Question

Note: this question contains numbers but it is possible to determine the correct answer without any calculations. A double slit experiment is carried out with the apparatus immersed in water (nW=1.33). The two slits are separated by d = 0.2 mm and the detector, which functions underwater, is a distance R = 4 m from the slits. When light of wavelength 400 nm in water is incident on the slits, the first-order bright fringes are observed at a distance y = 8 mm from the center of the central bright fringe. If the same experiment is carried out with the apparatus immersed in a transparent oil (nO=1.4), what can you say about the new positions ynew of the first-order bright fringes?

I have the answer being Ynew < 8mm but I do not understand why.

Answer 1

we know that the fringe width is y = m*L*D/d

   m order of the fringe
   L wavelength of the light
   D separtion of slit and screen
   d slit width

and wavelength of the ligh will change with refractive index L m = L/n_m

   so the ration of fringe width

   y oil/yw = L_o /L_w = (L/n_o)(n_w/L)

   y oil = n_w*yw/n_O

substituting the values
  
   y_oil = 1.33*8/1.4 = 7.6 mm

so the new fringe width is 7.6 mm which is less than 8 mm