Read the paragraph and chose the correct answer of the following questions In young experiment position of bright fringes is given $ x = n \lambda { D \over d } $ and the positon of dark fringes is given by $ x = (2n -1 ) { \lambda \over 2 } { D \over d } $ where n = 1,2,3........... for first, second, third bright / dark fringe. The center of the fringe pattern is bright (for n = 0). The width of each briht/dark fringe is $ \lambda = 5000 A ^\circ $ . With the light of wavelength $ 5000 A^\circ $ , If experiment were carried out under water of a n = 4 /3 the fringe width would be
When the experiment is carried out in a medium with refractive index \( n \), the wavelength of light changes to \( \lambda' = \frac{\lambda}{n} \). Therefore, the new fringe width \( \beta' \) is given by \( \beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{nd} \). Thus, the fringe width in water (where \( n = \frac{4}{3} \)) will be \( \frac{3}{4} \) times the fringe width in air.
