Solution:

We know that,

NA $=\sqrt{\frac{n_1^2-n_2^2}{n_0^2}}$

Given,

$\quad \mathrm{NA}=0.2$ in air $\left(n_0=1\right)$ and $n_2=1.59$

$ \begin{array}{ll}\\ \therefore & 0.2=\sqrt{n_1^2-(1.59)^2} \\\\ & n_1=1.602 . \end{array} $

Now we have to find the acceptance angle in water i.e.

$ n_o=1.33 $

$ \mathrm{NA}=\sin \alpha=\sqrt{\frac{n_1^2-n_2^2}{n_o^2}} \text { or } \sin \alpha=\frac{0.2}{1.33}=0.15\\ $

$\alpha=8.64^{\circ}$