Download PHYS 202 Notes, Week 10

PHYS 202 Notes, Week 10
Greg Christian
March 28 & 30, 2016
Last updated: 03/31/2016 at 12:28:48
This week we learn about optical instruments, interference, and thin
films.
Optical Instruments
Important points
Cameras
A camera is a device that is able to capture images and store them
either on film or in a digital medium. Figure 1 shows a diagram of how
a camera operates. The light coming off the object to be photographed
passes through a lens, creating a real and inverted image on the film (for
a traditional camera) or CCD array (for a digital camera). The image
is then un-flipped by developing the film or with digital processing
software, to create what you see when looking at photos.
In designing a camera, there are a couple of things which can be
altered to affect the image that forms. One is the intensity of light
reaching the film or CCD array, which can be adjusted by changing
the size of the aperture surrounding the lens. The size of this aperture
is usually given in terms of something called the f -number,
Focal length
f
f − number =
= ,
Aperture diameter
D
(1)
where f is the focal length and D is the aperture diameter.
• Cameras create a inverted, real image
on film or CCD.
• Eyes use biological fluids to create a
lens that focuses light onto the retina.
• Magnifiers change the size of images
using lenses; so do microscopes (2
lenses) and telescopes.
Important equations
• Angular magnification of a magnifier
M = θ θ0
= (25 cm) s
= (1/ f − 1/s0 ) × 25 cm
• Microscope
M = m1 M2
=
(25 cm) s10
f1 f2
Figure 1: Operation of a camera.
As you probably know, cameras can take images of objects at many
different distances. This is accomplished by moving the lens in or out.
This in turn adjusts the image distance. The goal is to place the lens
where the image distance matches the recording medium. Then the
phys 202 notes, week 10
camera is “in focus”. Most modern cameras do this automatically, but
older and/or high-end cameras still rely on manual focusing by the
photographer.
Eyes
Figure 2: Operation of an eye.
Figure 2 shows the operation of the human eye. Basically, the eye
works by forming a lens made up of biological material, focusing incoming light onto the retina where rods and cones sense the image and
transmit it to the brain. The parts of the eye forming the lens are (in
order):
• The aqueous humor, index of refraction n = 1.336.
• The crystalline lens, index of refraction n = 1.437.
• The vitreous humor, index of refraction n = 1.336.
In order to form a sharp image, the image point of the eye needs to
be exactly on the retina surface. Given that the retina is in a fixed location, the image distance, s0 is always fixed. In order to accommodate
different object distances, s, the eye changes its effective focal length. It
does that by contracting or relaxing a muscle called the ciliary muscle,
which has the effect of changing the radius of curvature, and thus the
focal length of the eye.
Contracting the ciliary muscle serves to decrease the radius of curvature, and thus focal lngth, of the eye. This allows up-close objects to be
in focus. Then the ciliary muscle is relaxed, the focal length increases,
creating focus on far-away objects. This process is called accomodation.
As people age, the ciliary muscle loses its ability to work, decreasing
the range in which the focal length can be changed. This is why eventually nearly everyone starts to have difficulty with seeing up-close
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objects without reading glasses. The term near point describes the closest distance upon which the eye can focus. For the average person this
decreases from about 7 cm at age 10 to 200 cm at age 60.
The eye also needs to accommodate different intensities of light. It
does this by dilating or contracting the pupils, to let more or less light
in.
Vision problems
Many people have eyes which are either too long or too short. This
causes light to be focused either in front of or behind the retina. The
first case is called myopia, or nearsightedness, because it causes images
of far-away objects to appear out of focus. The second is called hyperopia, or farsightedness, because it causes close-up images to be out of
focus. Drawings of these are shown in Figure 3.
Each of these vision problems can be fixed using lenses. In the case
of hyperiopia, we can use a converging lens to focus the image onto
the retina (Figure 4), thus creating a clear image. In the case of myopia,
we use a diverging lens to achieve the same result (Figure 5).
Figure 3: Myopia and hyperopia.
Figure 4: Hyperopia correction.
Magnifiers
Another optical instrument is a magnifying glass, or simply a magnifier.
This instrument uses a converging lens to form a virtual image that is
larger and farther away than the original. Magnifiers are typically
analyzed in terms of the angular size, or the total angle subtended by
the object, as shown in the left of Figure 6.
When dealing with magnifiers, we are concerned with the angular
magnification, or the change in angular size of the object. This is equal
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Figure 5: Myopia correction.
to
θ0
.
(2)
θ
For a magnifier with focal length f , we can calculate the angular
magnification as
M=
(25 cm)
s
1
1
=
− 0 × 25 cm,
f
s
M=
(3)
(4)
where
• s is the object distance; and
• We assume that the near point of the eye is 25 cm.
Figure 6: Operation of a magnifier.
Microscopes
To look at very small objects, a simple magnifier isn’t enough. Here
we need a microscope. The operation of a microscope is shown in
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Figure 7.
A microscope consists of two lenses:
1. The objective lens, which has focal length f 1 .
2. The eyepiece, which has focal length f 2 .
It works in the following way:
1. The objective lens creates a real, inverted image inside the focal point
F2 of the eyepiece.
