Answer
$482\;\rm nm$
Work Step by Step
As we see in the figure below, the two rays experience a $\pi$ phase change.
So, we have here two phase changes from the two reflected rays and we got a constructive interference for two wavelengths 525 nm, and 655 nm.
To find the thickness that allows for both colors to appear, we need to find the common thickness that works for both.
For $525$ nm: In this case, with a two-phase change, the path length difference for destructive interference is given by
$$2t= \left(m+\frac{1}{2}\right) \lambda_n$$
Thus, the film thickness is given by
$$ t=\dfrac{ \left(m+\frac{1}{2}\right)\lambda}{2n}$$
when $m=0$;
$$ t=\dfrac{\left(0+\frac{1}{2}\right) 525}{2\cdot 1.36}=\bf 96.5\;\rm nm$$
when $m=1$;
$$ t=\dfrac{\left(1+\frac{1}{2}\right) 525}{2\cdot 1.36}=\bf 289.5\;\rm nm$$
when $m=2$;
$$ t=\dfrac{\left(2+\frac{1}{2}\right) 525}{2\cdot 1.36}=\color{blue}{\bf 482.54}\;\rm nm$$
when $m=3$;
$$ t=\dfrac{\left(3+\frac{1}{2}\right) 525}{2\cdot 1.36}=\bf 675\;\rm nm$$
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For $655$ nm: In this case, with a two-phase change, the path length difference for constructive interference is given by
$$2t=m \lambda_n$$
Thus, the film thickness is given by
$$ t=\dfrac{ m\lambda}{2n}$$
when $m=1$;
$$ t=\dfrac{1\cdot 655}{2\cdot 1.36}=\bf 241\;\rm nm$$
when $m=2$;
$$ t=\dfrac{2\cdot 655}{2\cdot 1.36}=\color{blue}{\bf 481.62}\;\rm nm$$
when $m=3$;
$$ t=\dfrac{2\cdot 655}{2\cdot 1.36}=\bf 722\;\rm nm$$
when $m=4$;
$$ t=\dfrac{4\cdot 655}{2\cdot 1.36}=\bf 963\;\rm nm$$
Now we found that there are two approximately equal thickness, so we can take the average of both results.
$$t=\dfrac{482.54+481.62}{2}=482.08\;\rm nm$$
Therefore,
$$t=\color{red}{\bf 482}\;\rm nm$$
