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A thin sheet of a transparent material (mu = 1.60)is placed in the path of one of the interfering beams in a YDSE using sodium light, lamda = 5890 Å. The central fringe shifts to a position originally occupied by the 12th bright fringe. Calculate the thickness of the sheet. |
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Answer» Solution :`Delta X = (mu - 1)t= n LAMDA` . It is given that`mu = 1.60, n = 12,lamda = 5890 Å` ` therefore t = (12 xx 5890 xx 10^(-10))/(1.60 - 1)` ` = 1.18 xx 10^(-5) m = 12mu m` |
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