A fish at a depth of `sqrt(7)` cm bleow the surface of water sees the outside world through a circular horizon. What is the radius of the circular horizon? `[ ""_(a)mu_(w) = (4)/(3)]`

Options

A. 4 cm
B. 3 cm
C. `sqrt(7) cm`
D. 1 cm

Answer

To find the radius of the circular horizon that a fish sees while submerged at a depth of \(\sqrt{7}\) cm in water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Depth of the Fish**:
The fish is located at a depth of \(d = \sqrt{7}\) cm below the surface of the water.

2. **Understanding the Critical Angle**:
The critical angle (\(\alpha\)) can be determined using Snell's Law, which states:
\[ \mu_1 \sin \alpha = \mu_2 \sin 90^\circ \]
Here, \(\mu_1\) is the refractive index of water, \(\mu_2\) is the refractive index of air. Given that \(\mu_w = \frac{4}{3}\) and \(\mu_{air} = 1\), we can write:
\[ \frac{4}{3} \sin \alpha = 1 \]
Therefore,
\[ \sin \alpha = \frac{3}{4} \]

3. **Calculate the Critical Angle**:
To find \(\alpha\), we can use the inverse sine function:
\[ \alpha = \sin^{-1}\left(\frac{3}{4}\right) \]

4. **Using the Geometry of the Situation**:
The radius \(r\) of the circular horizon can be related to the depth \(d\) and the angle \(\alpha\) using the tangent function:
\[ \tan \alpha = \frac{r}{d} \]
Rearranging gives:
\[ r = d \tan \alpha \]

5. **Calculate \(\tan \alpha\)**:
From the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\), we can find \(\cos \alpha\):
\[ \sin^2 \alpha = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \implies \cos^2 \alpha = 1 - \frac{9}{16} = \frac{7}{16} \implies \cos \alpha = \frac{\sqrt{7}}{4} \]
Now we can find \(\tan \alpha\):
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{3}{\sqrt{7}} \]

6. **Substituting Values**:
Now substituting \(d = \sqrt{7}\) cm and \(\tan \alpha = \frac{3}{\sqrt{7}}\):
\[ r = \sqrt{7} \cdot \frac{3}{\sqrt{7}} = 3 \text{ cm} \]

### Final Answer:
The radius of the circular horizon that the fish sees is \(r = 3\) cm.

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