Question

The $xz$ plane separates two media $A$ and $B$ with refractive indices ${\mu }_1$ and ${\mu }_2$ respectively. A ray of light travels from $A$ to $B$. Its directions in the two media are given by the unit vectors, $\overrightarrow{r_A}=a\hat{i}+b\hat{j}$ & $\overrightarrow{r_B}=\alpha \hat{i}+\beta \hat{j}$ respectively where $\hat{i}$ & $\hat{j}$ are unit vectors in the $x$ & $y$ directions. Then:

(correct answer + 2, wrong answer - 0.50)

• A. ${\mu }_{1}a={\mu }_2\alpha$
• B. ${\mu }_1\alpha ={\mu }_{2}a$
• C. ${\mu }_{1}b={\mu }_2\beta$
• D. ${\mu }_1\beta ={\mu }_{2}b$

Answer 1

The correct option is A ${\mu }_{1}a={\mu }_2\alpha$

We have
$\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{ab}$
Similarly here
$\cos i=\frac{\overrightarrow{r_A}\hat{j}}{\left|r_A\right|.\left|1\right|}=\frac{(a\hat{i}+b\hat{j})\hat{j}}{\sqrt{a^2+b^2}}$
$=\frac{b}{\sqrt{a^2+b^2}}$
and
$\cos r=\frac{\overrightarrow{r_A}\hat{j}}{\left|r_B\right|.\left|1\right|}=\frac{(\alpha \hat{i}+\beta \hat{j})\hat{j}}{\sqrt{{\alpha }^2+{\beta }^2}}$
$=\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}}$
From snell's law ${\mu }_1\sin i={\mu }_2\sin r$
$\Rightarrow {\mu }_1\sqrt{1-{\cos }^{2}i}={\mu }_2\sqrt{1-{\cos }^{2}r}$
$\Rightarrow {\mu }_1\sqrt{1-{\cos }^{2}i}={\mu }_2\sqrt{1-{\cos }^{2}r}$
$\Rightarrow {\mu }_1\sqrt{1-\frac{b^2}{b^2+a^2}}={\mu }_2\sqrt{1-\frac{{\beta }^2}{{\alpha }^2+{\beta }^2}}$
$\Rightarrow {\mu }_1\frac{a}{\sqrt{b^2+a^2}}={\mu }_2\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}}$
Both $\overrightarrow{r_A}$ and $\overrightarrow{r_B}$ are unit vector so magnitudes of both vector is $1$
i.e., $\sqrt{b^2+a^2}=\sqrt{{\alpha }^2+{\beta }^2}=1$
So ${\mu }_{1}a={\mu }_2\alpha$