A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is `(4)/(3)`, and the diver's eyes are 15 cm below the surface of 3 water. Then the radius of the circle is:

Options

A. `15xx3xxsqrt5 cm`
B. `15xx3sqrt7 cm`
C. `(15xxsqrt7)/(3)cm`
D. `(15xx3)/(sqrt7)cm`

Explanation

To find the radius of the circular horizon seen by a diver looking up through the water, we can follow these steps:

### Step 1: Understand the scenario
The diver's eyes are located 15 cm below the surface of the water, and the refractive index of water is given as \( \mu = \frac{4}{3} \). The diver can only see a circular area due to the refraction of light at the water-air interface.

### Step 2: Identify the critical angle
The critical angle \( \theta_c \) for the water-air interface can be calculated using the formula:
\[ \theta_c = \sin^{-1}\left(\frac{1}{\mu}\right) \]
Substituting the value of \( \mu \):
\[ \theta_c = \sin^{-1}\left(\frac{1}{\frac{4}{3}}\right) = \sin^{-1}\left(\frac{3}{4}\right) \]

### Step 3: Set up the triangle
In the triangle formed by the diver's line of sight, the height from the diver's eyes to the water surface is 15 cm (the height of the diver's eyes below the surface). The radius \( r \) of the circular horizon can be related to the angle \( \theta_c \) using trigonometric relations.

### Step 4: Apply trigonometric relations
Using the sine function in the triangle:
\[ \sin(\theta_c) = \frac{r}{\sqrt{15^2 + r^2}} \]
We know that \( \sin(\theta_c) = \frac{3}{4} \) (from the critical angle calculation). Therefore, we can write:
\[ \frac{r}{\sqrt{15^2 + r^2}} = \frac{3}{4} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 4r = 3\sqrt{15^2 + r^2} \]
Squaring both sides results in:
\[ 16r^2 = 9(15^2 + r^2) \]
Expanding this gives:
\[ 16r^2 = 9 \cdot 225 + 9r^2 \]
\[ 16r^2 - 9r^2 = 2025 \]
\[ 7r^2 = 2025 \]

### Step 6: Solve for \( r^2 \)
Dividing both sides by 7:
\[ r^2 = \frac{2025}{7} \]

### Step 7: Calculate \( r \)
Taking the square root:
\[ r = \sqrt{\frac{2025}{7}} = \frac{\sqrt{2025}}{\sqrt{7}} = \frac{45}{\sqrt{7}} \]

### Step 8: Final answer
Thus, the radius of the circular horizon seen by the diver is:
\[ r = \frac{45}{\sqrt{7}} \text{ cm} \]