Answer to Question #118611 in Atomic and Nuclear Physics for hayley

Question

You are standing on the edge of a pond looking at a fish as shown in the figure above. The fish is located at a distance D = x1 + x2 from the bank.
Your eyes are located 1.5 m above the surface of the water at the edge of the water and your gaze is 25° below the horizontal.
Although it does not appear to be so from your point of view, the fish is actually 0.50 m below the surface of the water. How far away from the bank (D) is the fish? (to 2 s.f and in m)

Answer 1

We are looking 25° below the horizontal, which corresponds to the incidence angle $\theta_i$ of 90°-25°=65°.

Calculate the angle of reflection:

$$\theta_r=\text{arcsin}\frac{\text{sin}\theta_i}{n_w}=43^\circ.$$

From the inverted triangle with base x₂ and side of 0.5 m, we can calculate x₂:

$$x_2=0.5\cdot\text{tan}\theta_r=0.47\text{ m}.$$

Now it is easy to calculate x₂:

$$x_1=\frac{h}{\text{tan}25°}=\frac{1.5}{0.47}=3.22\text{ m}.$$

Therefore, the distance between the bank and the fish is

$$D=x_1+x_2=3.22+0.47=3.69\text{ m}.$$