Two wavelengths lambda_1 = 450 nm and lambda | vedclass

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(C) In Young's double slit experiment,the position of the $n$-th bright fringe for a wavelength $\lambda$ is given by $y_n = \frac{n D \lambda}{d}$.Fo

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$9$

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(C) In Young's double slit experiment,the position of the $n$-th bright fringe for a wavelength $\lambda$ is given by $y_n = \frac{n D \lambda}{d}$.For two wavelengths $\lambda_1$ and $\lambda_2$ to overlap,their positions must be equal: $y_{n_1} = y_{n_2}$.This implies $n_1 \lambda_1 = n_2 \lambda_2$,where $n_1$ and $n_2$ are the orders of the fringes.Substituting the given values: $n_1 (450 \ nm) = n_2 (650 \ nm)$.$\frac{n_1}{n_2} = \frac{650}{450} = \frac{13}{9}$.Since $n_1$ and $n_2$ must be integers,the minimum values are $n_1 = 13$ and $n_2 = 9$.Thus,the minimum order of the fringe produced by $\lambda_2$ is $n = 9$.

Two wavelengths $\lambda_1 = 450 \ nm$ and $\lambda_2 = 650 \ nm$ are used in Young's double slit experiment. The minimum order of fringe produced by $\lambda_2$ which overlaps with a fringe produced by $\lambda_1$ is $n$. The value of $n$ is . . . . . . .

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