A point source of light is `80.0cm` below the surface of a body of water. Find the diameter of the circle at the surface through which light emerges from the water.

Answer

To solve the problem of finding the diameter of the circle at the surface through which light emerges from the water, we can follow these steps:

### Step 1: Understand the Geometry
We have a point source of light located 80 cm below the surface of the water. The light will emerge from the water in a conical shape, and we need to find the diameter of the circle at the water's surface where the light can be seen.

### Step 2: Identify the Relevant Angles
The angle at which light can emerge from the water is determined by the critical angle (θ_c) for total internal reflection. The formula for the critical angle is given by:

\[ \sin(\theta_c) = \frac{n_2}{n_1} \]

where \( n_1 \) is the refractive index of water (approximately 4/3) and \( n_2 \) is the refractive index of air (approximately 1).

### Step 3: Calculate the Critical Angle
Using the refractive indices:

\[ \sin(\theta_c) = \frac{1}{\frac{4}{3}} = \frac{3}{4} \]

Now, we can find the critical angle:

\[ \theta_c = \sin^{-1}\left(\frac{3}{4}\right) \]

### Step 4: Relate the Diameter to the Critical Angle
The radius (r) of the circle at the surface can be related to the height (h) and the critical angle:

\[ \tan(\theta_c) = \frac{r}{h} \]

From this, we can express the radius as:

\[ r = h \cdot \tan(\theta_c) \]

Since we need the diameter (D), we have:

\[ D = 2r = 2h \cdot \tan(\theta_c) \]

### Step 5: Substitute the Known Values
We know that \( h = 80 \) cm. We will need to calculate \( \tan(\theta_c) \) using the critical angle we found.

### Step 6: Calculate the Diameter
Now, substituting the values into the diameter formula:

\[ D = 2 \cdot 80 \cdot \tan(\theta_c) \]

To find \( \tan(\theta_c) \), we can use the relationship between sine and tangent:

\[ \tan(\theta_c) = \frac{\sin(\theta_c)}{\sqrt{1 - \sin^2(\theta_c)}} \]

Calculating \( \tan(\theta_c) \):

1. Calculate \( \sin(\theta_c) = \frac{3}{4} \).
2. Find \( \sqrt{1 - \left(\frac{3}{4}\right)^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \).
3. Thus, \( \tan(\theta_c) = \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}} \).

Now substituting back into the diameter formula:

\[ D = 2 \cdot 80 \cdot \frac{3}{\sqrt{7}} = \frac{480}{\sqrt{7}} \approx 181.19 \text{ cm} \]

### Final Answer
The diameter of the circle at the surface through which light emerges from the water is approximately **182 cm**.