Understanding Fraunhofer Diffraction

# Understanding Fraunhofer Diffraction - Part A One reason...

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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 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 . =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 for which this equation gives criteria for dark bands? , , , Correct

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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
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Understanding Fraunhofer Diffraction - Part A One reason...

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