Question

A ray of light traveling 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 $15°$ 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 the above

Answer 1

The correct option is D

None of the above

Step 1: Given data

Equation of line $=x+y=1$

The angle of the refracted ray with $x$-axis$=15°$

Step 2: Formula used

The slope of the line $m=\tan \theta$

The point-slope form $=(y–y_1)=m(x–x_1)$

Step 3: Solution

The diagrammatic representation of the given data is:

As the slope of given line $x+y=1$ is $-1$,

Then $\theta =135°$, and in case if it's going straight, then the rest of the angle is $180°-135°=45°$

But it bends furthermore $15°$ about the x-axis, So total angle of inclination is $45°+15°=60°$

The slope of the line after bending, $m$ $=\tan (180°-60°)[Fromgivendiagram]$

$$\begin{gathered} =\tan 120° \\ =-\sqrt{3} \end{gathered}$$

Slope point form of the line at point $(1,0)$

$=(y–y_1)=m(x–x_1)$

$$\begin{gathered} \Rightarrow (y-0)=-\sqrt{3}(x-1) \\ \Rightarrow (y-0)=-\sqrt{3}x+\sqrt{3} \\ \Rightarrow y+\sqrt{3}x=+\sqrt{3} \end{gathered}$$

Hence, the correct option is $D$.