Download Part A One reason that Fraunhofer diffraction is relatively easy to

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 ?
Express your answer in terms of the slit width and the angle shown in the figure.
=a*sin(theta)Correct
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
light arriving at point
from the bottom of the slit equal
difference will give a phase difference of
from the top of the slit and
. What length of path
?
Express your answer in terms of the wavelength .
=lambdaCorrect
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
bands?
,
,
Correct
,
for which this equation gives criteria for dark
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.
arcsin(lambda/a),-arcsin(lambda/a)Correct
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 screen
corresponding to a particular value of .
For small angles,
yields
. Since
, the small-angle approximation
. 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
dark bands closest to the center of the diffraction pattern?
Express your answer in terms of , , and .
=2*x*lambda/aCorrect
between the two
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.
=3.80
Correct
?
Multislit Interference and Diffraction Gratings
Part A
Consider a pair of slits separated by
micrometers. What is the angle
interference maximum with
for red light with a wavelength of
Express your answer in degrees to three significant figures.
to the
nanometers?
=44.40 Correct
Part B
Consider the same pair of slits separated by
micrometers. What is the angle
the interference maximum with
for blue light with a wavelength of
Express your answer in degrees to three significant figures.
to
nanometers?
=23.60 Correct
Notice that the maxima for these two wavelengths are separated by over 20 degrees.
Purely on the basis of this information, it would appear that a pair of slits would be useful
for physically separating wavelengths for spectroscopy. However, these peaks of
intensity are rather wide, and so it is difficult to resolve similar colors, such as orange and
yellow. As you will see, the more slits, the better!
Part C
The intensity pattern from the two slits for a single wavelength looks like the one shown
on the left side of the figure.
If another slit, separated from one of the original slits by a distance , is added, how will
the intensity at the original peaks change?
By examining the phasors for light from the two slits, you can determine how the new slit
affects the intensity. Phasors are vectors that correspond to the light from one slit. The
length of a phasor is proportional to the magnitude of the electric field from that slit, and
the angle between a phasor and the previous slit's phasor corresponds to the phase
difference between the light from the two slits. Recall that at points of constructive
interference, light from the original two slits has a phase difference of , which
corresponds to a complete revolution of one phasor relative to the first.
Notice that, as shown in the figure, undergoing a complete revolution leaves the phasor
pointing in the same direction as the phasor from the other slit. Think about the phase
difference between the new slit and the closer of the two old slits and what this implies
about the direction of the phasor for the new slit.
The peak intensity increases.
Correct
Part D
Are there any points between the maxima of the original two slits where light from all
three slits interferes constructively? If so, what are they?
No
Correct
Since the original two slits do not interfere constructively anywhere except for the
original maxima, it would be impossible for all three slits to interfere constructively
somewhere between the original maxima. However, there are two places between the
maxima where the intensity is zero, instead of just one such minimum as in the case of
just two slits. This means that the intensity dies off more quickly as you move away from
one of the peaks than it would for two slits, and then stays low until the next peak is
reached.
Part E
What is the angle
spaced
to the second-order maximum from a diffraction grating with slits
apart, for light of wavelength
Express your answer in terms of
and
=arcsin(2*lambda_1/d_1)Correct
?
.
Part F
Recall that phasor diagrams give the magnitude of the electric field, and that the intensity
is related to the electric field squared. In going from two slits to three, the amount of
energy in the total interference pattern increases by
(three slits worth of light instead
of two), but the peak intensity of the interference maxima increases by
(from
to
, where is the electric field due to one slit).
What does this suggest about the intensity maxima for three slits?
They are narrower than the maxima for two slits because of conservation of energy.
Correct
Part G
By the same reasoning that worked with three slits, you can see that no matter how many
slits you have, the maxima will still fall at the same locations as the maxima for two slits.
The peak intensity of the maxima will be proportional to
of slits.
, where
is the total number
The energy at the screen is roughly equal to the product of the number of maxima, the
peak intensity of a maximum, and the width of a maximum. As increases, the number
and location of the maxima will not change, while the peak intensity of the maxima will
increase proportionally to . If the total energy available increases proportionally to
how does the width of the maxima change?
The width is proportional to
Correct
.
Now you can see why diffraction gratings have many thousands of slits. The more slits,
the narrower the maximum for a particular wavelength, and thus the more finely
wavelengths may be separated for analysis.
,