Download Chapter 32 Light: Reflection and Refraction

Chapter 33
Lenses and Optical
Instruments
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33-5 Cameras: Film and Digital
Basic parts of a camera:
• Lens
• Light-tight box
• Shutter
• Film or electronic
sensor
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33-5 Cameras: Film and Digital
Camera adjustments:
• Shutter speed: controls the amount of time
light enters the camera. A faster shutter speed
makes a sharper picture.
• f-stop: controls the maximum opening of the
shutter. This allows the right amount of light to
enter to properly expose the film, and must be
adjusted for external light conditions.
• Focusing: this adjusts the position of the lens
so that the image is positioned on the film.
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33-5 Cameras: Film and Digital
Example 33-8: Camera focus.
How far must a 50.0-mm-focallength camera lens be moved
from its infinity setting to sharply
focus an object 3.00 m away?
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33-6 The Human Eye; Corrective Lenses
The human eye resembles a camera in its
basic functioning, with an adjustable lens, the
iris, and the retina.
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33-6 The Human Eye; Corrective Lenses
Figure 33-26 goes
here.
Most of the refraction is
done at the surface of
the cornea; the lens
makes small
adjustments to focus at
different distances.
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33-6 The Human Eye; Corrective Lenses
Near point: closest distance at which eye can
focus clearly. Normal is about 25 cm.
Far point: farthest distance at which object can
be seen clearly. Normal is at infinity.
Nearsightedness: far point is too close.
Farsightedness: near point is too far away.
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33-6 The Human Eye; Corrective Lenses
Nearsightedness can be corrected with a
diverging lens.
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33-6 The Human Eye; Corrective Lenses
And farsightedness with a diverging lens.
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33-6 The Human Eye; Corrective Lenses
Example 33-12: Farsighted eye.
Sue is farsighted with a near point of 100 cm.
Reading glasses must have what lens power
so that she can read a newspaper at a
distance of 25 cm? Assume the lens is very
close to the eye.
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33-6 The Human Eye; Corrective Lenses
Example 33-13: Nearsighted
eye.
A nearsighted eye has near
and far points of 12 cm and
17 cm, respectively. (a)
What lens power is needed
for this person to see
distant objects clearly, and
(b) what then will be the
near point? Assume that the
lens is 2.0 cm from the eye
(typical for eyeglasses).
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33-6 The Human Eye; Corrective Lenses
Vision is blurry under water because light rays
are bent much less than they would be if
entering the eye from air. This can be avoided by
wearing goggles.
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33-7 Magnifying Glass
A magnifying glass (simple magnifier) is a
converging lens. It allows us to focus on
objects closer than the near point, so that
they make a larger, and therefore clearer,
image on the retina.
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33-7 Magnifying Glass
The power of a magnifying glass is described
by its angular magnification:
If the eye is relaxed (N is the near point distance
and f the focal length):
If the eye is focused at the near point:
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33-10 Aberrations of Lenses and Mirrors
Spherical aberration: rays far from the lens
axis do not focus at the focal point.
Solutions: compound-lens systems; use
only central part of lens.
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33-10 Aberrations of Lenses and Mirrors
Distortion: caused by variation in
magnification with distance from the lens.
Barrel and pincushion distortion:
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33-10 Aberrations of Lenses and Mirrors
Chromatic aberration: light of different
wavelengths has different indices of refraction
and focuses at different points.
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33-10 Aberrations of Lenses and Mirrors
Solution: Achromatic doublet, made of lenses of
two different materials
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Summary of Chapter 33
• Lens uses refraction to form real or
virtual image.
• Converging lens: rays converge at
focal point.
• Diverging lens: rays appear to diverge
from focal point.
• Power is given in diopters (m-1):
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Summary of Chapter 33
• Thin lens equation:
• Magnification:
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Summary of Chapter 33
• Camera focuses image on film or
electronic sensor; lens can be moved and
size of opening adjusted (f-stop).
• Human eye also makes adjustments, by
changing shape of lens and size of pupil.
• Nearsighted eye is corrected by diverging
lens.
• Farsighted eye is corrected by
converging lens.
