• 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?

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.