A pole of length `2.00 m` stands half dipped in a swimming pool with level `1 m` higher than bed (bottom). The refractive index of water is `4//3` and sunlight is coming at an angle of `37^(@)` with the vertical. Find the length of the shadow of the pole on the bed. Use `sin^(-1)(0.45)=26.8^(@), tan(26.8^(@))=0.5`.

Explanation

To solve the problem step by step, we will analyze the situation involving the pole, the water, and the sunlight coming at an angle.

### Step 1: Understand the Setup
We have a pole of length 2.00 m, which is half submerged in water. Therefore, the length of the pole above the water is 1.00 m, and the length below the water is also 1.00 m. The water level is 1 m above the bottom of the pool.

### Step 2: Identify the Angles
The sunlight is coming at an angle of 37° with the vertical. This means that the angle with the horizontal (let's call it θ) can be calculated as:
\[ \theta = 90° - 37° = 53° \]

### Step 3: Calculate the Length of the Shadow Above Water
Using the tangent function, we can find the length of the shadow (y) cast by the pole above the water:
\[ \tan(37°) = \frac{y}{1} \]
Given that \( \tan(37°) = 0.5 \):
\[ y = 1 \cdot \tan(37°) = 1 \cdot 0.5 = 0.5 \text{ m} \]

### Step 4: Apply Snell's Law for Refraction
When sunlight enters the water, it bends due to the change in medium. We can use Snell's Law:
\[ n_1 \sin(\theta) = n_2 \sin(r) \]
Where:
- \( n_1 = 1 \) (refractive index of air)
- \( n_2 = \frac{4}{3} \) (refractive index of water)
- \( \theta = 37° \)
- \( r \) is the angle of refraction in water.

Calculating \( \sin(37°) \):
\[ \sin(37°) = \frac{3}{5} \]

Now applying Snell's Law:
\[ 1 \cdot \frac{3}{5} = \frac{4}{3} \sin(r) \]
\[ \sin(r) = \frac{3}{5} \cdot \frac{3}{4} = \frac{9}{20} \]

### Step 5: Find the Angle of Refraction
Now, we can find the angle \( r \):
Using the inverse sine function:
\[ r = \sin^{-1}(0.45) \]
From the given information, we know:
\[ r = 26.8° \]

### Step 6: Calculate the Length of the Shadow Below Water
Now, we need to find the length of the shadow below the water (x):
Using the tangent function again:
\[ \tan(r) = \frac{x}{1} \]
Given that \( \tan(26.8°) = 0.5 \):
\[ x = 1 \cdot \tan(26.8°) = 1 \cdot 0.5 = 0.5 \text{ m} \]

### Step 7: Total Length of the Shadow
Finally, the total length of the shadow (d) on the bed is the sum of the lengths of the shadows above and below the water:
\[ d = x + y = 0.5 \text{ m} + 0.5 \text{ m} = 1.0 \text{ m} \]

### Final Answer
The length of the shadow of the pole on the bed is **1.0 m**.

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