In Young's double slit experiment,62 fringes | vedclass

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(D) The total width of the visible region $W$ remains constant for a given experimental setup.For $n$ fringes of wavelength $\lambda$,the width is giv

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$84$

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(D) The total width of the visible region $W$ remains constant for a given experimental setup.For $n$ fringes of wavelength $\lambda$,the width is given by $W = n 	imes \beta$,where $\beta = \frac{D\lambda}{d}$ is the fringe width.Thus,$W = n 	imes \frac{D\lambda}{d}$.Since $W$,$D$,and $d$ are constant,we have $n_1 \lambda_1 = n_2 \lambda_2$.Given $n_1 = 62$,$\lambda_1 = 5893 \ \mathring{A}$,and $\lambda_2 = 4358 \ \mathring{A}$.Substituting the values: $62 	imes 5893 = n_2 	imes 4358$.$n_2 = \frac{62 	imes 5893}{4358} \approx 83.83$.Rounding to the nearest whole number,we get $n_2 = 84$.

In Young's double slit experiment,$62$ fringes are seen in the visible region for sodium light of wavelength $5893 \ \mathring{A}$. If violet light of wavelength $4358 \ \mathring{A}$ is used in place of sodium light,then the number of fringes seen will be:

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