In Young's double slit arrangement,water is f | vedclass

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(D) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.When the space between the slits and the screen is filled with a me

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fringe width decreases and central bright fringe does not shift.

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(D) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$.When the space between the slits and the screen is filled with a medium of refractive index $\mu$,the wavelength of light in the medium becomes $\lambda' = \frac{\lambda}{\mu}$.Therefore,the new fringe width $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$.Since the refractive index of water $\mu > 1$,the fringe width $\beta'$ decreases.Regarding the shift,the path difference for the central fringe remains zero because the optical path length change is identical for both interfering waves traveling from the two slits to the center of the screen. Thus,the central bright fringe does not shift.

In Young's double slit arrangement,water is filled in the space between the screen and the slits. Then:

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In the figure shown,if a parallel beam of white light is incident on the plane of the slits,then the distance of the white spot on the screen from $O$ is [Assume $d << D, \lambda << d$].

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