In YDSE,the Y-coordinates of the central maxi | vedclass

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(C) The position of the central maxima is independent of the medium,so it remains at $Y = 2 \, cm$.The fringe width $\beta$ is given by $\beta = \frac

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$2 \, cm, 4 \, cm$

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(C) The position of the central maxima is independent of the medium,so it remains at $Y = 2 \, cm$.The fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$. When immersed in a liquid of refractive index $n = 1.5$,the wavelength becomes $\lambda' = \frac{\lambda}{n}$.Thus,the new fringe width $\beta'$ becomes $\beta' = \frac{\beta}{n} = \frac{\beta}{1.5}$.The initial distance between the central maxima and the $10^{th}$ maxima is $\Delta Y = 5 \, cm - 2 \, cm = 3 \, cm$.In the liquid,the new distance $\Delta Y'$ is $\Delta Y' = \frac{\Delta Y}{n} = \frac{3 \, cm}{1.5} = 2 \, cm$.Therefore,the new position of the $10^{th}$ maxima is $Y' = Y_{central} + \Delta Y' = 2 \, cm + 2 \, cm = 4 \, cm$.

In $YDSE$,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 $1.5$,what will be the corresponding $Y$-coordinates?

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