A small piece of wood is floating in the water of depth `1m` . The refractive index of water is `4//3` and the sum rays are incident at an angle of `53^(@)` to the vertical. Find the location of the shadow of the wood piece on the bottom of the ocean.

Answer

To find the location of the shadow of a small piece of wood floating in water, we can follow these steps:

### Step 1: Understand the Problem
We have a piece of wood floating in water at a depth of 1 meter. The refractive index of water is given as \( \frac{4}{3} \), and the sun's rays are incident at an angle of \( 53^\circ \) to the vertical. We need to find the location of the shadow of the wood piece on the bottom of the water body.

### Step 2: Set Up the Geometry
- The depth of the water is \( 1 \, \text{m} \).
- The angle of incidence \( i = 53^\circ \).
- The refractive index of air \( \mu_1 = 1 \) and that of water \( \mu_2 = \frac{4}{3} \).

### Step 3: Apply Snell's Law
Using Snell's Law:
\[ \mu_1 \sin i = \mu_2 \sin r \]
Substituting the known values:
\[ 1 \cdot \sin(53^\circ) = \frac{4}{3} \cdot \sin r \]

### Step 4: Calculate \( \sin(53^\circ) \)
Using the sine value:
\[ \sin(53^\circ) \approx \frac{4}{5} \]
So we have:
\[ \frac{4}{5} = \frac{4}{3} \sin r \]

### Step 5: Solve for \( \sin r \)
Rearranging gives:
\[ \sin r = \frac{4}{5} \cdot \frac{3}{4} = \frac{3}{5} \]

### Step 6: Find the Angle of Refraction \( r \)
Now, we can find \( r \):
\[ r = \arcsin\left(\frac{3}{5}\right) \approx 37^\circ \]

### Step 7: Calculate the Displacement \( d \)
Using the triangle formed by the depth and the displacement, we can use the tangent function:
\[ \tan r = \frac{d}{1} \]
Where \( d \) is the horizontal displacement. Since \( \tan(37^\circ) = \frac{3}{4} \):
\[ d = 1 \cdot \tan(37^\circ) = 1 \cdot \frac{3}{4} = 0.75 \, \text{m} \]

### Step 8: Conclusion
The shadow of the wood piece will be located \( 0.75 \, \text{m} \) away from the point directly below the center of the piece of wood on the bottom of the water body.

### Final Answer
The location of the shadow of the wood piece on the bottom of the ocean is \( 0.75 \, \text{m} \) away from the point directly below the center of the wood.

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