= Solution
Write $\lambda_0$ for the vacuum <wavelength>. Initially the identical air-filled boxes add equal optical paths. When the <refractive index> in one box falls from $n_{\rm air}$ to $n(P)$, the extra path in the box that remains full is
$$
\Delta(P)=l[n_{\rm air}-n(P)].
$$
Identical fixed window contributions cancel. At the screen centre the geometric slit-path difference is zero, so, neglecting unequal transmission changes,
$$
I(0,P)=I_{\max}\cos^2\left(\frac{\pi l[n_{\rm air}-n(P)]}{\lambda_0}\right).
$$
Each extra optical path of one <wavelength> produces one full intensity period and one new maximum. Taking the stated count in the usual sense of 20 completed fringe periods during evacuation, \b[excluding the initial maximum], and using $n(0)=1$ for vacuum gives
$$
N=\frac{l(n_{\rm air}-1)}{\lambda_0},\qquad\boxed{n_{\rm air}=1+\frac{N\lambda_0}{l}}.
$$
Taking $N=20$, vacuum <wavelength> $\lambda_0\simeq460\ \mathrm{nm}$ and $l=30\ \mathrm{mm}$,
$$
\boxed{n_{\rm air}-1\simeq3.07\times10^{-4},\qquad n_{\rm air}\simeq1.00031}.
$$
If the quoted $460\ \mathrm{nm}$ instead denotes the <wavelength> $\lambda_{\rm air}$ in ambient air, use $\lambda_0=n_{\rm air}\lambda_{\rm air}$ and the exact relation $n_{\rm air}=[1-N\lambda_{\rm air}/l]^{-1}$. It gives $1.00030676$, differing by less than $10^{-7}$ from the vacuum-wavelength estimate, well below one-fringe counting resolution.
The count convention matters: 20 maxima including both the initial and final maxima span only 19 periods, giving approximately $1.000291$. If the final fractional fringe is not measured, an integer count alone has a resolution of order $\lambda_0/l\simeq1.53\times10^{-5}$ in <refractive index>. The standard 20-period estimate should therefore not be interpreted as an arbitrarily precise index measurement.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-resit-young-fringe-count.png]
{title=Optical-path phase shift and 20 completed centre-intensity fringe periods}
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