Question

One of the contests at the school carnival is to throw a spear at an underwater target lying flat on the bottom of a pool. The water is 1.0 m deep. You’re standing on a small stool that places your eyes 3.0 m above the bottom of the pool. As you look at the target, your gaze is 30° below horizontal. At what angle below horizontal should you throw the spear in order to hit the target? Your raised arm brings the spear point to the level of your eyes as you throw it, and over this short distance you can assume that the spear travels in a straight line rather than a parabolic trajectory.

Short answer

The angle below horizontal at which we should throw the spear in order to hit the target is 35^o

Step-by-step solution

Step 1. Given information is :Depth of the pool = 1.0 mDistance between eyes and the bottom of the pool = 3.0 mEyes make 30o angle below the horizontal when they look at the target.Refractive index of water n1 = 1.33Refractive index of air n2 = 1

We need to find out the angle below horizontal at which we should throw the spear in order to hit the target.

Step 2. Applying Snell's Law

Let${\theta }_2$be the angle of incidence of light ray on the water-air interface as shown below.

Therefore,

$$\begin{gathered} {\theta }_2=90°-30° \\ {\theta }_2=60° \end{gathered}$$

Hence $\tan {\theta }_2$will be represented as,

$$\begin{gathered} \tan {\theta }_2=\frac{x_2}{2.0m} \\ {x}_2=(2.0m)\times \tan {\theta }_2 \\ {x}_2=2\sqrt{3}m \end{gathered}$$

Applying snell's Law,

$$\begin{gathered} n_1\sin {\theta }_1=n_2\sin {\theta }_2 \\ \sin {\theta }_1=\frac{n_2}{n_1}\sin {\theta }_1 \\ {\theta }_1={\sin }^{-1}\left(\frac{n_2}{n_1}\sin {\theta }_1\right) \\ {\theta }_1={\sin }^{-1}\left(\frac{1.00}{1.33}\sin 60°\right) \\ {\theta }_1=40.628° \end{gathered}$$

Tangent of ${\theta }_1$will be expressed as shown below :

$$\begin{gathered} \tan {\theta }_1=\frac{x_1}{1.0m} \\ {x}_1=\tan 40.628° \\ {x}_1=0.85796m \end{gathered}$$

Tangent of angle$\varphi$as shown below will be calculated as :

$$\begin{gathered} \tan \varphi =\frac{3.0m}{x_1+x_2} \\ \varphi ={\tan }^{-1}\left(\frac{3.0m}{x_1+x_2}\right) \\ \varphi ={\tan }^{-1}\left(\frac{3.0m}{2\sqrt{3}m+0.85796m}\right) \\ \varphi =35° \end{gathered}$$