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Summary of Chapter 33
• Magnification of simple magnifier:
• Telescope: objective lens or mirror
plus eyepiece lens. Magnification:
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Chapter 34
The Wave Nature of Light;
Interference
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Units of Chapter 34
• Waves versus Particles; Huygens’ Principle
and Diffraction
• Huygens’ Principle and the Law of Refraction
• Interference – Young’s Double-Slit Experiment
• Intensity in the Double-Slit Interference
Pattern
• Interference in Thin Films
• Michelson Interferometer
• Luminous Intensity
Copyright © 2009 Pearson Education, Inc.
34-1 Waves versus Particles;
Huygens’ Principle and Diffraction
Huygens’ principle:
every point on a wave
front acts as a point
source; the wave front
as it develops is
tangent to all the
wavelets.
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34-1 Waves versus Particles;
Huygens’ Principle and Diffraction
Huygens’ principle is consistent with
diffraction:
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34-2 Huygens’ Principle and the Law
of Refraction
Huygens’ principle can also explain the law of
refraction.
As the wavelets propagate from each point,
they propagate more slowly in the medium of
higher index of refraction.
This leads to a bend in the wave front and
therefore in the ray.
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34-2 Huygens’ Principle and the Law
of Refraction
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34-2 Huygens’ Principle and the Law
of Refraction
The frequency of the light does not change, but
the wavelength does as it travels into a new
medium:
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34-2 Huygens’ Principle and the Law
of Refraction
Highway mirages are due to a gradually
changing index of refraction in heated air.
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34-3 Interference – Young’s DoubleSlit Experiment
If light is a wave, interference effects will be
seen, where one part of a wave front can
interact with another part.
One way to study this is to do a double-slit
experiment:
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34-3 Interference – Young’s DoubleSlit Experiment
If light is a wave,
there should be
an interference
pattern.
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ConcepTest 34.1
Superposition
If waves A and B are
superposed (that is, their
amplitudes are added) the
resultant wave is
1)
2)
3)
4)
ConcepTest 34.1
Superposition
If waves A and B are
superposed (that is, their
amplitudes are added) the
resultant wave is
1)
The amplitudes of
waves A and B have to
2)
be added at each
point!
3)
4)
34-3 Interference – Young’s DoubleSlit Experiment
The interference occurs because each point on
the screen is not the same distance from both
slits. Depending on the path length difference,
the wave can interfere constructively (bright
spot) or destructively (dark spot).
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34-3 Interference – Young’s DoubleSlit Experiment
We can use geometry to find the conditions for
constructive and destructive interference:
and
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34-3 Interference – Young’s DoubleSlit Experiment
Between the maxima and the minima, the
interference varies smoothly.
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34-3 Interference – Young’s DoubleSlit Experiment
Conceptual Example 34-1: Interference
pattern lines.
(a) Will there be an infinite number of
points on the viewing screen where
constructive and destructive interference
occur, or only a finite number of points?
(b) Are neighboring points of constructive
interference uniformly spaced, or is the
spacing between neighboring points of
constructive interference not uniform?
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34-3 Interference – Young’s DoubleSlit Experiment
Example 34-2: Line spacing for double-slit
interference.
A screen containing two slits 0.100 mm apart is 1.20
m from the viewing screen. Light of wavelength λ =
500 nm falls on the slits from a distant source.
Approximately how far apart will adjacent bright
interference fringes be on the screen?
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34-3 Interference – Young’s DoubleSlit Experiment
Conceptual Example 34-3: Changing the
wavelength.
(a) What happens to the interference
pattern in the previous example if the
incident light (500 nm) is replaced by
light of wavelength 700 nm? (b) What
happens instead if the wavelength stays
at 500 nm but the slits are moved farther
apart?
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34-3 Interference – Young’s DoubleSlit Experiment
Since the position of the maxima (except the
central one) depends on wavelength, the firstand higher-order fringes contain a spectrum of
colors.
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34-4 Intensity in the Double-Slit
Interference Pattern
The electric fields at
the point P from the
two slits are given by
.
where
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34-4 Intensity in the Double-Slit
Interference Pattern
The two waves
can be added
using phasors, to
take the phase
difference into
account:
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34-4 Intensity in the Double-Slit
Interference Pattern
The time-averaged intensity is proportional
to the square of the field:
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34-4 Intensity in the Double-Slit
Interference Pattern
This plot shows the intensity as a
function of angle.
