Solution:

We know that the numerical aperture $N A$ is given by,

$ N A=\sqrt{\frac{n_1^2-n_2^2}{n_0^2}}=\frac{(N A)_{\text {ait }}}{n_0} $

For air $n_0=1$. Given $n_2=1.50$ and $\mathrm{NA}=0.15$

$ \therefore \quad 0.15=\sqrt{n_1^2-(1.50)^2} $

$ n_=1.5075 \text {, the refractive index of the core. } $

Also, Given,

$n_0=1.30$ for water and $(N A)_{\text {sit }}=0.15$

We know, $\quad N A=\frac{(N A)_{\text {air }}}{n_0}$

$ \therefore \quad(N A)_{\text {water }}=\frac{0.15}{1.30}=0.1154 $