Question

In this problem, use 473 nm for the wavelength of blue light and the index of refraction of oil to be 1.40. Calculate the minimum thickness (in nanometers) of an oil slick on water that appears blue when illuminated by white light that is incident perpendicular to the oil's surface.

Answer 1

di 473 mm wervelength of klere light, the index of refraction of all, no 2.40 When, the light is reflected from i the oil surface, the path difference 25 given by path difference, (opfical path difference) + Again the optical bath diffenence on teams of thickness of the Chonkeling to lhickness of thinkslon optreed path of ffenonef= ant So, equation 0 given by ( - enten har behom of 2 this film becmes. gott X2

Fon cedructive confenference, the path defense, captal puth als frente) should be equal to mi m), 80 where einlegen m=positive 2nd +-ml mst, 2,3 the values m = 0, 1, 2, take Chaud if com anda ml - (m.b) Replacing in by mts, so the above equation becomes. 2nd - Hom+1-12) - (m+la) of thickness of the thon bils of oil A (m +0.5) t = t (m +0.5) an for thickness to be minemum, t Admin todo m = 0 and = (0.5)( 473 nm) 0.5to 20 so Amin = 84.464 nm 2(1.40) Amin = 84.46hnm

Comment

The minimum thickness of the oil thin-film was found to be be 84.464 nm