2. The eyepiece uses the image I as its object and creates an enlarged,
virtual image I 0 which is still inverted.
The most important quantity for a microscope is the overall angular
magnification. This is given by the following equation,
M = m1 M2
=
(25
cm) s10
,
f1 f2
(5)
(6)
Figure 7: Operation of a microscope.
where
• m1 is the lateral magnification of the objective lens (focal length f 1 ).
• M2 is the angular magnification of the eyepiece (focal length f 2 ).
Telescopes
A telescope allows viewing of objects that are large, but very far away.
One type of telescope is called the astronomical telescope, and it has a
very similar design to a microscope, namely a combination of objective
lens and eyepiece. An example is shown in Figure 8.
Figure 8: Operation of a astronomical
telescope.
Because the object to be viewed is so far away, the first image is
formed nearly at the second focal point of the objective lens ( f 1 ). This
image is then the object for the eyepiece. Ideally, we want this to form
a final image at infinity, which means that the first image needs to be
located at the focal point of the eyepiece ( f 2 ). This means that the total
length of the telescope will be equal to the sum of the focal lengths,
f1 + f2 .
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The overall angular magnification for a telescope can be worked
out from the geometry shown in Figure 8. The result is that it’s simply
proportional the the ratio of focal lengths (with a negative sign),
M=−
f1
.
f2
(7)
Note that due to the negative sign, the telescope forms an inverted
image. This can be corrected using a device called prism binoculars,
shown in Figure 9. These are placed between the objective and the
eyepiece and serve to flip the image (making it upright) by way of
four internal reflections.
Figure 9: Prism binoculars, to make the
image formed by a telescope upright.
For practical reasons, it’s often more feasible to construct telescopes
out of reflecting mirrors rather than lenses. These devices are called
reflecting telescopes. Some schemes for constructing reflecting telescopes
are shown in Figure 10.
In one scheme, (part a in the figure), there is a “cage” inside the telescope which receives the reflected light, typically in a camera. This design is most practical for very large telescopes because the cage blocks
incoming light rays.
Figure 10: Examples of reflecting telescopes.
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Other designs (parts b and c in the figure) use a secondary mirror to
reflect light either along the direction of incoming light (which requires
a hole in the primary mirror), or out the side of the telescope (part c).
Optical Interference
So far, we’ve been working with the ray model of light, which treats
light as something traveling along a straight line path. Now we’ll move
into talking about some phenomena which rely on the wave nature of
light.Recall that light is a form of electromagnetic wave, consisting of
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oscillating E and B fields.
Interference is the phenomena where two or more waves overlap in
space. Related to this is something called the Principle of Superposition:
• Constructive interference occurs
when waves are in phase.
• Destructive interference occurs when
waves are 1/2 cycle out of phase.
• Double slits and thin films can both
form interference patterns.
Important equations
• Interference
When two or more waves overlap, the resulting displacement at any
point and any time is found by adding the instantaneous displacements
that would be produced at that place and time if the individual waves
were each present alone.
r2 − r1 = mλ (constructive)
r2 − r1 = (m + 1/2) λ (destructive)
• Double-slit experiment
In other words, you can basically just sum up the individual displacements to get the total result. In the case of light, what we mean by “displacement” is the magnitude of the electric or magnetic field forming
the wave.
One concept that we need to introduce is that of phase. Phase basically tells whether or not the peaks and troughs making up a wave
are lined up or not. If two waves are “in phase”, the displacements
are in step with each other. If they are “one cycle out of phase”, the
displacements are exactly counter-aligned: the peak of one wave lines
up with the trough of the other. Examples are shown in Figure 11.
Depending on the phase of two waves, they can interfere either constructively or destructively. Constructive interference (or “reinforcement”) happens when the waves are in phase, resulting in an increase
in amplitude. The condition for constructive interference to happen
is that the path difference from the two sources must line up with an
integer number of wavelengths:
r2 − r1 = mλ (m = 0, ± 1, ± 2, ± 3, . . .),
Important points
d sin θ = mλ (bright)
d sin θ = (m + 1/2) λ (dark)
ym = Rmλ d
(bright)
ym = R (m + 1/2) λ d (dark)
• Thin film interference
2t = mλ (constructive)
2t = (m + 1/2) λ (destructive)
λ = λ0 /nfilm
(8)
with
• The resulting amplitude is the sum of the amplitudes of wave 1 and
wave 2;
• r1 and r2 are the distances from the wave source to the point of
overlap; and
• Each of the waves has wavelength λ.
Figure 11: Example of in- and out-ofphase waves.
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Figure 12: Wave interference.
An example of this is shown in part (b) of Figure 12.
The opposite phenomena, destructive interference (or “cancellation”), happens when the waves are exactly a half cycle out of phase.
This results in a reduced amplitude. For destructive interference to
happen, the path differences align with a half integer multiple of the
wavelength,
r2 − r1 = (m + 1/2) λ (m = 0, ± 1, ± 2, ± 3, . . .),
(9)
with the following being true:
• If the two waves have the same amplitude, they completely cancel
each other.
