Explanation:
: Given, \(\lambda=600 \mathrm{~nm}, \mu=1.5\)
We know that,
For constructive interference-
\(2 \mu \mathrm{t}=(2 \mathrm{n}-1) \frac{\lambda}{2}\)
When, \(\mathrm{n}=0,1,2,3 \ldots\).
For minimum thickness, \(\mathrm{n}=0\)
\(2 \mu t=\frac{\lambda}{2}\)
Thicknes
\(\operatorname{ss}(\mathrm{t}) =\frac{\lambda}{4 \mu}=\frac{600}{4 \times 1.5}=\frac{600}{6}\)
\(=100 \mathrm{~nm}\)