Answer
Please see the work below.
Work Step by Step
(a) We know that
$ t=\frac{(m+\frac{1}{2})\lambda}{2n}$
for $ m=0$
$ t=\frac{(0.5)(590\times 10^{-9}m)}{2(1.33)}=111.0nm $
for $ m=1$
$ t=\frac{(1.5)(590\times 10^{-9}m)}{2(1.33)}=333nm $
Thus, the two minimum thicknesses are $111nm $ and $333nm $.
(b) We can obtain the two minimum thicknesses that destructively interfere as follows:
$ t=\frac{m\lambda}{2n}$
for $ m=1$
$ t=\frac{590\times 10^{-9}m}{2(1.33)}$
$ t=222nm $
and for $ m=2$
$ t=\frac{2(590\times 10^{-9}m)}{2(1.33)}$
$ t=444nm $
Thus, the two minimum thicknesses that reflect constructively are $222nm $ and $444nm $.
