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Coherent Sources
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Wavefront splitting Interferometer
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Young’s Double Slit Experiment
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Young’s double slit © SPK
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Path difference:
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For a bright fringe, For a dark fringe, m: any integer
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For two beams of equal irradiance (I0)
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Visibility of the fringes (V)
Maximum and adjacent minimum of the fringe system
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Photograph of real fringe pattern for Young’s double slit
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The two waves travel the same distance Therefore, they arrive in phase
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The upper wave travels one wavelength farther
Therefore, the waves arrive in phase S' S
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This is destructive interference
The upper wave travels one-half of a wavelength farther than the lower wave. This is destructive interference S' S
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Uses for Young’s Double Slit Experiment
Young’s Double Slit Experiment provides a method for measuring wavelength of the light This experiment gave the wave model of light a great deal of credibility.
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Phase Changes Due To Reflection
An electromagnetic wave undergoes a phase change of 180° upon reflection from a medium of higher index of refraction than the one in which it was traveling Analogous to a reflected pulse on a string μ2 μ1
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Phase shift
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Fresnel double mirror P1 P2 © SPK
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Problem In a Fresnel mirror the angle between the mirrors a=12’. The distance r= 10 cm and b=130 cm. Find The fringe width on the screen and the number of possible maxima. (b) the shift of the interference pattern on the screen when the slit S is displaced by dl=1 mm along the arc of radius r about the center O. (c) The maximum width of the source slit at which the fringe pattern on the screen can still be observed sufficiently sharp.
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Fresnel biprism © SPK
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Lloyd’s mirror © SPK
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Billet’s split lens © SPK
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Wavefront splitting interferometers
Young’s double slit Fresnel double mirror Fresnel double prism Lloyd’s mirror
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Division of Amplitude
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Optical beam splitter
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Fringes of equal inclination
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t A B C D n1 nf n2 C D B t i A d
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Optical path difference for the first two reflected beams
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Condition for maxima Condition for minima
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Fringes of equal thickness
Constant height contour of a topographial map
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Wedge between two plates
1 2 glass D t glass air x Path difference = 2t Phase difference = 2kt - (phase change for 2, but not for 1) Maxima 2t = (m + ½) o/n Minima 2t = mo/n Fizeau Fringes
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Newton’s Ring Ray 1 undergoes a phase change of 180 on reflection, whereas ray 2 undergoes no phase change R= radius of curvature of lens r=radius of Newton’s ring
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Reflected Newton’s Ring
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Newton’s Ring
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Types of localization of fringes
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Interference fringes Real Virtual Localized Non-localized
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Localized fringe Observed over particular surface Result of extended source
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Non-localized fringe Exists everywhere Result of point/line source
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POHL’S INTERFEROMETER
Real Non-localized Virtual Localized Refer Hecht for details
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Problem The width of a certain spectral line at 500 nm is 2×10-2 nm. Approximately what is the largest path difference for which the interference fringes produces by the light are clearly visible?
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