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 Jul 30th 2009, 02:24 AM #1 Senior Member   Join Date: Feb 2009 Location: Singapore Posts: 206 Interference a) the thickness of a mica sheet is 4.8micrometers. the refractive index of mica for light of wavelength 512nm is 1.60. calculate the ratio of the thickness os the sheet of mica to: 1) the wavelength of the 512nm light in air, 2)the wavelength of the same light in mica. (i think it should be the speed instead of wavelength.) b)the figure shows the slits and the screen in an optical two-slits experiment. light of wavelength 512nm is used, the separation of the two slits is 0.42mm and the perpendicular distance from the line S1S2 to the screen at O is 75cm. P marks the center of the third bright fringe from O. Calculate the distance OP. Write down the value of the distance PS2-PS1. The mica sheet from (a)is mounted normally across the light path from S1 without cutting the light path from S2. explain why the center of the pattern on the screen moves from O to a new O'. calculate the distance OO'. how would the distance OO' be affected by rotating the mica sheet about an axis parallel to S1? I have completed everything up to the part when the mica sheet is placed across the light path from S1. i cant picture it and there are no other figures given in the book for the question other than one that shows the refraction of light on an air-mica interface. i will appreciate any help in picturing how the mica sheet is mounted across the light path. and also the very last part about rotating the mica sheet. i think if i can picture it i should be able to solve the question myself. thanks!
Jul 30th 2009, 02:27 AM   #2
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oops! forgot to put up the figure
its attched
Attached Images
 Fig 19D (9).bmp (907.1 KB, 4 views)

Jul 30th 2009, 06:30 PM   #3
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 Originally Posted by arze I have completed everything up to the part when the mica sheet is placed across the light path from S1. i cant picture it and there are no other figures given in the book for the question other than one that shows the refraction of light on an air-mica interface. i will appreciate any help in picturing how the mica sheet is mounted across the light path. and also the very last part about rotating the mica sheet. i think if i can picture it i should be able to solve the question myself. thanks!
When there is no mica sheet, path difference between S1O and S2O is the same. When the mica sheet is placed in the path of S1O, distance S1O decreases, because the path of the light in mica shortens. Because of the path difference between S1O and S2O the central bright fringe shifts.

 Jul 30th 2009, 07:43 PM #4 Senior Member   Join Date: Feb 2009 Location: Singapore Posts: 206 so i find the phase change which is refractive index times thickness of material. how do i proceed from here? i know that path difference S1O'-S2O'=0.

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