Physics - Geometrical Optics Question with Solution | TestHub

Ans. $\vec{r}=\frac{3}{5\sqrt{2}}\hat{i}+\frac{2\sqrt{2}}{5}\hat{j}-\frac{1}{\sqrt{2}}\hat{k}$ (angle of incidence $=60^0;r=45^0$ ) From the figure, it can be seen that normal to the boundary separating the two media will be in z -direction, as incident ray, refracted ray and normal are in the same plane The angle made by the incident ray with the normal is given as

$\cos i=\frac{A_z}{A}=\frac{10}{\sqrt{(6\sqrt{3}{)}^2+(8\sqrt{3}{)}^2+100^2}}=\frac{10}{\sqrt{108+192+100}}=\frac{10}{20}$ $\Rightarrow \cos \mathrm{i}=\frac{-1}{2}\Rightarrow \mathrm{i}={\cos }^1\left(\frac{1}{2}\right)\Rightarrow \mathrm{i}=60^{\circ }$ Applying Snell's law of refraction at the interface of two media ${\mathrm{n}}_1\mathrm{sini}={\mathrm{n}}_2\mathrm{sinr}$ $\sqrt{2}\sin 60^{\circ }=\sqrt{3}\sin \mathrm{r},\sin \mathrm{r}=\frac{\sqrt{2}\times \sqrt{3}}{\sqrt{3}\times 2}=\frac{1}{\sqrt{2}}$ $\Rightarrow \mathrm{r}={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)\Rightarrow \mathrm{r}=45^{\circ }$ As the $\mathrm{X},\mathrm{Y}$ components remain unchanged, so unit vector along direction of refracted ray is $\hat{R}=\frac{6\sqrt{3}\hat{i}+8\sqrt{3}\hat{j}}{\sqrt{(6\sqrt{3}{)}^2+(8\sqrt{3}{)}^2}}\sin 45+\cos 45(-\hat{k})$ $\hat{\mathrm{R}}=\frac{3}{5\sqrt{2}}\hat{\mathrm{i}}+\frac{4}{5\sqrt{2}}\hat{\mathrm{j}}-\frac{1}{\sqrt{2}}\hat{\mathrm{k}},\hat{\mathrm{R}}=\frac{1}{5\sqrt{2}}(3\hat{\mathrm{i}}+4\hat{\mathrm{j}}-5\hat{\mathrm{k}})$ Hence, the answer is $\frac{1}{5\sqrt{2}}(3\hat{\mathrm{i}}+4\hat{\mathrm{j}}-5\hat{\mathrm{k}})$