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

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(C) The position of the central maximum $(CM)$ is given by the path difference $\Delta x = 0$. Since the path difference is independent of the medium,

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$2 \, cm, 4 \, cm$

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(C) The position of the central maximum $(CM)$ is given by the path difference $\Delta x = 0$. Since the path difference is independent of the medium,the position of the $CM$ remains unchanged at $y = 2 \, cm$.The fringe width $\beta$ in air is given by $\beta = \frac{\lambda D}{d}$. The position of the $n^{th}$ maximum is $y_n = y_{CM} + n\beta$.In air,the $10^{th}$ maximum is at $y_{10} = 5 \, cm$. Thus,$10\beta = 5 - 2 = 3 \, cm$,which means $\beta = 0.3 \, cm$.When immersed in a liquid of refractive index $\mu = 1.5$,the new wavelength becomes $\lambda' = \frac{\lambda}{\mu}$. Consequently,the new fringe width becomes $\beta' = \frac{\beta}{\mu} = \frac{0.3}{1.5} = 0.2 \, cm$.The new position of the $10^{th}$ maximum is $y'_{10} = y_{CM} + 10\beta' = 2 + 10(0.2) = 2 + 2 = 4 \, cm$.Therefore,the new coordinates are $2 \, cm$ and $4 \, cm$.

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 corresponding $y$-coordinates?

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