# Rutgers Analytical Physics 750:228, Spring 2016 ( RUPHY228S16 )

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2 2 of 14 2/22/ :28 PM 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. Hint 1. Phase differences for multiple slits The phase difference between the original two slits is not affected by the presence of a third slit. Similarly, the phase difference between the new slit and the closest of the old slits, slit 2 in the figure, can be determined in the same way that you would determine the phase difference for a regular two-slit interference problem. Recall that in two-slit interference, phase shift is proportional to. How do and for slit 2 and the new slit compare to the values of and for slit 1 and slit 2? The peak intensity increases. The peak intensity remains constant. The peak intensity decreases. 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 Yes, halfway between the original maxima Yes, at intervals one-third of the distance between the original maxima Yes, halfway between every second pair of maxima

3 3 of 14 2/22/ :28 PM 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. Hint 1. Equation for diffraction gratings Recall that you have shown that the maxima for a diffraction grating are the same as the maxima for a two-slit setup. Therefore, the locations of the maxima are given by the equation. 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 wider than the maxima for two slits, because they contain more energy. They are narrower than the maxima for two slits because of conservation of energy. They have a more blue color than the maxima for two slits, because they have more energy. They are the same width and color as the maxima for two slits. 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?

4 4 of 14 2/22/ :28 PM Hint 1. Writing an equation If you take the information above and put it into an equation, you obtain, where is the energy at the screen, is the number of maxima, is the peak intensity, and is the width of the maxima. You know that is constant, is proportional to, and is proportional to. You are asked to determine the power of to which must be be proportional so that both sides of the equation remain equal as varies. The width does not change. The width is proportional to. The width is proportional to. The width is proportional to. The width is proportional to. 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. ± A Diffraction Grating Spectrometer Description: ± Includes Math Remediation. The angular separation of two diffracted spectral lines is found, and then the ability of the same grating to resolve two similar lines is explored. Suppose that you have a reflection diffraction grating with grating and is diffracted onto a distant screen. 120 lines per millimeter. Light from a sodium lamp passes through the 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? Express your answer in degrees to two significant figures. Hint 1. Find reflection angle of 498 spectral line Find the angle of reflection for the first maximum of the 498 spectral line. Recall that the angles to the intensity maxima from a diffraction grating are determined by the equation, where is any integer and is the distance between lines on the grating. Express your answer in degrees to three significant figures. Hint 1. Find What is the separation millimeter. between lines on this diffraction grating? Recall that 120 is the number of lines per Express your answer in nanometers to three significant figures. 8330

5 5 of 14 2/22/ :28 PM 3.43 Hint 2. Find reflection angle of 569 line Find the angle of reflection for the 569 line. Recall that the angles to the intensity maxima from a diffraction grating are determined by the equation, where is any integer. Express your answer in degrees to three significant figures Part B 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 ( )? Express your answer in millimeters to two significant figures. Hint 1. Find the necessary spectral resolving power What is the chromatic resolving power necessary to resolve these two lines? Express your answer to two significant figures. Hint 1. Definition of chromatic resolving power Recall that the chromatic resolving power is defined as where is the smaller of the two wavelengths and is the difference between the two wavelengths you are attempting to resolve. Note that in practice, the two wavelengths are usually close enough together that you could use either for with only minor change in the computed resolving power., 1000 Hint 2. Two expressions for resolving power Recall that resolving power is defined by For a diffraction grating spectrometer, chromatic resolving power may be found from the equation, where is

6 6 of 14 2/22/ :28 PM the order of the maxima being used and is the number of lines on the diffraction grating. Hint 3. Relation between and the grating width. Since is the total number of lines on the grating, width of grating number of lines per unit length. 4.2 Most diffraction gratings are much wider than this. What you are actually finding is the width of the part of the grating that must be illuminated. The parts of the grating that no light shines onto obviously can't affect the way the light diffracts. To see why illuminating more of the grating gives better resolving power, recall that the width of an interference maximum decreases as you add more slits to the screen that the light passes through. Illuminating more of the grating has the same effect as adding slits to the screen. If the interference maxima for different wavelengths are narrower, the wavelengths can be resolved more finely. Using X-ray Diffraction Description: Use X-ray diffraction to measure the spacing between the atoms in a crystal. Next, find the angles for the secondand third-order Bragg maxima. Grazing angle convention. When an x-ray beam is scattered off the planes of a crystal, the scattered beam creates an interference pattern. This phenomenon is called Bragg scattering. For an observer to measure an interference maximum, two conditions have to be satisfied: 1. The angle of incidence has to be equal to the angle of reflection. 2. The difference in the beam's path from a source to an observer for neighboring planes has to be equal to an integer multiple of the wavelength; that is, The path difference can be determined from the diagram. The second condition is known as the Bragg condition.. Part A An x-ray beam with wavelength is directed at a crystal. As the angle of incidence increases, you observe the first strong interference maximum at an angle What is the spacing between the planes of the crystal? Express your answer in nanometers to four significant figures.

