Explanation

(c)
Here, in the figure it is shown that a ray of light passing through the point $\mathcal{Q}(1,\ 2)$ and reflected from a point $P(\alpha,\ 0)$ on $x$-axis towards point $R(5,\ 3)$.
[image:10_image9.emf]
$\therefore$ slope of incident ray $(ie,\$before reflection) is given by
$\tan(\pi - \theta) = \frac{0 - 2}{\alpha - 1}$
$\Rightarrow \ \tan\theta = \frac{2}{\alpha - 1}\ \ \ldots(i)$
Similarly, slope of reflected ray $(ie$, after reflection ) is given by
$\Rightarrow \ \tan\theta = \frac{3}{5 - \alpha}\ \ \ldots(ii)$
From Eq. (i) and (ii),
$\frac{2}{\alpha - 1} = \frac{3}{5 - \alpha}$
$\Rightarrow 10 - 2\alpha = 3\alpha - 3\ \Rightarrow \ \alpha = \frac{13}{5}$