For two structure namely $S_1$ with $n_1=\sqrt{45}/4$ and $S_2$ with $n_1=8/5$ and $n_2=7/5$ and taking the refractive index of water to be $4/3$ and that of air to be $1$, the correct option(s) is (are)

$Case-1$

$S_1$ $\text{immersed in }\text{ air }$

$n_1=\frac{\sqrt{45}}{4}{n}_2=\frac{7}{5}$

$\text{ (1) }\sin 1=\frac{\sqrt{45}}{4}\sin \gamma \to {\sin }^{2}i={\left(\frac{45}{4}\right)}^2\left(1-{\cos }^2\gamma \right)$ 

$\left(\frac{\sqrt{45}}{4}\right)\cos \gamma ⩾\left(\frac{7}{5}\right)1-\frac{16}{45}{\sin }^{2}i={\cos }^{2}x-(1)\text{ Squaring }$

$\begin{array}{l}\frac{45}{141^2}{\cos }^2\gamma ⩾\frac{49}{25}\\ \frac{45}{(4{)}^2}\left(1-\frac{16}{45}{\sin }^{2}i\right)⩾\frac{49}{25}\\ \frac{45}{(4{)}^2}-\frac{4{\sin }^{2}i}{4}⩾\frac{49}{25}\\ \frac{45}{16}-\frac{49}{25}⩾{\sin }^{2}i\end{array}$

$$\begin{gathered} \text{ Case }-\text{ (2) } \\ {\text{ S }}_2\text{ immersed in water. } \end{gathered}$$

$\begin{array}{l}\frac{4}{3}\sin i=\frac{\sqrt{45}}{4}\sin \gamma \\ \frac{16}{9}{\sin }^{2}i=\frac{45}{16}{\sin }^2\gamma \Rightarrow 1-\frac{(16{)}^2}{9(45)}{\sin }^{2}1={\cos }^2\gamma \\ \frac{45}{16}{\cos }^2\gamma ⩾\frac{49}{25}\\ \frac{45}{16}\left[1-\frac{(16{)}^2}{9(45)}{\sin }^{2}1\right]⩾,\frac{49}{25}\Rightarrow \frac{45}{16}-\frac{49}{25}⩾{\sin }^{2}i\\ \text{ So option }\text{ c is correct }\end{array}$