• # question_answer The xz plane separates two media A and B with refractive indices ${{\mu }_{1}}$ & ${{\mu }_{2}}$ respectively. A ray of light travels from A to B. Its directions in the two media are given by the unit vectors, ${{\vec{r}}_{A}}=a\,\hat{i}+b\,\hat{j}$ & ${{\vec{r}}_{B}}=\alpha \,\hat{i}+\beta \,\hat{j}$ respectively where $\hat{i}$ & $\hat{j}$ are unit vectors in the x & y directions. Then: A) ${{\mu }_{1}}a={{\mu }_{2}}\,\alpha$ B) ${{\mu }_{1}}\,\alpha ={{\mu }_{2}}\,a$ C) ${{\mu }_{1}}\,b={{\mu }_{2}}\,\beta$ D) ${{\mu }_{1}}\,\beta ={{\mu }_{2}}\,b$

 We have $\cos \theta =\frac{\vec{A}\cdot \vec{B}}{ab}$ Similarly here $\cos i=\frac{{{{\vec{r}}}_{A}}\cdot \hat{j}}{|{{r}_{A}}|\cdot |1|}=\frac{(a\hat{i}+b\hat{j})\cdot \hat{j}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$    and $\cos r=\frac{{{{\vec{r}}}_{B}}\cdot \hat{j}}{|{{r}_{B}}|\cdot |1|}$$=\frac{(\alpha \hat{i}+\beta \hat{j})\cdot \hat{j}}{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}}=\frac{\beta }{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}}$
 From shell's law${{\mu }_{1}}\sin i={{\mu }_{2}}\sin r$$\Rightarrow$${{\mu }_{1}}\sqrt{1-{{\cos }^{2}}i}={{\mu }_{2}}\sqrt{1-{{\cos }^{2}}r}$$\Rightarrow$${{\mu }_{1}}\sqrt{1-\frac{{{b}^{2}}}{{{b}^{2}}+{{a}^{2}}}}={{\mu }_{2}}\sqrt{1-\frac{{{\beta }^{2}}}{{{\alpha }^{2}}+{{\beta }^{2}}}}$ $\Rightarrow$${{\mu }_{1}}\frac{a}{\sqrt{{{b}^{2}}+{{a}^{2}}}}={{\mu }_{2}}\frac{\alpha }{\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}}$ both ${{\vec{r}}_{A}}$ & ${{\vec{r}}_{B}}$ are unit vectors so magnitudes of both vectors is 1 i.e.            $\sqrt{{{b}^{2}}+{{a}^{2}}}=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}=I$ So ${{\mu }_{1}}a={{\mu }_{2}}\alpha$