A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is `4//3` and the fish is `12cm` below the surface, the radius of this circle in cm is

Options

A. `36sqrt5`
B. `4sqrt5`
C. `36sqrt7`
D. `36//sqrt7`

Explanation

To solve the problem of finding the radius of the circular horizon that a fish sees when looking up through the water, we can follow these steps:

### Step 1: Understand the Geometry
The fish is located 12 cm below the surface of the water. When it looks up, it sees a circular horizon due to the refraction of light at the water-air interface.

### Step 2: Identify the Refractive Index
The refractive index (μ) of water is given as \( \frac{4}{3} \).

### Step 3: Use the Critical Angle
The critical angle (c) can be calculated using the formula:
\[ \sin c = \frac{1}{\mu} \]
Thus,
\[ \sin c = \frac{1}{\frac{4}{3}} = \frac{3}{4} \]

### Step 4: Calculate the Critical Angle
Using the inverse sine function:
\[ c = \sin^{-1}\left(\frac{3}{4}\right) \]

### Step 5: Relate the Radius (R) and Depth (H)
Using the geometry of the situation, we can relate the radius of the circular horizon (R) to the depth (H) of the fish:
\[ \frac{R}{H} = \tan c \]
Where \( H = 12 \, \text{cm} \).

### Step 6: Express Tan in Terms of Sin and Cos
Using the identity for tangent:
\[ \tan c = \frac{\sin c}{\cos c} \]
We can express this in terms of the refractive index:
\[ \cos c = \sqrt{1 - \sin^2 c} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]
Thus,
\[ \tan c = \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}} \]

### Step 7: Substitute and Solve for R
Now substituting back into the equation:
\[ \frac{R}{12} = \frac{3}{\sqrt{7}} \]
Thus,
\[ R = 12 \cdot \frac{3}{\sqrt{7}} = \frac{36}{\sqrt{7}} \, \text{cm} \]

### Final Answer
The radius of the circular horizon that the fish sees is:
\[ R = \frac{36}{\sqrt{7}} \, \text{cm} \]

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