Question
A transparent sheet of paper of refractive index 'n' and thickness 't' is pasted on one of the slits in YDSE which uses monochromatic light of wavelength 'x' . How many fringes will cross through the centre if the paper is removed ?
Joshi sir comment
total path difference created by the paper = (n-1)t
geometrical path difference = yd/D
so position of central maxima will be obtained by eq. (n-1)t = yd/D
here y is position of central bright, d is slit appeture and D is distance between slits and screen.
so y = (n-1)tD/d
width of maxima = xD/d , x is wavelength of light
so no. of fringes shifted = [(n-1)tD/d]/[xD/d] = (n-1)t/x