Answer
a) See the answer below.
b) $290\;\rm nm$
Work Step by Step
a) The thickness of the oil film at A is too small and when the first reflected light beam from the upper surface of the oil film experience a $180^\circ$ phase change while the second reflected light beam will experience no phase change. The two beams will interact destructively since the thickness of the film at A is assumed to be negligible in comparison to the light wavelengths.
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b) At $t$, we have a constructibe interference for the yellow light which has a wavelength of 580 nm.
This means that the path length difference which is twice the thickness of the film when we got a constructive interference with one phase change is given by
$$2t=\left(m+\frac{1}{2}\right) \lambda_n $$
Thus,
$$ t=\left(m+\frac{1}{2}\right) \dfrac{\lambda_n }{2}=\left(m+\frac{1}{2}\right) \dfrac{\lambda }{2n_{oil}}$$
Now we can see that we have two yellow fringes and our point is the second one. We know that at the first one $m=0$, then for the second yellow fringe, $m=1$.
$$ t= \left(1+\frac{1}{2}\right) \dfrac{\lambda }{2n_{oil}}$$
Plugging the known;
$$ t= \left( \frac{3}{2}\right) \dfrac{580 }{2\cdot 1.5}=\color{red}{\bf 290}\;\rm nm$$
