White light reflected at perpendicular incidence from a soap film has, in the visible specturm, an interference amximum at 6000Å and a minimum at 4500Å with no minimum in between. If `mu=4//3` for film, what is the film thickness ?

White light reflected at normal incidence from a soap film has minima at 6500Å and maxima at 7500Å in the visible region without minimum in between. If mu is (5)/(3) for the thin film, thickness of the film is

White light reflected at normal incidence from a soap film has minima at 6500Å and maxima at 7500Å in the visible region without minimum in between. If mu is (5)/(3) for the thin film, thickness of the film is

White light incident perpendicular on a soap film gets reflected . It has an interference maximum at 5800 Å and a minimum at 4200 Å in a visible spectrum. If there is no minimum in between the two bands, find the thickness of the soap film. Take refractive index of film = 4//3 .

White light reflected from a soap film (Refractive index =1.5 ) has a maxima at 600 nm and a minima at 450 nm at with no minimum in between. Then, the thickness of the film is

White light reflected from a soap film (Refractive index =1.5 ) has a maxima at 600 mm and a minima at 450 nm at with no minimum in between. Then, the thickness of the film is

White light reflected from a soap film (Refractive Index =1.5) has a maxima at 600 nm and a minima at 450 nm with no minimum in between. Then the thickness of the film is..... xx 10^-7m .

Light of wavelength 6000 Å is reflected at nearly normal incidence from a soap films of refractive index 1.4 The least thickness of the fringe then will appear black is