A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in visible light with wavelength 638 nm in vacuum. Assuming that the visible spectrum extends from 380 to 750 nm, what is the longest visible wavelength (in vacuum) for which the film will appear bright due to constructive interference? Number i Unitslet refractive index of oil Isn. wavelength of Light 1 = 638 mm = 638x 10-9m condition for destructive interference 2nt = m * = thickness of film m= order of spectrum so , for least possible thickness m = 1 hence , 2nt = 1 Zht = 638 nmwavelength range of visible spectrum is soonm to 7sonm condition for constructive interference - 2nt = ( m+ 1/2)d 1 = 2 nt /( m + 1 / 2 ) 1 = ( 638 hm ) /(m+ 1 /2 ) ( from ep D, 2nt = 630hm) here , m = 0 1 1 , 2 13 if m = 0 1 = 630 = 12 76 hmif mel, A = 6 38 638 x 2 3 = 425. 34 mm = 4 25. 34 x10 - m This is the only possible wavelength in visible range - required wavelength = 425. 34 hm. = 425. 34 x10 - mlongest visible wave ength ( in vaccum for which the film will appear bright due to constructive interference = 425. 34 nm