InterviewSolution
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InterviewSolution
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Using the expression `2d sin theta = lambda`, one calculates the values of `d` by measuring the corresponding angles `theta` in the range `0 to [email protected]`. The wavelength `lambda` is exactly known and error in `theta` is constant for all values of `theta`. As `theta` increases from `[email protected]`A. the absolute error in `d` remains constant.B. the absolute error in `d` increses.C. the fractional error in `d` remains constant.D. the fractional error in `d` decreases. |
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Answer» Correct Answer - D `2d sintheta=lambda` `d=(lambda)/(2 sin theta)` differntiate `del(d)=(lambda)/(2) del (cosectheta)` `del(d)=(lambda)/(2)(-cos esthetacottheta)deltheta` `del(d)=(-lambdacostheta)/(2sin^(2)theta)deltheta` as `theta= "increases", (lambdacostheta)/(2sin^(2)theta) , "decreases"` Alternate solution `d=(lambda)/(2sintheta)` `ln d=ln lambda-ln2-ln sintheta` `(Delta(d))/(d)=0-0-(1)/(sin theta)xxcostheta(Deltatheta)` Fraction error `|+(d)|=|cotthetaDeltatheta|` Absoulute error `Deltad=(dcottheta)Deltatheta` `(d)/(2sintheta)xx(costheta)/(sintheta)` `Deltad=(costheta)/(sin^(2)theta)` |
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