Answer
$325\;\rm nm$
Work Step by Step
As we see in the figure below, the first ray experiences a $\pi$ phase change while the second one does not.
So, we have here one phase change from the two reflected rays and we got a constructive interference for two wavelengths 390 nm, and 650 nm.
To find the minimum thickness that allows for these both colors to appear, we need to find the common thickness that works for both.
In our case, the thickness is given by
$$t=(m+\frac{1}{2})\dfrac{\lambda_n}{2}$$
$$\boxed{t=(m+\frac{1}{2})\dfrac{\lambda}{2n}}$$
For $\lambda=390$ nm:
when $m=0$;
$$t=(0+\frac{1}{2})\dfrac{390}{2\cdot 1.5}=\bf 65\;\rm nm$$
when $m=1$;
$$t=(1+\frac{1}{2})\dfrac{390}{2\cdot 1.5}=\bf 195\;\rm nm$$
when $m=2$;
$$t=(2+\frac{1}{2})\dfrac{390}{2\cdot 1.5}=\color{red}{\bf 325}\;\rm nm$$
when $m=3$;
$$t=(3+\frac{1}{2})\dfrac{390}{2\cdot 1.5}=\bf 455\;\rm nm$$
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For $\lambda=650$ nm:
when $m=0$;
$$t=(0+\frac{1}{2})\dfrac{650}{2\cdot 1.5}=\bf 108\;\rm nm$$
when $m=1$;
$$t=(1+\frac{1}{2})\dfrac{650}{2\cdot 1.5}=\color{red}{\bf 325}\;\rm nm$$
when $m=2$;
$$t=(2+\frac{1}{2})\dfrac{650}{2\cdot 1.5}=\bf 541\;\rm nm$$
when $m=3$;
$$t=(3+\frac{1}{2})\dfrac{650}{2\cdot 1.5}=\bf 758\;\rm nm$$
Therefore, the minimum thickness is 325 nm, as you see above.