Answer:
Fringe width in Young's double slit experiment, \[\beta =\frac{D\lambda }{d}\] When apparatus is immersed in liquid, only wavelength of light \[(\lambda )\] changes. Wavelength of light in liquid, \[\lambda '=\frac{\lambda }{n}'\] where n = refractive index of medium. Initial fringe width, \[\beta =\frac{D\lambda }{d}\] ?(i) Fringe width in liquid, \[\beta '=\frac{D\lambda '}{d}\] ?(ii) On dividing Eq. (ii) by Eq. (i), we get \[\frac{\beta '}{\beta }=\frac{\lambda '}{\lambda }\] or \[\frac{\beta '}{\beta }=\frac{\frac{\lambda }{n}}{\lambda }=\frac{1}{n}\] or \[\beta '=\frac{\beta }{n}\] The new fringe width, \[\beta '=\frac{2.0}{1.33}mm=1.5\,mm\] Location of maxima on the screen is given by \[{{Y}_{n}}=\frac{D}{d}(n\lambda )\] Where symbols have their standard meanings. When apparatus is immersed in the liquid, wavelength of light decreases. Thus, maxima gets closer to centre of the screen.
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