7 7 of 14 2/22/ :28 PM Part B Find the angle at which you will find a second maximum. Express your answer in degrees to three significant figures Part C Will you observe a third maximum? Choose the most complete answer. Hint 1. What is the sine of the angle at the third maximum? At a third maximum, is equal to. What must the value of be? Is this a physical value? Express your answer numerically to three significant figures Yes, because all crystals have at least three planes. Yes, because the diffraction pattern has an infinite number of maxima. No, because the angle of a third maximum is greater than. No, because the existence of such a maximum produces an unphysical result such as the sine of an angle being greater than one. ± Resolving Pixels on a Computer Screen Description: ± Includes Math Remediation. Given the size of pixels on a computer screen and the maximum distance from the screen at which you can still resolve individual pixels, calculate the effective diameter of the pupil. 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

8 8 of 14 2/22/ :28 PM nanometers as a characteristic optical wavelength. Express your answer in millimeters to three significant figures. Hint 1. Rayleigh's criterion Rayleigh's criterion states that two points can be resolved if the center of one point's diffraction pattern coincides with the first minimum of the second point's diffraction pattern. This corresponds to two points having an angular separation such that Part B Assuming that the screen is sufficiently bright, at what distance can you no longer resolve two pixels on diagonally opposite corners of the screen, so that the entire screen looks like a single spot? Note that the size (0.360 meters) quoted for a monitor is the length of the diagonal. Express your answer in meters to three significant figures. Hint 1. How to approach this problem The separation between the two pixels at opposite corners is given by the size of the screen (i.e., ). You can use Rayleigh's criterion of resolvability to find the distance at which the two points are "just resolved." This will be the answer to the needed precision, because any slight increase beyond that distance (including changes small enough to be unnoticed at the given precision) will make the two unresolvable. You will need the diameter of the pupil that you found in Part A. distance 1660 Also accepted: 1670 Polarization of Light and Malus's Law Description: Practice identifying the polarization angle of a beam of light from a figure and using Malus's law to calculate light intensity of a beam after passing through one or more polarizing filters. Learning Goal: To understand polarization of light and how to use Malus's law to calculate the intensity of a beam of light after passing through one or more polarizing filters. The two transverse waves shown in the figure both travel in the +z direction. The waves differ in that the top wave oscillates horizontally and the bottom wave oscillates vertically. The direction of oscillation of a wave is called the polarization of the wave. The upper wave is described as polarized in the +x direction whereas the lower wave is polarized in the +y direction. In general, waves can be polarized along any direction. Recall that electromagnetic waves, such as visible light, microwaves, and X rays, consist of oscillating electric and magnetic fields. The polarization of an electromagnetic wave refers to the oscillation direction of the electric field, not the magnetic field. In this problem all figures depicting light waves illustrate only the electric field. A linear polarizing filter, often just called a polarizer, is a device that only

9 9 of 14 2/22/ :28 PM transmits light polarized along a specific transmission axis direction. The amount of light that passes through a filter is quantified in terms of its intensity. If the polarization angle of the incident light matches the transmission axis of the polarizer, 100 of the light will pass through, so the transmitted intensity will equal the incident intensity. More generally, the intensity of light emerging from a polarizer is described by Malus's law: where is the intensity of the polarized light beam just before entering the polarizer, is the intensity of the transmitted light beam immediately after passing through the polarizer, and is the angular difference between the polarization angle of the incident beam and the transmission axis of the polarizer. After passing through the polarizer, the transmitted light is polarized in the direction of the transmission axis of the polarizing filter. In the questions that follow, assume that all angles are measured counterclockwise from the +x axis in the direction of the +y axis., Part A A beam of polarized light with intensity and polarization angle strikes a polarizer with transmission axis. What angle should be used in Malus's law to calculate the transmitted intensity? This process is illustrated in the figure, where the polarization of the light wave is visually illustrated by a magenta double arrow oriented in the direction of polarization, the transmission axis of the polarizer is represented by a blue double arrow, and the direction of motion of the wave is illustrated by a purple arrow. Part B What is the polarization angle of the light emerging from the polarizer?

