Question

The $X-Z$ plane separates two media $A$ and $B$ of refractive indices ${\mu }_A=1.5$ and ${\mu }_B=2.$ A ray of light travels from $A$ to $B$, its direction in two media are given by unit vectors. ${\hat{u}}_A=a\hat{i}+b\hat{j}$ and ${\hat{u}}_B=c\hat{i}+d\hat{j}$ respectively, 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 $\frac{a}{c}=\frac{4}{3}$

Given that,

Refractive indices ${\mu }_A=1.5$

Refractive indices ${\mu }_B=2$

Unit vector ${\overrightarrow{u}}_A=a\hat{i}+b\hat{j}$

Unit vector ${\overrightarrow{u}}_B=c\hat{i}+d\hat{j}$

We know that,

As $a\hat{i}+b\hat{j}$ and $c\hat{i}+d\hat{j}$ are unit vectors their x components represents $sin\theta$

So, by Snell’s law

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

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

$a\times 1.5=c\times 2$

$\frac{a}{c}=\frac{2}{1.5}$

$\frac{a}{c}=\frac{4}{3}$

This is the answer.