In a Young's double slit experiment using monochromatic light of wavelength 620 nm , a transparent paper (refractive index = 1.45 ) with a thickness of 0.02 mm is pasted over one of the slits. Determine the number of fringes that will cross through the centre upon removal of the paper.

Step-by-step solution

The path difference introduced by the transparent paper is given by x = ( - 1)t . The number of fringes shifted across the centre upon removal of the paper is n = x = ( - 1)t . Substituting the given values: n = (1.45 - 1) 0.02 10⁻³ 620 10⁻⁹ n = 0.45 2 10⁻⁵ 6.2 10⁻⁷ n = 0.9 10⁻⁵ 6.2 10⁻⁷ n = 90 6.2 14.5 Thus, approximately 14.5 fringes will cross through the centre.