Answer
a. 80.9 nm
b. 243 nm.
Work Step by Step
See figure 35.11. Exactly one ray undergoes a $180^{\circ}$ phase change on reflection. There is a net phase difference introduced.
a. Since there is a half-cycle phase shift at just one of the interfaces, the condition for constructive interference is $2t=(m+\frac{1}{2})\lambda$.
For the thinnest film, m=0.
$$t=\frac{\lambda}{4} $$
Note that $\lambda=\frac{\lambda_0}{1.70}$.
$$t=\frac{\lambda_0}{4(1.70)}=\frac{550nm}{4(1.70)}=80.9 nm $$
b. The next smallest thickness for constructive interference is with another half wavelength thickness added.
$$t=\frac{3 \lambda_0}{4(1.70)}=\frac{3(550nm)}{4(1.70)}=243 nm $$
