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

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Will decrease

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(B) The fringe width $\beta$ in 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 experiment is performed in a medium with refractive index $\mu$,the wavelength of light changes to $\lambda' = \frac{\lambda}{\mu}$.Since the refractive index of water $(\mu \approx 1.33)$ is greater than that of air $(\mu \approx 1.0)$,the wavelength of light decreases in water.Consequently,the fringe width $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$.Since $\mu > 1$,the fringe width $\beta'$ will be less than $\beta$. Therefore,the fringe width will decrease.

Young's double-slit experiment is performed in air and then performed in water. How does the fringe width change?

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