question_answer

In Young's experiment, the width of fringes obtained with light of wavelength \[6000\text{ }\overset{{}^\circ }{\mathop{A}}\,\] is 2 mm. What will be the fringe width, if the entire apparatus is immersed in a liquid of refractive index 1.33? Also, explain how location of the maxima will be affected?

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.