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

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

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$2\, cm, 4\, cm$

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(C) In $YDSE$,the position of the $n^{th}$ maximum is given by $y_n = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Initially,the central maximum is at $y_0 = 2\, cm$ and the $10^{th}$ maximum is at $y_{10} = 5\, cm$.The distance between them 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 $\lambda' = \frac{\lambda}{\mu}$.Consequently,the fringe width becomes $\beta' = \frac{\beta}{\mu} = \frac{\beta}{1.5}$.The new distance between the central maximum and the $10^{th}$ maximum is $\Delta y' = \frac{\Delta y}{\mu} = \frac{3\, cm}{1.5} = 2\, cm$.The position of the central maximum remains unchanged at $y_0' = 2\, cm$.The new position of the $10^{th}$ maximum is $y_{10}' = y_0' + \Delta y' = 2\, cm + 2\, cm = 4\, cm$.

In Young's double slit experiment,the $y$-coordinates of the central maximum and the $10^{th}$ maximum 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 new $y$-coordinates?

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