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34-4 Intensity in the Double-Slit
Interference Pattern
Example 34-5: Antenna intensity.
Two radio antennas are located close to each other,
separated by a distance d. The antennas radiate in phase
with each other, emitting waves of intensity I0 at
wavelength λ. (a) Calculate the net intensity as a function
of θ for points very far from the antennas. (b) For d = λ,
determine I and find in which directions I is a maximum
and a minimum. (c) Repeat part (b) when d = λ/2.
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ConcepTest 34.3a
Double Slits I
In a double-slit experiment,
1) spreads out
when the wavelength of the light
2) stays the same
is increased, the interference
3) shrinks together
pattern
4) disappears
ConcepTest 34.3a
Double Slits I
In a double-slit experiment,
1) spreads out
when the wavelength of the light
2) stays the same
is increased, the interference
3) shrinks together
pattern
d sin  = m
If  is increased and d does
not change, then  must
increase, so the pattern
spreads out.
4) disappears
34-5 Interference in Thin Films
Another way path lengths can differ, and
waves interfere, is if they travel through
different media. If there is a very thin film of
material – a few wavelengths thick – light will
reflect from both the bottom and the top of
the layer, causing interference. This can be
seen in soap bubbles and oil slicks.
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34-5 Interference in Thin Films
The wavelength of the
light will be different
in the oil and the air,
and the reflections at
points A and B may or
may not involve
phase changes.
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34-5 Interference in Thin Films
A similar effect takes place when a shallowly
curved piece of glass is placed on a flat one.
When viewed from above, concentric circles
appear that are called Newton’s rings.
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34-5 Interference in Thin Films
A beam of light reflected
by a material with index
of refraction greater than
that of the material in
which it is traveling,
changes phase by 180°
or ½ cycle.
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34-5 Interference in Thin Films
Example 34-6: Thin film of air,
wedge-shaped.
A very fine wire 7.35 x 10-3 mm
in diameter is placed between
two flat glass plates. Light
whose wavelength in air is 600
nm falls (and is viewed)
perpendicular to the plates
and a series of bright and dark
bands is seen. How many light
and dark bands will there be in
this case? Will the area next to
the wire be bright or dark?
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34-5 Interference in Thin Films
Example 34-7:
Thickness of soap
bubble skin.
A soap bubble appears
green (λ = 540 nm) at
the point on its front
surface nearest the
viewer. What is the
smallest thickness the
soap bubble film could
have? Assume n = 1.35.
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34-5 Interference in Thin Films
Problem Solving: Interference
1. Interference occurs when two or more waves
arrive simultaneously at the same point in
space.
2. Constructive interference occurs when the
waves are in phase.
3. Destructive interference occurs when the
waves are out of phase.
4. An extra half-wavelength shift occurs when
light reflects from a medium with higher
refractive index.
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34-5 Interference in Thin Films
Example 34-8:
Nonreflective coating.
What is the thickness of
an optical coating of
MgF2 whose index of
refraction is n = 1.38 and
which is designed to
eliminate reflected light
at wavelengths (in air)
around 550 nm when
incident normally on
glass for which n = 1.50?
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ConcepTest 34.6d
Two identical microscope slides in
air illuminated with light from a
laser are creating an interference
pattern. The space between the
slides is now filled with water (n =
1.33). What happens to the
interference fringes?
Parallel Slides IV
1) spaced farther apart
2) spaced closer together
3) no change
ray 1 ray 2
ray 3
t
ConcepTest 34.6d
Two identical microscope slides in
air illuminated with light from a
laser are creating an interference
pattern. The space between the
slides is now filled with water
(n=1.33). What happens to the
interference fringes?
Parallel Slides IV
1) spaced farther apart
2) spaced closer together
3) no change
The path difference between ray 2 and
ray 3 is 2t (in addition, ray 3
experiences a phase change of 180°).
Thus, the dark fringes will occur for:
2t = mwater where water = air/n
Thus, the water has decreased the
wavelength of the light.
ray 1 ray 2
ray 3
t