IIT JEE , AIEEE Forum - Physics , Optics

rakesh61

Interference fringes are produced in youngs double slit experiment using light of wavelength 5000A When a film of material 2.5 * 10^-3 m thick was placed over one of the slits the fringe pattern shifted by a distance equal to 20 fring width

the refractive index of material of the film is

Ans 1.33 Please give detailed solution rates assured for satisfactory ans

pannaguma

fringe width = lambda*D/d = 20Y

optical path difference = (myu-1)2.5*10^-3 = dsin@ = dY/D
gives Y=(myu-1)2.5*10^-3 D/d

equate both.

Ask & Discuss Questions with Community & Experts