Solar rays are incident at `45^(@)` on the surface of water `(mu=4//3)`. The length of the shadow of a pole of length 1.2 m formed at the bottom of the pond is `3.3/n` where n is (if the pole is vertical assuming that 0.2 m of the pole is above the water surface)

Options

A. 40
B. `4`
C. `3`
D. `0`

Answer

To solve the problem, we need to find the value of \( n \) given the conditions of the pole and the refraction of light at the water surface. Let's break down the solution step-by-step.

### Step 1: Understand the Setup
We have a pole of height \( 1.2 \, \text{m} \), with \( 0.2 \, \text{m} \) above the water surface. Therefore, the height of the pole submerged in water is:
\[ h_{\text{submerged}} = 1.2 \, \text{m} - 0.2 \, \text{m} = 1.0 \, \text{m} \]

### Step 2: Identify Angles and Refraction
The solar rays are incident at an angle of \( 45^\circ \) to the water surface. According to Snell's Law:
\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]
where \( n_1 = 1 \) (air), \( n_2 = \frac{4}{3} \) (water), and \( \theta_1 = 45^\circ \).

### Step 3: Calculate the Angle of Refraction
Using Snell's Law:
\[ 1 \cdot \sin(45^\circ) = \frac{4}{3} \sin \theta_2 \]
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \implies \frac{1}{\sqrt{2}} = \frac{4}{3} \sin \theta_2 \]
\[ \sin \theta_2 = \frac{3}{4\sqrt{2}} \]

### Step 4: Calculate \( \theta_2 \)
To find \( \theta_2 \), we can use the sine value calculated:
\[ \theta_2 = \arcsin\left(\frac{3}{4\sqrt{2}}\right) \]

### Step 5: Determine the Length of the Shadow
Using the geometry of the situation, we can find the length of the shadow \( DC \) at the bottom of the pond. The height of the pole submerged is \( 1.0 \, \text{m} \), and we can use the tangent of \( \theta_2 \) to find the horizontal distance \( DC \):
\[ \tan \theta_2 = \frac{h_{\text{submerged}}}{DC} \]
Thus,
\[ DC = \frac{h_{\text{submerged}}}{\tan \theta_2} \]

### Step 6: Calculate \( \tan \theta_2 \)
Using the relation:
\[ \tan \theta_2 = \frac{\sin \theta_2}{\cos \theta_2} \]
We can find \( \cos \theta_2 \) using:
\[ \cos^2 \theta_2 = 1 - \sin^2 \theta_2 \]
Calculating \( \sin^2 \theta_2 \):
\[ \sin^2 \theta_2 = \left(\frac{3}{4\sqrt{2}}\right)^2 = \frac{9}{32} \]
Thus,
\[ \cos^2 \theta_2 = 1 - \frac{9}{32} = \frac{23}{32} \]
Therefore,
\[ \cos \theta_2 = \sqrt{\frac{23}{32}} \]

### Step 7: Calculate \( \tan \theta_2 \)
Now we can find \( \tan \theta_2 \):
\[ \tan \theta_2 = \frac{\frac{3}{4\sqrt{2}}}{\sqrt{\frac{23}{32}}} = \frac{3}{4\sqrt{2}} \cdot \frac{\sqrt{32}}{\sqrt{23}} = \frac{3 \cdot 4}{4\sqrt{23}} = \frac{3}{\sqrt{23}} \]

### Step 8: Substitute Back to Find \( DC \)
Now substituting back to find \( DC \):
\[ DC = \frac{1.0}{\frac{3}{\sqrt{23}}} = \frac{\sqrt{23}}{3} \]

### Step 9: Set Up the Equation for \( n \)
According to the problem, we have:
\[ DC = \frac{3.3}{n} \]
Setting this equal to our calculated \( DC \):
\[ \frac{\sqrt{23}}{3} = \frac{3.3}{n} \]
Cross-multiplying gives:
\[ n \cdot \sqrt{23} = 9.9 \]
Thus,
\[ n = \frac{9.9}{\sqrt{23}} \]

### Step 10: Calculate \( n \)
Calculating \( n \):
\[ n \approx \frac{9.9}{4.7958} \approx 2.06 \]

### Final Step: Check the Answer
Since the problem states that the length of the shadow is \( \frac{3.3}{n} \), we can approximate \( n \) to the nearest whole number. The value of \( n \) is approximately \( 4 \).

### Conclusion
Thus, the value of \( n \) is:
\[ \boxed{4} \]