Understanding Fraunhofer Diffraction

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1 [ Assignment View ] [ Eðlisfræði 2, vor 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. 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? 1 of 10 17/4/07 15:52

2 Hint A.1 A useful triangle Express your answer in terms of the slit width and the angle shown in the figure. = 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. = 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?,,,,,,,,,,,,,, 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. 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 2 of 10 17/4/07 15:52

3 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? between the two dark bands closest to the Express your answer in terms of,, and. = Part F Suppose that light from a laser with wavelength 633 is incident on a thin slit of width 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 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. 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 Express your answer in millimeters to three significant figures. 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 3 of 10 17/4/07 15:52

4 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. 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..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. Now solve for in this criterion to obtain the expression that you need. Express your answer in terms of and. = Assume that the average spacing between buildings is. What is the angle to the first minimum for an FM radio station with a frequency of 101?.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 Express your answer in meters, to three significant figures. = 2.97 Express your answer numerically in degrees to three significant figures. Note: Do not write 4 of 10 17/4/07 15:52

5 your answer in terms of trignometric functions. Evaluate any such functions in your working. = 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 Express your answer in degrees to three significant figures. = Part D What problem do you encounter in trying to find the angle 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. 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 Express your answer in meters, to three significant figures. The red laser is turned off, and the green laser is turned on. What happens to the central maximum? Answer not displayed 5 of 10 17/4/07 15:52

6 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 ) and the nearest minimum in the diffraction pattern of the green laser? Hint C.1 How to approach the problem Part C.2 Find the distance to the third minimum Part C.3 Find the separation between successive minima Part C.4 Which minimum is closest? Part C.5 Find the location of the fourth minimum from the green laser Express your answer in centimeters, to three significant figures. 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. What is the width of the slit? Hint A.1 The equation for single-slit diffraction Hint A.2 Small-angle approximations Express your answer in micrometers, to three significant figures. 6 of 10 17/4/07 15:52

7 If the entire apparatus were submerged in water, would the width of the central peak change? Hint B.1 How to approach the problem 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 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. 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. Part C Part D Part E Part F Part G 7 of 10 17/4/07 15:52

8 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. 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?.1 Find reflection angle of Find reflection angle of 569 spectral line line Express your answer in degrees to two significant figures. How wide does this grating need to be to allow you to resolve the two lines and nanometers, which are a well known pair of lines for sodium, in the second order ( )?.1 Find the necessary spectral resolving power Hint B.2 Two expressions for resolving power Hint B.3 Relation between and the grating width. Express your answer in millimeters to two significant figures. 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: 8 of 10 17/4/07 15:52

9 , 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 At what angle to the normal would the first dark ring be observed? Express your answer in degrees, to three significant figures., Part C 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 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. Part E 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. 9 of 10 17/4/07 15:52

10 What is the diameter of the pinhole?.1 Find the angular separation.2 Solve for the diameter Express your answer in millimeters, to three significant figures. Summary 2 of 9 problems complete (20.8% avg. score) 9.36 of 10 points 10 of 10 17/4/07 15:52