In a Young's double-slit experiment,the fring | vedclass

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(A) The fringe width in air is given by $\beta_{air} = \frac{\lambda D}{d} = 0.8 \, mm$.When the entire apparatus is immersed in a liquid of refractiv

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$0.5$

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(A) The fringe width in air is given by $\beta_{air} = \frac{\lambda D}{d} = 0.8 \, mm$.When the entire apparatus is immersed in a liquid of refractive index $\mu$,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.Consequently,the new fringe width $\beta_{liquid}$ is given by $\beta_{liquid} = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta_{air}}{\mu}$.Substituting the given values: $\beta_{liquid} = \frac{0.8 \, mm}{1.6} = 0.5 \, mm$.

In a Young's double-slit experiment,the fringe width is $0.8 \, mm$ when light of wavelength $6000 \, \mathring{A}$ is used,and the screen is at a distance of $2.5 \, m$. If the experiment is performed in a liquid with a refractive index of $1.6$,the new fringe width will be.....$mm$.

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