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Eðlisfræði 2, vor 2007
36. Diffraction
Assignment is due at 2:00am on Wednesday, January 17, 2007
Credit for problems submitted late will decrease to 0% after the deadline has passed.
The wrong answer penalty is 2% per part. Multiple choice questions are penalized as described in the online help.
The unopened hint bonus is 2% per part.
You are allowed 4 attempts per answer.
Diffraction from a single slit: Introduction and increasingly difficult problems with some interesting applications
Understanding Fraunhofer Diffraction
Learning Goal: To understand the derivations of, and be able to use, the equations for Fraunhofer diffraction.
Diffraction is a general term for interference effects related to edges or apertures. Diffraction is more familiar in
waves with longer wavlengths than those of light. For example, diffraction is what causes sound to bend around
corners or spread as it passes through a doorway. Water waves spread as they pass between rocks near a rugged coast
because of diffraction. Two different regimes for diffraction are usually identified: Fresnel and Fraunhofer.
Fresnel diffraction is the regime in which the diffracted waves are observed close (as compared to the size of the
object causing the diffraction) to the place where they are diffracted. Fresnel diffraction is usually very complicated
to work with. The other regime, Fraunhofer diffraction, is much easier to deal with. Fraunhofer diffraction applies
to situations in which the diffracted waves are observed far from the point of diffraction. This allows a number of
simplifying approximations to be used, reducing diffraction to a very manageable problem.
An important case of Fraunhofer diffraction is the pattern formed by light shining through a thin slit onto a distant
screen (see the figure).
Notice that if the light from the top of the slit and the light from the bottom of
the slit arrive at a point on the distant screen with a phase difference of ,
then the electric field vectors of the light from each part of the slit will cancel
completely, resulting in a dark fringe. To understand this phenomenon, picture
a phasor diagram for this scenerio (as show in the figure).
A phasor diagram consists of vectors (phasors) with magnitude proportional to
the magnitude of the electric field of light from a certain point in the slit. The
angle of each vector is equal to the phase of the light from that point. These
vectors are added together, and the resultant vector gives the net electric field due to light from all points in the slit.
In the situation described above, since the magnitude of the electric field vectors is the same for light from any part
of the slit and the angle of the phasors changes continuously from to , the phasors will make a complete circle,
starting and ending at the origin. The distance from the origin to the endpoint of the phasor path (also the origin) is
zero, and so the magnitude of the electric field at point is zero.
Part A
One reason that Fraunhofer diffraction is relatively easy to deal with is that the large distance from the slit to the
screen means that the light paths will be essentially parallel.
Therefore, the distance marked in the figure is the entire path-length
difference between light from the top of the slit and light from the bottom of
the slit. What is the value of ?
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Hint A.1 A useful triangle
Hint not displayed
Express your answer in terms of the slit width
ANSWER:
and the angle
shown in the figure.
=
Part B
As described in the problem introduction, a criterion for a dark band to appear at point is that the phase
difference between light arriving at point from the top of the slit and light arriving at point from the bottom
of the slit equal . What length of path difference will give a phase difference of ?
Express your answer in terms of the wavelength
ANSWER:
.
=
Combining your answers from Parts A and B gives the criterion for a dark band in the diffraction pattern as
.
Part C
Consider the phasor diagram from the introduction. The magnitude of the electric field at a point will equal zero as
long as the endpoint for the phasor diagram is the origin. Thus, a point with a phasor diagram that goes around a
circle twice, for example, ending at the origin, will be another location for a dark band. This idea can be used to
modify the equation for the location of a dark band by introducing a variable :
.
What is the complete set of values of
ANSWER:
for which this equation gives criteria for dark bands?
, , ,
, , , ,
,
, ,
,
,
, ,
any rational number
The value
corresponds to
, which is the center of the diffraction pattern. The center of the diffraction
pattern is a bright band. To see why, notice that if the phase difference from top to bottom is zero, then the
phasor diagram will just be a straight line segment pointing away from the origin. This gives the maximum
possible intensity in the diffraction pattern.
Part D
What are the angles for the two dark bands closest to the central maximum.