• r1 and r2 are the same as in Eq. (8).
An example of this is shown in part (c) of Figure 12.
Double-Slit Interference
One of the first experiments showing interference of light waves was
the double-slit experiment performed in 1800 by Thomas Young.1 A
graphical representation of Young’s experiment is shown in Figure 13.
The experiment consists of the following steps:
1. Monochromatic light passes through a single slit S0 , creating a cylindrical wave front on the other side (as a consequence of Huygen’s
principle).
2. The wave front travels to two separate slits S1 and S2 which are
at different vertical distances from S0 . This creates two separate
cylindrical wave fronts on the other side.
3. These two wavefronts travel to a screen where they are observed.
Often referred to as “Young’s DoubleSlit Experiment”.
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Figure 13: Thomas Young’s double-slit
experiment.
The result, observed on the screen in the last step, is a series of
bright and dark bands. The bright bands occur when the light from
S1 and S2 traveled some integer multiple of the wavelength, satisfying
Eq. (8) and giving constructive interference. Likewise, the dark bands
occur when the distances satisfy Eq. (9) (half integer multiple of λ) and
give destructive interference.
The path lengths in the bottom of Figure 13 can be analyzed to figure out the angles for which constructive and destructive interference
occur. From part (c) in the figure, the difference in path lengths is
given by
r2 − r1 = d sin θ,
(10)
where d is the distance from the slits to the screen. Combining with
Eq. (8), we can figure out the angles for which constructive interference occurs. These are given by
d sin θ = mλ (m = 0, ± 1, ± 2, ± 3, . . .).
(11)
Similarly, we can work out the location of destructive interference
from Eqns. (10) and (9):
d sin θ = (m + 1/2) λ (m = 0, ± 1, ± 2, ± 3, . . .).
(12)
Figure 14: Fringes deom the double slit
experiment.
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The bands that form on the screen are often called fringes. A picture
of what these look like is shown in Figure 14. Sometimes it’s useful
to have an equation for the vertical positions of these fringes, ym . This
can be worked out from Eq. (11) and the geometry of the problem,
mλ
(m = 0, ± 1, ± 2, ± 3, . . .),
(13)
d
where R is the distance from the slits to the screen and d is the distance
between the slits.
Note that Eq. (13) gives the position of the bright fringes. Similarly,
the position of the dark fringes is given by
ym = R
ym = R
(m + 1/2) λ
(m = 0, ± 1, ± 2, ± 3, . . .),
d
(14)
Thin Film Interference
Another way to create interference patterns is to pass light through a
thin film of refracting material, as shown in the top of Figure 15. When
this happens, part of the light is reflected directly from the top of the
film to the observer’s eye, while the rest is refracted; reflected at the
bottom of the film; and then refracted to the observer. This process
creates a path difference resulting in interference.
Since the path length is fixed, we get either constructive or destructive interference depending on the wavelength of the light. The formation of constructive or destructive interference is described by the
following equations:
Constructive
2t = mλ (m = 0, ± 1, ± 2, ± 3, . . .)
(15)
Destructive
2t = (m + 1/2) λ (m = 0, ± 1, ± 2, ± 3, . . .),
(16)
where t is the thickness of the film. Note that the wavelength λ in
Eqns. (15) and (16) is the wavelength in the film medium. This is
related to the incoming wavelength λ0 by
λ = λ0 /nfilm .
(17)
This is why rainbow-like patterns form on a thin film, such as the
oil slick shown in Figure 15(b). In this case, the incoming light is made
up of many different colors (each a different wavelength). Some of
the colors have the right wavelength to give constructive interference;
others give destructive interference. Those colors interfering constructively are the ones that we see as the bright rainbow spots.
In the example above, we discussed thin film interference of light
of all colors. Figure 16 shows another situation where thin film interference can occur on monochromatic (single color/wavelength) light. It
Figure 15: Thin film interference.
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involved stacking two refracting materials close together with an angled gap. The thickness of the gap at any point is labeled t. The series
of events leading to interference patterns is:
1. Some of the incoming light is reflected back to the observer by the
top surface.
2. The rest travels through thickness t before reflecting off the bottom
surface.
3. In the second reflection, the light undergoes a half-cycle phase
shift. This happens whenever light reflects off a surface with nb >
na .
4. The reflected light travels back to the observer, with path length
difference 2t and phase shift of 1/2.
5. Interference patterns occur, but the patterns are the opposite of what
we would expect if there wasn’t a phase change. In other words,
Eqns. (15) and (16) are interchanged in terms of whether they refer
to constructive or destructive interference.
Note that in general, whether Eqns. (15) and (16) apply to constructive or destructive interference depends on whether or not the first and
second reflections induce a phase shift. To summarize,
• Neither reflection induces a phase shift
– Eq. (15) constructive
– Eq. (16) destructive
• Only one reflection induces a phase shift
– Eq. (16) constructive
– Eq. (15) destructive
• Both reflections induce a phase shift
– Eq. (15) constructive
– Eq. (16) destructive.
Figure 16:
Thin film reflection of
monochromatic light due to a gap between refracting materials.