91Ó°ÊÓ

Consider a spotlight that is mounted beside a swimming pool sends a beam of light at an angle into the pool. The light hits water 2.4m from the edge of the pool. The light is mounted 1.2m above the surface of the water.

The rays of the light bend inward or outward at the surface between the air and the other medium. The bending of rays of light between the interface of two materials is called refraction.

(a)

Consider the Figure 23.125 given in the question. Consider the name of the spots, spotlight is A and the surface of the water is B. The point where the light hits the water is C. Assume this triangle $\mathrm{Δ}ABC$ to find $\mathrm{θ}$as follows,

$$\begin{gathered} \tan \mathrm{θ}=\frac{BC}{BA} \\ \tan \mathrm{θ}=\frac{1.2\text{m}}{2.4\text{m}} \end{gathered}$$

$$\begin{gathered} \tan \mathrm{θ}=0.5 \\ \mathrm{θ}={\tan }^{-1}\left(0.5\right) \\ \mathrm{θ}={27}^0 \end{gathered}$$

From the value of the $\mathrm{θ}$, it can be understood that the angle of the beam from the normal in the air is ,

$$\begin{gathered} {\mathrm{θ}}_1=90°-27° \\ {\mathrm{θ}}_1=63° \end{gathered}$$

Therefore, the angle of the beam from the normal in the air is, $\mathrm{θ}$.