Given formula: $2n_{oil}t = (2m-1)\frac{\lambda}{2}$
Substitute $n_{oil} = 1.5$, $t = 0.40 \mu m$, and $m = 1$ into the formula.
We get: $2(1.5)(0.40) = (2(1)-1)\frac{\lambda}{2}$
$2.4 = \frac{\lambda}{2}$
$4.8 = \lambda$
So, for the first reflection at the
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