Question

A ray of light travelling along the line $x+y=1$ is incident on the x-axis and after refraction it enters the other side of the x-axis by turning $\frac{\pi }{6}$ away from positive direction of x-axis. The equation of the line along which the refracted ray travels is

• A. $x+(2-\sqrt{3})y=1$
• B. $(2-\sqrt{3})x+y=1$
• C. $y+(2+\sqrt{3})x=2+\sqrt{3}$
• D. none of these

Answer 1

Answer: (A), (C)

The correct options are
A $x+(2-\sqrt{3})y=1$
C $y+(2+\sqrt{3})x=2+\sqrt{3}$

Incident ray along $x+y=1$ can approach $x-$axis either from above or from below as shown in figure.
Accordingly the refracted ray will be along $l_1$ or $l_2$ after $\frac{\pi }{6}$ rotation away from positive direction of $x-$axis as shown.
Both $l_1$ and $l_2$ will pass through $(1,0)$ on $x-$axis
From the figure, slope of $l_1$ is $\tan ({135}^o-{30}^o)$
$=\frac{-1-\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=-\frac{\sqrt{3}+1}{\sqrt{3}-1}=-(2+\sqrt{3})$
Slope of $l_2$ is $\tan ({135}^o+{30}^o)$
$=\frac{-1+\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=-\frac{\sqrt{3}-1}{\sqrt{3}+1}=-(2-\sqrt{3})$
.
Equation of $l_1$ is $y-0=(x-1)[-(2+\sqrt{3}\left)\right]$
$\Rightarrow y+(2+\sqrt{3})x=2+\sqrt{3}$
Multiplying both sides by $(2-\sqrt{3})$,
$\Rightarrow y(2-\sqrt{3})+x=1$
.
Equation of $l_2$ is $y-0=(x-1)[-(2-\sqrt{3}\left)\right]$
$\Rightarrow y+(2-\sqrt{3})x=2-\sqrt{3}$, which is not in the options.
Hence, only option A and C are correct.