Question

A ray of light is sent along the line $x-2y+5=0$; upon reaching the line $3x-2y+7=0,$ the ray is reflected from it. Find the equation of the line containing the reflected ray.

Answer 1

We know that incident ray and reflected ray are equally inclined to the normal to the surface.

Given that the incident ray is along the line $x-2y+5=0⟹y=\frac{1}{2}x+\frac{5}{2}$ ------(1)

Therefore, slope of the incident ray is $m_1=\frac{1}{2}$

Surface is given by the line $3x-2y+7=0⟹y=\frac{3}{2}x+\frac{7}{2}$ -----(2)

Therefore, the slope of the surface is $\frac{3}{2}$

From the figure, the incident ray meets the surface at $P$

The coordinates of $P$ are found by substituting (1) in (2)

we get $3(2y-5)-2y+7=0⟹6y-15-2y+7=0⟹4y=8⟹y=2$

$x-2\left(2\right)+5=0⟹x=-1$

Therefore, the coordinates of P are $(-1,2)$.

Normal to the surface (i.e. perpendicular) will have slope $m=-\frac{2}{3}$.

Let the slope of reflected ray be $m_2$ and passes through the point $P(-1,2)$.

Hence, its equation is $y-2=m_2(x+1)$

normal $(m=-2/3)$ is equally incline to incident ray $(m_1=\frac{1}{2})$ and the reflected ray $m_2$

Therefore, $\frac{-\frac{2}{3}-\frac{1}{2}}{1+(-\frac{2}{3})\frac{1}{2}}=\frac{-\frac{2}{3}-m_2}{1+(-\frac{2}{3})m_2}$

$⟹\frac{7}{4}=\frac{3m_2+2}{2m_2-3}$

$⟹(14-12)m_2=8+21$

$⟹m_2=\frac{29}{2}$

Therefore, the reflected ray is $y-2=\frac{29}{2}(x+1)⟹29x-2y+33=0$