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In a Young's double slit experiment using red and blue lights of wavelengths \mathrm{600 \mathrm{~nm}} and \mathrm{480 \mathrm{~nm}} respectively, the value of \mathrm{n} for which the \mathrm{\mathrm{n}^{\text {th }}} red fringe coincides with (\mathrm{n}+1)^{\mathrm{th}} blue fringe is:

Option: 1

5


Option: 2

4


Option: 3

3


Option: 4

2


Answers (1)

best_answer

Let \mathrm{\mathrm{n}^{\text {th }}} red coincides with \mathrm{\mathrm{m}^{\text {th }}} blue, then

\mathrm{ \begin{aligned} & \frac{\mathrm{n} \lambda_{\mathrm{R}} \mathrm{D}}{\mathrm{d}}=\frac{\mathrm{m} \lambda_{\mathrm{B}} \mathrm{D}}{\mathrm{d}} \\ & \therefore \frac{\mathrm{n}}{\mathrm{m}}=\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{R}}}=\frac{480}{600}=\frac{4}{5} \end{aligned} }

i.e., \mathrm{4^{\text {th }}} red coincides with \mathrm{5^{\text {th }}} blue. Next \mathrm{8^{\text {th }}} red will coincide with \mathrm{10^{\text {th }}} blue and so on. Thus,\mathrm{ \mathrm{n}=4} and

\mathrm{ \mathrm{m}=\mathrm{n}+1=5}

Posted by

sudhir.kumar

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