10 10 of 14 2/22/ :28 PM Part C If 20.0, 25.0, and 40.0, what is the transmitted intensity? Express your answer numerically in watts per square meter If the polarization axis of the incident light and the transmission axis of the filter are aligned (so ) or if they point in exactly opposite directions (so ), then in Malus's law, indicating that 100 of the incident intensity will be transmitted. Part D One way to produce a beam of polarized light with intensity and polarization angle would be to pass unpolarized light with intensity through a polarizer whose transmission axis is oriented such that. How large must be if the transmitted light is to have intensity? Express your answer as a decimal number times the symbol. For example, if, enter 0.25 * I. Part E A beam of unpolarized light with intensity falls first upon a polarizer with transmission axis then upon a second polarizer with transmission axis, where (in other words the two axes are perpendicular to one another). What is the intensity of the light beam emerging from the second polarizer? Express your answer as a decimal number times the symbol. For example, if, enter 0.25 * I_0.

11 11 of 14 2/22/ :28 PM Hint 1. How to approach the problem Consider doing the problem in two steps. First, determine the intensity and polarization angle of the light emerging from the first polarizer. Then calculate the intensity of the light emerging from the second polarizer based on those answers as well as on the transmission axis angle of the second polarizer,. Hint 2. Determine the intensity of light between the two polarizers What is the intensity polarizer? of the light emerging from the first polarizer, before it attempts to pass through the second Express your answer as a decimal number times the symbol. For example, if, enter 0.25 * I_0. Now think about what the polarization angle of this light will be when it reaches the second polarizer. Hint 3. Determine the angle to be used in Malus's law When Malus's law is applied to calculate the intensity of light that emerges from the second polarizer, what angle be used? Recall that all angles are measured counterclockwise from the +x axis in the direction of the +y axis. should Hint 1. Find the orientation of the light between the two polarizers What is, the polarization of the light after it has passed through the first polarizer? Express your answer numerically in degrees. 0 Also accepted: degrees 45 degrees 90 degrees 180 degrees Adjacent polarizers with perpendicular transmission axes are said to be "crossed." No light can get through both polarizers, regardless of the light's initial polarization state.

12 12 of 14 2/22/ :28 PM Part F Notice that a polarizer modifies the light intensity according to Malus's law and also reorients the polarization angle of the beam to match its own transmission axis. Hence it is possible for light to pass through a pair of crossed polarizers if a third polarizer is inserted between them with an intermediate transmission axis direction. What is the new intensity of the light emerging from the final polarizer in Part E if a third polarizer (Polarizer A in the figure ), whose transmission axis is offset 45 from each of the others, is inserted between the original two polarizers? Express your answer as a decimal number times the symbol. For example, if, enter 0.25 * I_0. Hint 1. How to approach the problem Besides the intensities,, and used in Part E, notice in the figure the new intensity immediately following the inserted polarizer. Use the rules learned in the previous parts to relate intensities across each polarizer individually. First relate to, then to, and finally to. Combine your expressions to relate to directly. Hint 2. Relate to What is the value of in terms of? Express your answer as a decimal number times the symbol. For example, if, enter 0.25 * I_0. Hint 3. Relate to What is the value of in terms of? Express your answer as a decimal number times the symbol. For example, if, enter 0.25 * I_1. Also accepted: Hint 4. Relate to What is the value of in terms of? Express your answer as a decimal number times the symbol. For example, if, enter 0.25 * I_A.

13 13 of 14 2/22/ :28 PM Also accepted: Also accepted: Polarizing filters for visible light are made of Polaroid, which contains long molecular chains that have been aligned by stretching the material during production. Polaroid is commonly used in sunglasses because it reduces the intensity of unpolarized sunlight light by 50. Glare is often at least partially polarized, so Polaroid sunglasses, when properly oriented, can selectively reduce glare by even more than 50. ± Circular Diffraction Patterns Description: ± Includes Math Remediation. The problem describes a diffraction pattern caused by a pinhole, and the student is asked to calculate the size of the hole. 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. Part A What is the diameter of the pinhole? Express your answer in millimeters, to three significant figures. Hint 1. Find the angular separation What is the formula for Rayleigh's criterion of resolvability? Recall that this formula relates the angular separation between the center of a diffraction pattern and its first dark ring. Express, where is the angular separation, in terms of the diameter of the pinhole and the wavelength.

14 14 of 14 2/22/ :28 PM Hint 2. Solve for the diameter Using your answer from part A.1 and small-angle approximations, find the diameter of the hole. Express your answer in terms of the distance to the screen, the wavelength, and the distance center of the diffraction pattern and the first dark ring. between the Copyright 2016 Pearson. All rights reserved. Legal Notice Privacy Policy Permissions Support

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