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(B) The incident ray passes through the origin $(0, 0)$ with slope $m = 	an 30^{\circ} = \frac{1}{\sqrt{3}}$.Its equation is $y = \frac{1}{\sqrt{3}}x

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The incident ray passes through the origin $(0, 0)$ with slope $m = 	an 30^{\circ} = \frac{1}{\sqrt{3}}$.Its equation is $y = \frac{1}{\sqrt{3}}x$.The intersection point $P$ of this ray with the line $x + y = 1$ is found by substituting $y = \frac{x}{\sqrt{3}}$ into $x + y = 1$:$x + \frac{x}{\sqrt{3}} = 1$ $\Rightarrow x(1 + \frac{1}{\sqrt{3}}) = 1$ $\Rightarrow x = \frac{\sqrt{3}}{\sqrt{3} + 1}$.Thus,$P = \left(\frac{\sqrt{3}}{\sqrt{3} + 1}, \frac{1}{\sqrt{3} + 1}ight)$.The line $x + y = 1$ has slope $m_1 = -1$,so its angle is $135^{\circ}$.The incident ray makes $30^{\circ}$ with the $x$-axis. The angle of incidence with the normal is $135^{\circ} - 30^{\circ} = 105^{\circ}$ (relative to the line). The reflected ray makes an angle of $135^{\circ} + (135^{\circ} - 30^{\circ}) = 240^{\circ}$ or $60^{\circ}$ with the positive $x$-axis.The slope of the reflected ray is $	an 60^{\circ} = \sqrt{3}$.The equation of the reflected ray passing through $P$ is $y - \frac{1}{\sqrt{3} + 1} = \sqrt{3}(x - \frac{\sqrt{3}}{\sqrt{3} + 1})$.Setting $y = 0$ to find the $x$-intercept $Q$:$-\frac{1}{\sqrt{3} + 1} = \sqrt{3}x - \frac{3}{\sqrt{3} + 1}$ $\Rightarrow \sqrt{3}x = \frac{3 - 1}{\sqrt{3} + 1} = \frac{2}{\sqrt{3} + 1}$.$x = \frac{2}{\sqrt{3}(\sqrt{3} + 1)} = \frac{2}{3 + \sqrt{3}}$.

$A$ light ray emits from the origin making an angle $30^{\circ}$ 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

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