Physics II Lab w answers_Part_58

Physics II Lab w answers_Part_58 - rm2 = (2R-t)t where rm...

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127 r m 2 = (2R-t)t where r m is the radius of the m th order dark ring .(Note: The dark ring is the m th dark ring excluding the central dark spot). Now R is the order of 100 cm and t is at most 1 cm. Therefore R>>t. Hence t ) t R 2 ( r R r ) t R ( 2 m 2 2 m 2 - = = + - (neglecting the t 2 term ), giving R r t 2 2 m Putting the value of “ 2 t” in eq(1) gives R m r R r m 2 m 2 m λ , m =0,1,2,3… … ----------------(3) and eq (2) gives (for the radius r m of m th order bright ring ) R 2 1 m r ) 2 1 m ( R r 2 m 2 m ± ² ³ ´ µ + = + = ----------------------(4) Hence for dark rings R m r m = --------------------------(5) while for bright rings R 2 1 m r m ± ² ³ ´ µ + = ; m =0,1,2,3… … ------------------------(6) With the help of a traveling microscope we can measure the diameter of the m th ring order dark ring = D m . Then r m = 2 D m and hence, R m 4 D 2 m = -----------------------(7) So if we know the wavelength
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This note was uploaded on 11/26/2011 for the course PHY 2053 taught by Professor Lind during the Fall '09 term at FSU.

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Physics II Lab w answers_Part_58 - rm2 = (2R-t)t where rm...

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