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(D) The fringe width is given as $\beta = 0.002 	ext{ cm}$.For a Young's double-slit interference pattern,the distance of the $n^{	ext{th}}$ dark fr

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None of the above

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(D) The fringe width is given as $\beta = 0.002 	ext{ cm}$.For a Young's double-slit interference pattern,the distance of the $n^{	ext{th}}$ dark fringe from the central bright fringe is given by the formula $x_n = (n - 0.5) \beta$,where $n$ is the order of the dark fringe $(n = 1, 2, 3, \dots)$.For the $5^{	ext{th}}$ dark fringe,$n = 5$.Substituting the values: $x_5 = (5 - 0.5) 	imes 0.002 	ext{ cm}$.$x_5 = 4.5 	imes 0.002 	ext{ cm} = 0.009 	ext{ cm}$.Converting to scientific notation: $0.009 	ext{ cm} = 0.9 	imes 10^{-2} 	ext{ cm}$.Comparing this result with the given options,none of the options match the calculated value.Therefore,the correct choice is $D$.

$A$ fringe width of a certain interference pattern is $\beta = 0.002 \text{ cm}$. What is the distance of the $5^{\text{th}}$ dark fringe from the center?

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