of text on this page
if the centre of the lens was placed:
(a) 4.0 cm above the page
(b) 3.0 cm above the page.
23. A teacher is using a slide projector but the
image on the screen is smaller than the screen.
What needs to be done to produce a clear
image on the full screen?
24. Calculate the distance the photographic film
needs to be from the centre of a camera lens
of focal length 5.0 cm in order to take a sharp
photograph of a family group 15.0 m away.
25. (a) To appear invisible you need to become
transparent. What must your refractive index
be if your movement is not to be detected?
(b) The retina of your eye is a light-absorbing
screen. What does that imply about your
own vision if you are to remain invisible?
(Hint: If you are invisible all light passes
through you.)
More of a challenge
26. Calculate the angle of deviation at a glass–air
interface for an angle of incidence of 65° and
refractive index of glass of 1.55.
27. Calculate the sideways deflection as a ray of
light goes through a parallel-sided plastic
block (n = 1.4) with sides 5.0 cm apart, as in
the figure below.
30°
22. A 35 mm slide is placed in a slide projector. A
sharp image is produced on a screen 4.0 m away.
The focal length of the lens system is 5.0 cm.
(a) How far is the slide from the centre of the
lens?
(b) What is the size of the image?
(i)
(a)
(ii)
(b)
(iii)
(c)
n = 1.4
(iv)
(d)
(v)
(e)
5 cm
(vi)
(f)
(g)
CHAPTER 2 REFRACTING LIGHT
47
CHAPTER REVIEW
time delay between the two rays after one
internal reflection. Do you think this could
be a problem in an optical fibre? If so,
when? How could the problem be overcome?
Jac Phys 1 2E - 02 Page 48 Tuesday, October 21, 2003 1:43 PM
28. Calculate the angle of deviation as the light ray
goes through the triangular prism shown in
the figure below.
n = 1.5
60°
?
40°
29. A ray of light enters a glass sphere (n = 1.5), as
in the figure below. What happens to the ray?
glass sphere
30°
CHAPTER REVIEW
centre of
circle
30. (a) Model a heat mirage on a spreadsheet (see
the example at the base of the page) as a
set of air layers whose refractive indices
each decrease by 0.000 02, starting at
1.0003. As the light ray travels through
each layer, the angle of refraction at the
layer’s top surface becomes the angle of
incidence at its bottom surface. Total
internal reflection will occur when the
spreadsheet tries to find an angle whose
sine is greater than 1 and displays #NUM.
(b) If the initial angle of incidence is 89°, in
what layer does total internal reflection
occur? Experiment with different initial
angles of incidence, initial refractive
indices and different rates of change in
the refractive index from layer to layer.
31. In the diagram of Fata Morgana (figure 2.12,
page 30), describe how the temperature and
refractive index vary with height.
32. A hair 1.0 mm in diameter is looked at
through a microscope. The objective lens has a
focal length of 3.0 mm, the eyepiece has a
48
WAVE-LIKE PROPERTIES OF LIGHT
focal length of 25 mm and the lenses are
18.0 cm apart. The hair is placed just outside
the focus of the objective lens at 3.5 mm.
(a) Calculate the location of the image produced by this lens. Using this image as the
object for the eyepiece lens, calculate the
location of the final image.
(b) Calculate the size of the image produced
by the objective lens, then calculate the
size of the final image produced by the
microscope.
33. A screen captures the image of a small globe
from a convex lens. The screen is then taken
away. Where should you place your eye and
where should you look to see the image?
34. A hollow 15 cm tube contains a convex lens of
unknown focal length but greater than 15 cm.
However, you do not know where the lens is in
the tube. You have a light source, a screen and
a ruler. How would you determine the focal
length of the lens and its position in the tube?
35. Ray tracing can be used to locate images produced by a concave lens. The only difference
from ray tracing with convex lenses is that the
principal focus is on the opposite side of the
concave lens from the object. Draw ray tracing
diagrams for objects at different distances and
describe the properties of images produced by
a concave lens.
36. The closer an object is to your eye, the more
detail you can see. The object casts a larger
visual angle at the eye. But this is only true up
to your near point, which is the limit of your
eye’s focusing ability. This is an important consideration in the design of magnifying glasses
and microscopes. For the most detailed image,
the device should produce an image located at
the user’s near point.
(a) A slide viewer is a small magnifying glass in
a frame which holds a slide at a distance of
5.0 cm from the magnifying lens. What
should be the focal length of the lens if
the average near point is 25 cm and in
normal use the slide viewer is held up to
the eye?
(b) A person with a near point of 50 cm asks
you to design a magnifying glass
to enable the person to read a
book. What focal length convex
lens
should
you
use?
(Remember to estimate object
and image distances.)