Express your answers in terms of
and . Separate the two angles with a comma.
ANSWER:
Part E
The equation for the angles to dark bands is valid for any angle from
to . In practice, the bright bands at
large angles are usually so dim that the diffraction pattern appearing on a screen is invisible for such angles. For
small angles, it is easy to find the distance from the center of the diffraction pattern to the dark band on the
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screen corresponding to a particular value of .
For small angles,
. Since
, the small-angle approximation yields
. By solving the
dark-band criterion, you obtain
. Setting the two expressions for
equal gives the formula for the
position (i.e., distance from the center of the diffraction pattern) of dark bands:
,
or equivalently,
.
Assuming that the angle between them is small, what is the distance
center of the diffraction pattern?
Express your answer in terms of
ANSWER:
between the two dark bands closest to the
, , and .
=
Part F
Suppose that light from a laser with wavelength 633
is incident on a thin slit of width 0.500 . If the
diffracted light projects onto a screen at distance 1.50 , what is the distance from the center of the diffraction
pattern to the dark band with
?
Express your answer in millimeters to two significant figures.
ANSWER:
= 3.80
Resolving Pixels on a Computer Screen
A standard
-inch ( -meter) computer monitor is
pixels wide and
pixels tall. Each pixel is a square
approximately micrometers on each side. Up close, you can see the individual pixels, but from a distance they
appear to blend together and form the image on the screen.
Part A
If the maximum distance between the screen and your eyes at which you can just barely resolve two adjacent
pixels is
meters, what is the effective diameter of your pupil? Assume that the resolvability is
diffraction-limited. Furthermore, use
nanometers as a characteristic optical wavelength.
Hint A.1 Rayleigh's criterion
Hint not displayed
Express your answer in millimeters to three significant figures.
ANSWER:
= Answer not displayed
Part B
Part not displayed
Why You Can Still Receive AM Radio in a City
When radio waves try to pass through a city, they encounter thin vertical slits: the separations between the
buildings. This causes the radio waves to diffract. In this problem, you will see how different wavelengths refract as
they pass through a city and relate this to reception for radios and cell phones. You will use the angle from the
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center of the central intensity maximum to the first intensity minimum as a measure of the width of the central
maximum (where nearly all of the diffracted energy is found).
Consider radio waves of wavelength entering a city where the buildings have an average separation of .
Part A
Find the angle to the first minimum from the center of the central maximum.
Hint A.1 The equation for intensity
The equation for intensity as a function of angle for diffraction from a slit is
.
Part A.2 A criterion for the first minimum
Which of the following is a correct (exact) criterion for the location of the first intensity minimum of the
diffraction pattern?
Hint A.2.a Finding the first intensity minimum
The equation for intensity has a minimum value of zero. Therefore, to find the first intensity minimum, you
must look for the smallest value of that gives
. You are therefore looking for the smallest angle at which
the numerator of the squared term is zero, as long as the denominator of the squared term is not zero. If both
numerator and denominator equal zero, then you will have to evaluate the limit with L'Hopital's rule to find the
value of the intensity function.
ANSWER:
Now solve for in this criterion to obtain the expression that you need.
Express your answer in terms of
ANSWER:
and .
=
Assume that the average spacing between buildings is
.
Part B
What is the angle
to the first minimum for an FM radio station with a frequency of 101
?
Part B.1 Find the wavelength
Find the wavelength for a radio wave with a frequency of 101
. Recall that radio waves are electromagnetic
waves, and therefore travel at the speed of light,
meters per second.
Hint B.1.a Relating wavelength and frequency
Hint not displayed
Express your answer in meters, to three significant figures.
ANSWER:
= 2.97
Express your answer numerically in degrees to three significant figures. Note: Do not write
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your answer in terms of trignometric functions. Evaluate any such functions in your
working.
ANSWER:
= 8.54
Part C
What is the angle
for a cellular phone that uses radiowaves with a frequency of 900
?
Part C.1 Find the wavelength
Part not displayed
Express your answer in degrees to three significant figures.
