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.
Top of Form
=a*sin(theta)Correct
Bottom of Form
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 .
Top of Form
=lambdaCorrect
Bottom of Form
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 for which this equation gives criteria for dark bands?
Top of Form
, , ,
Correct
Bottom of Form
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.
Top of Form
arcsin(lambda/a),-arcsin(lambda/a)Correct
Bottom of Form
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, . 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 between the two dark bands closest to the center of the diffraction pattern?
Express your answer in terms of , , and .
Top of Form
=2*x*lambda/aCorrect
Bottom of Form
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.
Top of Form
=3.80Correct
Bottom of Form
Multislit Interference and Diffraction Gratings
Part A
Consider a pair of slits separated by micrometers. What is the angle to the interference maximum with for red light with a wavelength of nanometers?
Express your answer in degrees to three significant figures.
Top of Form
=44.40Correct
Bottom of Form
Part B
Consider the same pair of slits separated by micrometers. What is the angle to the interference maximum with for blue light with a wavelength of nanometers?
Express your answer in degrees to three significant figures.
Top of Form
=23.60Correct
Bottom of Form
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.
Top of Form
The peak intensity increases.
Correct
Bottom of Form
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?
Top of Form
No
Correct
Bottom of Form
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 to the second-order maximum from a diffraction grating with slits spaced apart, for light of wavelength ?
Express your answer in terms of and .
Top of Form
=arcsin(2*lambda_1/d_1)Correct
Bottom of Form
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?
Top of Form
They are narrower than the maxima for two slits because of conservation of energy.
Correct
Bottom of Form
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 , where is the total number of slits.
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?
Top of Form
The width is proportional to .
Correct
Bottom of Form
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.