Question

The X-Z plane separates two media A and B of refractive indices ${\mu }_1=1.5$ and ${\mu }_2=2.A$ A ray of light travels from A to B. Its directions in the two media are given by unit vectors $\overrightarrow{[}{u}_1]=a\hat{i}+b\hat{j}$ and $\overrightarrow{[}{u}_1]=c\hat{i}+d\hat{j}$ Then -

• A. $\frac{a}{c}=\frac{4}{3}$
• B. $\frac{a}{c}=\frac{3}{4}$
• C. $\frac{b}{d}=\frac{4}{3}$
• D. $\frac{b}{d}=\frac{3}{4}$

Answer 1

The correct option is A $\frac{a}{c}=\frac{4}{3}$

$\begin{array}{c}\text{given}\\ \begin{array}{c}{\mu }_A=1.5\\ {\mu }_B=2\end{array}\\ \text{unid vector}\begin{array}{cc}{\text{u}}_A^{\prime }& =a\hat{i}+b\hat{j}\\ u_B^{\prime }& =c\hat{r}+d\hat{j}\end{array}\\ \text{we know that}\end{array}$

As $a\hat{i}+b\hat{j}$ and $c\hat{r}+d\hat{j}$ are unit vecter their

$x$ component represents sinQ

$\text{So by snell'e law}$

${\mu }_A\sin i={\mu }_B\sin r$

${\mu }_A\left(\frac{a}{\sqrt{a^2+b^2}}\right)={\mu }_B\left(\frac{c}{\sqrt{c^2+d^2}}\right)$

$\begin{array}{c}a\times 1.5=c\times 2\\ \frac{a}{c}=\frac{4}{3}\\ \text{hence option}A\text{is coracet.}\end{array}$