Question

A ray of light travelling along the line $x+y=1$ is inclined on the $x$-axis and after refraction it enters the other side of the $x$-axis by turning ${15}^{\circ }$ away from the $x$-axis. The equation of the line along which the refraction ray travels is

• A. $\sqrt{3}y-x+1=0$
• B. $\sqrt{3}y+x+1=0$
• C. $\sqrt{3}y+x-1=0$
• D. None of these

Answer 1

The correct option is D None of these

Slope of the refracted ray is $m=\tan (180-60)=\tan 120=-\sqrt{3}$

We have the point where $x+y=1$ meets $x$ axis i.e.$(1,0)$, the refracted line will also pass through the same point.

So, we have the standard point slope form given as

$(y-y_1)=m(x-x_1)$

We have the line as

$y=-\sqrt{3}(x-1)$

$y+\sqrt{3}x=\sqrt{3}$