Question

= 498.4 nm in vacuum) when viewed at normal incidence in white light. Determine the two smallest, non-zero thicknesses for the soap film if its index of refraction is 1.40. 125 nm and 375 nm O 97 nm and 179 nm 89 nm and 267 nm O 100 nm and 310 nm 158 nm and 346 nm

Answer 1

Sol- given = da 498.4 nm The formula that gives the maximum For interference in a soup bubble tilmis 2 x refretive Index x Film Thickness C (m+2) wavelength (m= 0,1,2) -9 201.4 x Hom thickness =(m+/2) X 498x10 m For the first possible thickness m=0 Him thickness = (498-4 X169 241.482 89 nm any m=1 For second possible thickness thickness so Flim (1+2) *498.42109 2x1-4 3 x 498.4 x 159 2 2X2 X3-4 267 hm Ang meo so 89 nm 267 nm Ang mal