Question

Angle made by light ray with the normal in medium of refractive index $\sqrt{2}$ is

• A. ${\sin }^{-1}\sqrt{\frac{3}{8}}$
• B. ${\sin }^{-1}\sqrt{\frac{3}{4}}$
• C. ${\sin }^{-1}\sqrt{\frac{3}{6}}$
• D. ${\sin }^{-1}\sqrt{\frac{3}{10}}$

Answer 1

The correct option is A ${\sin }^{-1}\sqrt{\frac{3}{8}}$

we will apply shell's law for

the medium of refractive index

1.6 & $\sqrt{3}$

$n_1\sin {30}^{\circ }=n_2\sin \gamma$

$\sin {\gamma }_1=\frac{n_1}{n_2}\times \sin {30}^{\circ }=\frac{1}{1.6}\times \frac{\sqrt{3}}{2}$

$\sin {\gamma }_1=\frac{\sqrt{3}}{3\cdot 2}$

$\begin{array}{c}\text{Now, we will apply shell's law for}\\ \text{the medium of refractive index}\\ 1.6\text{\&}\sqrt{2}\text{-}\\ \begin{array}{c}{n}_2\sin {\gamma }_1=n_3\sin {\gamma }_2\\ \Rightarrow 1.6\times \frac{\sqrt{3}}{3.2}=\sqrt{2}\sin {\gamma }_2\\ \Rightarrow \sin {\gamma }_2=\frac{\sqrt{3}}{2\sqrt{2}}=\sqrt{\frac{3}{8}}\\ \text{:}\text{. Required answer}=\sqrt{\frac{3}{8}}\end{array}\end{array}$