Question

The $x$ - $z$ plane separates two media $A$ and $B$ of refractive indices ${\mu }_A=2$ and ${\mu }_B=3$. A ray of light travels from $A$ to $B$ and its direction in the two media is given by the unit vectors. $\widehat{U_A}=a\hat{i}+b\hat{j}$ and ${\hat{U}}_B=c\hat{i}+d\hat{j}$ respectively. Then, which of the following is true?

• A. $\frac{a}{c}=\frac{3}{2}$
• B. $\frac{a}{c}=\frac{2}{3}$
• C. $\frac{b}{d}=\frac{3}{2}$
• D. $\frac{b}{d}=\frac{2}{3}$

Answer 1

The correct option is A $\frac{a}{c}=\frac{3}{2}$

Light ray is getting refracted from rarer to denser medium because ${\mu }_A<{\mu }_B$.

Rarer medium $({\mu }_A=2)$
Denser medium $({\mu }_B=3)$

$\widehat{U_A}=a\hat{i}+b\hat{j}$
It represents unit vector in the direction of incident ray.
$\Rightarrow |\widehat{U_A}|=1$
$\sqrt{a^2+b^2}=1-----\left(1\right)$
$\Rightarrow \sin i=\frac{(\widehat{U_A}{)}_x}{\sqrt{a^2+b^2}}=\frac{a}{\sqrt{a^2+b^2}}$

Similarly $\widehat{U_B}=c\hat{i}+d\hat{j}$
$\Rightarrow \sin r=\frac{(\widehat{U_B}{)}_x}{\sqrt{c^2+d^2}}=\frac{c}{\sqrt{c^2+d^2}}$

Applying Snell's law at the interface of $A$ and $B$
${\mu }_A\sin i={\mu }_B\sin r$
$\Rightarrow \frac{2a}{\sqrt{a^2+b^2}}=\frac{3c}{\sqrt{c^2+d^2}}----\left(2\right)$

Since $\widehat{U_A}$ and $\widehat{U_B}$ represents unit vector along incident and refracted ray direction.
$\Rightarrow \sqrt{a^2+b^2}=1$
$\Rightarrow \sqrt{c^2+d^2}=1$
Substituting in Eq. $\left(2\right)$
$2a=3c$
$\Rightarrow \frac{a}{c}=\frac{3}{2}$

Why this Question? Concept: Importance of unit vectors in finding directional sines and cosines is represented here.