ANSWER:
= 0.955
Part D
What problem do you encounter in trying to find the angle
ANSWER:
for an AM radio station with frequency 1000
?
The angle becomes zero.
The angle can be given only in radians.
To find the angle it would be necessary to take the arcsine of a negative number.
To find the angle it would be necessary to take the arcsine of a number greater than one.
This problem indicates that there is not an intensity minimum for the wavelength of AM radio. The maximum
for cell-phone signals is far narrower than the maximum for FM radio waves. Therefore, while you are likely to
encounter dead zones for cell phones in a city (unless you are in an area with many cell-phone towers), you
should expect less trouble with FM radio, and you should have no trouble listening to AM radio. Note also
that some buildings have no roads between them, making for slits with much smaller width . These slits give
broad central maxima for FM radio waves, but still have relatively narrow central maxima for cell-phone
signals. You can estimate the separation of such buildings and calculate for yourself how this affects
transmissions.
Overlapping Diffraction Patterns
Two lasers, one red (with wavelength
nanometers) and the other green (with wavelength
nanometers), are
mounted behind a
-millimeter slit. On the other side of the slit is a white screen. When the red laser is turned
on, it creates a diffraction pattern on the screen.
Part A
The distance from the center of the pattern to the location of the third diffraction minimum of the red laser is
centimeters. How far is the screen from the slit?
Hint A.1 A criterion for dark fringes
Hint not displayed
Express your answer in meters, to three significant figures.
ANSWER:
= Answer not displayed
Part B
The red laser is turned off, and the green laser is turned on. What happens to the central maximum?
ANSWER: Answer not displayed
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Part C
With both lasers turned on, the screen shows two overlapping diffraction patterns. The central maxima of the two
patterns are at the same position. What is the distance between the third minimum in the diffraction pattern of
the red laser (from Part A) and the nearest minimum in the diffraction pattern of the green laser?
Hint C.1 How to approach the problem
Hint not displayed
Part C.2 Find the distance to the third minimum
Part not displayed
Part C.3 Find the separation between successive minima
Part not displayed
Part C.4 Which minimum is closest?
Part not displayed
Part C.5 Find the location of the fourth minimum from the green laser
Part not displayed
Express your answer in centimeters, to three significant figures.
ANSWER:
= Answer not displayed
Single-Slit Diffraction
You have been asked to measure the width of a slit in a piece of paper. You mount the paper
centimeters from a
screen and illuminate it from behind with laser light of wavelength
nanometers (in air). You mark two of the
intensity minima as shown in the figure, and measure the distance between them to be
millimeters.
Part A
What is the width of the slit?
Hint A.1 The equation for single-slit diffraction
Hint not displayed
Hint A.2 Small-angle approximations
Hint not displayed
Express your answer in micrometers, to three significant figures.
ANSWER:
= Answer not displayed
Part B
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If the entire apparatus were submerged in water, would the width of the central peak change?
Hint B.1 How to approach the problem
Hint not displayed
ANSWER: Answer not displayed
Easy development of diffraction gratings from multislit interference and a challenging application problem
Multislit Interference and Diffraction Gratings
Learning Goal: To understand multislit interference and how it leads to the design of diffraction gratings.
Diffraction gratings are used in modern spectrometers to separate the wavelengths of visible light. The working of a
diffraction grating may be understood through multislit interference, which can be understood as an extension of
two-slit interference. In this problem, you will follow the progression from two-slit to many-slit interference to
arrive at the important equations describing diffraction gratings.
A typical diffraction grating consists of a thin, opaque object with a series of very closely spaced slits in it. (There
are also reflection gratings, which use a mirror with nonreflecting lines etched into it to provide the same effects.)
To see how a diffraction grating can separate different wavelengths within a spectrum, we will first consider a
"grating" with only two slits.
Recall that the angles for constructive interference from a pair of slits are given by the equation
is the separation between the slits, is the wavelength of the light, and is an integer.
, where
Part A
Consider a pair of slits separated by
micrometers. What is the angle
for red light with a wavelength of nanometers?
to the interference maximum with
Express your answer in degrees to three significant figures.
