Aspire Faculty ID #13228 · Topic: JEE Main 2023 (29 January Morning Shift) · Just now
JEE Main 2023 (29 January Morning Shift)

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:

Solution

$y=\tan30^\circ,x=\dfrac{x}{\sqrt3}$ hits the mirror $x+y=1$ at $P\left(\dfrac{\sqrt3}{\sqrt3+1},,\dfrac{1}{\sqrt3+1}\right)$. 
The mirror’s normal is along $(1,1)$, so reflecting the unit direction $u=(\cos30^\circ,\sin30^\circ)=\left(\dfrac{\sqrt3}{2},\dfrac12\right)$ about the line gives $u'=u-2(u\cdot \hat n)\hat n=\left(-\dfrac12,-\dfrac{\sqrt3}{2}\right)$, 
i.e. slope $m'=\sqrt3$. 
The reflected ray through $P$ is $y-y_0=\sqrt3(x-x_0)$. 
Intersecting $y=0$ gives $x=x_0-\dfrac{y_0}{\sqrt3} $
$=\dfrac{\sqrt3}{\sqrt3+1}-\dfrac{1}{\sqrt3(\sqrt3+1)}$
$=\dfrac{2}{3+\sqrt3}$

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