A light ray leaves point $\mathrm{A}$in FIGURE EX34.5, reflects from the mirror, and reaches point $\mathrm{B}$. How far below the top edge does the ray strike the mirror?

According to the law of reflection,

The incident and reflected rays are parallel to the surface and in the same plane.

The incidence angle is the same as the reflection angle: ${\theta }_i={\theta }_r$

As seen in FIGURE 34.5, a light beam travels from point A to point B after reflecting off a mirror.

The distance between the top of the mirror and the point where the ray strikes the mirror must be calculated.

We can see from the green-shaded triangle that the tangent of angle ${\theta }_1$is

$\tan {\theta }_1=\frac{x}{10\mathrm{cm}}$

We can see from the yellow shaded triangle that the tangent of angle${\theta }_2$ is

$\tan {\theta }_2=\frac{10\mathrm{cm}-x}{15\mathrm{cm}}$

${\theta }_1={\theta }_2$, according to the law of reflection, therefore

$\tan {\theta }_1=\tan {\theta }_2$

$\frac{x}{10\mathrm{cm}}=\frac{10\mathrm{cm}-x}{15\mathrm{cm}}$

Rearrange the equations and solve for $x$.

$(15\mathrm{cm})x=(10\mathrm{cm})(10\mathrm{cm}-x)$

$(15\mathrm{cm})x=100{\mathrm{cm}}^2-(10\mathrm{cm})x$

$(15\mathrm{cm}+10\mathrm{cm})x=100{\mathrm{cm}}^2$

$$\begin{gathered} x=\frac{100{\mathrm{cm}}^2}{25\mathrm{cm}} \\ =4.0\mathrm{cm} \end{gathered}$$

The figure's geometry and the distance from top edge, where the light impacts the mirror

$d=5.0\mathrm{cm}+x$

$=5.0\mathrm{cm}+4.0\mathrm{cm}$

$=9.0\mathrm{cm}$