What is the minimum (non-zero) thickness of a soap film having index of refraction 1.46, with air on both sides, that transmits virtually 100% of 500 nm light incident along the normal? a. 345 nm. b. 86 nm. c. Any thickness transmits 100%. d. 172 nm. I used 5.0x10^-9m/(1.46x20 and got 1.71x10^-9 which would be d, 172 nm, but not according to sig figs, so did I do this right. I know the equation is for constructive interference but is that the same? Thanks for the help if you can!