A light ray emits from the origin making an angle 30 ° with the positive x -axis. After getting reflected by the line x + y = 1 , if this ray intersects the x -axis at Q, then the abscissa of Q is [2023]

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Q.
A light ray emits from the origin making an angle $30 °$ with the positive $x$ -axis. After getting reflected by the line $x + y = 1,$ if this ray intersects the $x$ -axis at Q, then the abscissa of Q is [2023]

1 $\frac{2}{(\sqrt{3} - 1)}$

2 $\frac{\sqrt{3}}{2 (\sqrt{3} + 1)}$

3 $\frac{2}{3 + \sqrt{3}}$

4 $\frac{2}{3 - \sqrt{3}}$

Ans.
(3)

$Given Slope of incident ray = \tan 30 ° = \frac{1}{\sqrt{3}}$

$So, slope of incident ray = slope of reflected ray = \frac{1}{\sqrt{3}}$

$Now, Equation of line making an angle 30 ° with positive x-axis and passing through origin is,$ $y = \frac{x}{\sqrt{3}}$

$So, line y = \frac{x}{\sqrt{3}} and x + y = 1 intersect at (\frac{\sqrt{3}}{\sqrt{3} + 1}, \frac{1}{\sqrt{3} + 1})$

$So, equation of reflected ray is$

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

$Put y = 0 \Rightarrow x = \frac{2}{3 + \sqrt{3}}$

Solution

Q.
A light ray emits from the origin making an angle $30 °$ with the positive $x$ -axis. After getting reflected by the line $x + y = 1,$ if this ray intersects the $x$ -axis at Q, then the abscissa of Q is [2023]

1 $\frac{2}{(\sqrt{3} - 1)}$

2 $\frac{\sqrt{3}}{2 (\sqrt{3} + 1)}$

3 $\frac{2}{3 + \sqrt{3}}$

4 $\frac{2}{3 - \sqrt{3}}$

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Solution

Q. A light ray emits from the origin making an angle 30° with the positive x-axis. After getting reflected by the line x+y=1, if this ray intersects the x-axis at Q, then the abscissa of Q is [2023]

1 2(3-1)

2 32(3+1)

3 23+3

4 23-3

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Q.
A light ray emits from the origin making an angle $30 °$ with the positive $x$ -axis. After getting reflected by the line $x + y = 1,$ if this ray intersects the $x$ -axis at Q, then the abscissa of Q is [2023]

1 $\frac{2}{(\sqrt{3} - 1)}$

2 $\frac{\sqrt{3}}{2 (\sqrt{3} + 1)}$

3 $\frac{2}{3 + \sqrt{3}}$

4 $\frac{2}{3 - \sqrt{3}}$