ANSWER:
= Answer not displayed
Part B
Consider the same pair of slits separated by
micrometers. What is the angle
with
for blue light with a wavelength of
nanometers?
to the interference maximum
Express your answer in degrees to three significant figures.
ANSWER:
= Answer not displayed
Part C
Part not displayed
Part D
Part not displayed
Part E
Part not displayed
Part F
Part not displayed
Part G
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Part not displayed
A Diffraction Grating Spectrometer
Suppose that you have a reflection diffraction grating with
970 lines per millimeter. Light from a sodium lamp
passes through the grating and is diffracted onto a distant screen.
Part A
Two visible lines in the sodium spectrum have wavelengths 498
and 569 . What is the angular separation
of the first maxima of these spectral lines generated by this diffraction grating?
Part A.1 Find reflection angle of 498
spectral line
Part not displayed
Part A.2 Find reflection angle of 569
line
Part not displayed
Express your answer in degrees to two significant figures.
ANSWER:
= Answer not displayed
Part B
How wide does this grating need to be to allow you to resolve the two lines 589.00 and 589.59 nanometers,
which are a well known pair of lines for sodium, in the second order (
)?
Part B.1 Find the necessary spectral resolving power
Part not displayed
Hint B.2 Two expressions for resolving power
Hint not displayed
Hint B.3 Relation between
and the grating width.
Hint not displayed
Express your answer in millimeters to two significant figures.
ANSWER: Answer not displayed
Introduction and good problem on circular diffraction
Understanding Circular-Aperture Diffraction
Learning Goal: To use the formulas for the locations of the dark bands and understand Rayleigh's criterion of
resolvability.
An important diffraction pattern in many situations is diffraction from a circular aperture. A circular aperture is
relatively easy to make: all that you need is a pin and something opaque to poke the pin through. The figure shows
a typical pattern. It consists of a bright central disk, called the Airy disk, surrounded by concentric rings of dark and
light.
While the mathematics required to derive the equations for circular-aperture
diffraction is quite complex, the derived equations are relatively easy to use.
One set of equations gives the angular radii of the dark rings, while the other
gives the angular radii of the light rings. The equations are the following:
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,
,
where is the wavelength of light striking the aperture, is the diameter of the aperture, and is the angle
between a line normal to the screen and a line from the center of the aperture to the point of observation. There are
more alternating rings farther from the center, but they are so faint that they are not generally of practical interest.
Consider light from a helium-neon laser (
nanometers) striking a pinhole with a diameter of 0.125
.
Part A
At what angle
to the normal would the first dark ring be observed?
Express your answer in degrees, to three significant figures.
ANSWER:
= Answer not displayed
Part B
Part not displayed
Part C
Part not displayed
Diffraction due to a circular aperture is important in astronomy. Since a telescope has a circular aperture of finite
size, stars are not imaged as points, but rather as diffraction patterns. Two distinct points are said to be just
resolved (i.e., have the smallest separation for which you can confidently tell that there are two points instead of
just one) when the center of one point's diffraction pattern is found in the first dark ring of the other point's
diffraction pattern. This is called Rayleigh's criterion for resolvability.
Consider a telescope with an aperture of diameter 1.01 .
Part D
What is the angular radius of the first dark ring for a point source being imaged by this telescope? Use
nanometers for the wavelength, since this is near the average for visible light.
Express your answer in degrees, to three significant figures.
ANSWER:
= Answer not displayed
Part E
Part not displayed
Circular Diffraction Patterns
Monochromatic light of wavelength
nanometers is incident on a small pinhole in a piece of paper. On a screen
meters from the pinhole, you observe the diffraction pattern shown in the figure. You carefully measure the
diameter of the central maximum to be
millimeters, as shown in the figure.
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Part A
What is the diameter
of the pinhole?
Part A.1 Find the angular separation
Part not displayed
Part A.2 Solve for the diameter
Part not displayed
Express your answer in millimeters, to three significant figures.
ANSWER:
Summary
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2 of 9 problems complete (20.8% avg. score)
9.36 of 10 points
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