Question

spotlight mounted beside a swimming pool sends a beam of light at an angle into the pool, as shown in Figure 23.125. The light hits the water 2.4m from the edge of the pool. The pool is 2m deep and the light is mounted 1.2m above the surface of the water. (The diagram is not drawn to scale.) The indices of refraction are: water, 1.33; air, 1.00029.

Figure 23.125

(a) What is the angle of the beam from the normal in the air?

(b) What is the angle of the beam from the normal in the water?

(c) How far from the edge of the pool is the spot of light on the bottom of the pool?

Short answer

(a)${\theta }_1=63°$

Step-by-step solution

Explain the given information

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.

Explain the bending of light.

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.

What is the angle of the beam from the normal in the air?

(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 $\Delta ABC$ to find $\theta$as follows,

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

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

From the value of the $\theta$, it can be understood that the angle of the beam from the normal in the air is ,

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

Therefore, the angle of the beam from the normal in the air is, $\theta$.