Question

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)

• A. $NA$ of $S_1$ immersed in water is the same as that of $S_2$ immersed in a liquid of refractive index $\frac{16}{3\sqrt{15}}$
• B. $NA$ of $S_1$ immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $S_2$ immersed in water
• C. $NA$ of $S_1$ placed in air is the same as that of $S_2$ immersed in liquid of refractive index $\frac{4}{\sqrt{15}}$
• D. $NA$ of $S_1$ placed in air is the same as that of $S_2$ laced in water

Answer 1

The correct option is C $NA$ of $S_1$ placed in air is the same as that of $S_2$ immersed in liquid of refractive index $\frac{4}{\sqrt{15}}$

$Case-1$

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

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

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

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

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

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

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