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MasteringPhysics: Print View with Answers http://session.masteringphysics.com/myct/assignmentPrin... Part D Part E Part F 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. 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: 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 . ANSWER: , , , , , , , , , , , , , , any rational number , Also accepted: , = 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. 8 of 11 2/7/13 1:01 PM , .
MasteringPhysics: Print View with Answers http://session.masteringphysics.com/myct/assignmentPrin... ANSWER: ± Single-Slit Diffraction Description: ± Includes Math Remediation. The problem describes a typical single-slit diffraction setup, including a picture of the diffraction pattern produced. The student is asked to calculate (numerically) the slit width. 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 Part B = 3.80 What is the width of the slit? Express your answer in micrometers, to three significant figures. Hint 1. The equation for single-slit diffraction The equation for the angle between a line perpendicular to the screen and a line connecting the slit to the th dark fringe is , where is a nonzero integer, is the wavelength of the light, and is the width of the slit. Use this equation, together with some small-angle approximations, to solve this problem. Hint 2. Small-angle approximations Recall that for a small angle , . For this problem, find the value of , in terms of the distance to the screen and the distance from the central maximum to the dark fringe, and then substitute this expression for in the equation for location of the dark fringes. ANSWER: = 170 If the entire apparatus were submerged in water, would the width of the central peak change? Hint 1. How to approach the problem If you are puzzled, look at each variable in your equations and ask yourself whether it would be changed by placing the whole apparatus underwater. If you identify any of the variables as changing, think about how such a change would affect the width of the central fringe (if at all). 9 of 11 2/7/13 1:01 PM
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