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

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(A) The fringe width in a Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$.Here,$\lambda$ is the wavelength of lig

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$0.30 \, mm$

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(A) The fringe width in a Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$.Here,$\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.When the apparatus is immersed in a medium of refractive index $\mu$,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.Since $D$ and $d$ remain unchanged,the new fringe width $\beta'$ is given by $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$.Given $\beta = 0.4 \, mm$ and $\mu = 4/3$,we have $\beta' = \frac{0.4}{4/3} = 0.4 	imes \frac{3}{4} = 0.3 \, mm$.

In a Young's double slit experiment,the fringe width is found to be $0.4 \, mm$. If the whole apparatus is immersed in water of refractive index $4/3$ without disturbing the geometrical arrangement,the new fringe width will be

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