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  • A ray of light is sent along the line . After refracting across the line <img alt="\ma

A ray of light is sent along the line \mathrm{x-6 y=8}. After refracting across the line \mathrm{x+y= 1} it enters the opposite side after turning by \mathrm{15^{\circ}} away from the line \mathrm{x+y=1}. The equation of the line along which the refracted ray travels is \mathrm{(70-37 \sqrt{3}) x-13 y-153 +\mathrm{k} \sqrt{3}=0} where \mathrm{\mathrm{k}=}
 

Option: 1

\mathrm{k=72}


Option: 2

\mathrm{k=-72}


Option: 3

\mathrm{k=73}


Option: 4

\mathrm{k=74}


Answers (1)

best_answer

The point of intersection of \mathrm{x-6 y=8\: and \: x+y=1} is \mathrm{A \equiv(2-1)} Let the refracted ray have the slope \mathrm{=m} then

\tan 15^{\circ}=\left|\frac{\mathrm{m}-\frac{1}{6}}{1+\frac{\mathrm{m}}{6}}\right|=\left|\frac{6 \mathrm{~m}-1}{6+\mathrm{m}}\right|=\frac{\sqrt{3}-1}{\sqrt{3}+1}
\mathrm{\therefore \quad \frac{6 m-1}{6+m}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \: or\: \: \frac{1-\sqrt{3}}{1+\sqrt{3}}}
\mathrm{\Rightarrow \quad \frac{6 m-1}{6+\mathrm{m}}=2-\sqrt{3} \: or \: \sqrt{3}-2}
\mathrm{\Rightarrow \quad \mathrm{m}=\frac{70-37 \sqrt{3}}{13} \: \: or \: \: \mathrm{m}=\frac{37 \sqrt{3}-70}{61}}
Let the angle between \mathrm{x+y=1} and the line through \mathrm{A}(2,-1) with the slope \frac{70-37 \sqrt{3}}{13} be \alpha then
\tan \alpha=\left|\frac{\frac{70-37 \sqrt{3}}{13}-(-1)}{1-\left(\frac{70-37 \sqrt{3}}{13}\right)}\right|

Let the angle between \mathrm{x+y=1} and the line through \mathrm{A(2,-1)} with the slope \mathrm{\frac{70-37 \sqrt{3}}{13}} be \mathrm{\alpha} then
\begin{array}{r} \tan \alpha=\left|\frac{\frac{70-37 \sqrt{3}}{13}-(-1)}{1-\left(\frac{70-37 \sqrt{3}}{13}\right)}\right| \\ \end{array}

\begin{array}{r} =\left|\frac{83-37 \sqrt{3}}{37 \sqrt{3}-57}\right|=\frac{83-37 \sqrt{3}}{37 \sqrt{3}-57} \end{array}
and if angle between \mathrm{x+y=1} and the line through \mathrm{\mathrm{A}(2,-1)} with the slope \mathrm{\frac{37 \sqrt{3}-70}{61}} be \mathrm{\beta} then
\tan \beta=\left|\frac{\frac{37 \sqrt{3}-70}{61}-(-1)}{1-\left(\frac{37 \sqrt{3}-70}{61}\right)}\right|

\begin{aligned} & =\left|\frac{37 \sqrt{3}-9}{131-37 \sqrt{3}}\right| \\ & =\frac{37 \sqrt{3}-9}{131-37 \sqrt{3}} \\ \because & \tan \alpha>\tan \beta \quad \therefore \alpha>\beta \end{aligned}
therefore the slope of the refracted ray =\frac{70-37 \sqrt{3}}{13}
\therefore the equation of the refracted ray is \mathrm{y}+1=\frac{70-37 \sqrt{3}}{13}(\mathrm{x}-2)

\begin{aligned} & \Rightarrow \quad 13 \mathrm{y}+13=(70-37 \sqrt{3}) \mathrm{x}-140+74 \sqrt{3} \\ \end{aligned}

\begin{aligned} & \Rightarrow \quad(70-37 \sqrt{3}) \mathrm{x}-13 \mathrm{y}-153+74 \sqrt{3}=0 \end{aligned}

Hence option 4 is correct

Posted by

Nehul

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