Answer
a) $180\;\rm nm$
b) $361\;\rm nm$, $541\;\rm nm$
c) See the answer below.
Work Step by Step
a) The soap film is surrounded by air on both sides, so the second reflected light beam will reflect without phase change while the first light beam will reflect with a $180^\circ$ (or $\pi$) phase change.
See the figure below.
This means that we have destructive interference from the film.
This means that for the smallest thickness of a soap film, its thickness is half the length of the wavelength.
$$ t=\dfrac{\lambda_n}{2}=\dfrac{\lambda}{2n}=\dfrac{480}{2\cdot 1.33}=\color{red}{\bf180.45}\;\rm nm$$
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b) The two other thicknesses that gave us black interference are $1.5\lambda$ and $2.5\lambda$.
$$t=\dfrac{m\lambda_n}{2}\tag{$m=1,2,3,..$}$$
Thus, when $m=1$, we got the smallest thickness above.
when $m=2$
$$t_1=\dfrac{2\lambda}{2n}=\dfrac{2\cdot 480}{2\cdot 1.33}=\color{red}{\bf 360.9}\;\rm nm$$
And when $m=3$;
$$t_2=\dfrac{3\lambda}{2n}=\dfrac{3\cdot 480}{2\cdot 1.33}=\color{red}{\bf 541.35}\;\rm nm$$
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c) If the thickness of the film was much less than the wavelength, the two reflected rays will still have a phase change difference of $\pi$ which gave us a destructive interference.
The thickness will give a very small phase change that is too small to affect the final interaction of the two reflected rays.