In Young's double slit experiment,the y-coord | vedclass

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(C) The position of the $n^{th}$ maxima in $YDSE$ is given by $y_n = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Initially,the ce

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$2 \, cm, 4 \, cm$

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(C) The position of the $n^{th}$ maxima in $YDSE$ is given by $y_n = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Initially,the central maxima is at $y_0 = 2 \, cm$ and the $10^{th}$ maxima is at $y_{10} = 5 \, cm$.The distance between the central maxima and the $10^{th}$ maxima is $\Delta y = y_{10} - y_0 = 5 \, cm - 2 \, cm = 3 \, cm$.When the apparatus is immersed in a liquid of refractive index $\mu = 1.5$,the wavelength $\lambda'$ becomes $\lambda / \mu$,and consequently,the fringe width $\beta'$ becomes $\beta / \mu$.Since the central maxima is formed at the path difference of zero,its position remains unchanged at $y_0' = 2 \, cm$.The new distance between the central maxima and the $10^{th}$ maxima becomes $\Delta y' = \frac{\Delta y}{\mu} = \frac{3 \, cm}{1.5} = 2 \, cm$.Therefore,the new position of the $10^{th}$ maxima is $y_{10}' = y_0' + \Delta y' = 2 \, cm + 2 \, cm = 4 \, cm$.Thus,the coordinates are $2 \, cm$ and $4 \, cm$.

In Young's double slit experiment,the $y$-coordinates of the central maxima and the $10^{th}$ maxima are $2 \, cm$ and $5 \, cm$ respectively. When the $YDSE$ apparatus is immersed in a liquid of refractive index $\mu = 1.5$,what will be the corresponding $y$-coordinates?

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