The light intensity 10 m from a point source is 1000 W/`m^(2)` . The intensity 100 m from the same source is

Options

A. `1000 W//m^(2)`
B. `100 W//m^(2)`
C. `10 W//m^(2)`
D. `1 W//m^(2)`

Answer

To find the intensity of light at a distance of 100 m from a point source when the intensity at 10 m is given as 1000 W/m², we can use the inverse square law of light intensity. The formula for intensity (I) at a distance (r) from a point source is given by:

\[ I = \frac{P}{4\pi r^2} \]

Where:
- \( I \) is the intensity,
- \( P \) is the power of the source,
- \( r \) is the distance from the source.

### Step-by-Step Solution:

1. **Identify the Known Values**:
- Intensity at 10 m, \( I_1 = 1000 \, \text{W/m}^2 \)
- Distance 1, \( r_1 = 10 \, \text{m} \)
- Distance 2, \( r_2 = 100 \, \text{m} \)

2. **Set Up the Ratio of Intensities**:
Since the power \( P \) remains constant for the same point source, we can set up the ratio of the intensities at the two distances:
\[ \frac{I_1}{I_2} = \frac{r_2^2}{r_1^2} \]

3. **Substitute the Known Values**:
Substitute \( I_1 \), \( r_1 \), and \( r_2 \) into the equation:
\[ \frac{1000}{I_2} = \frac{100^2}{10^2} \]

4. **Calculate the Right Side**:
Calculate \( \frac{100^2}{10^2} \):
\[ \frac{10000}{100} = 100 \]

5. **Set Up the Equation**:
Now we have:
\[ \frac{1000}{I_2} = 100 \]

6. **Solve for \( I_2 \)**:
Rearranging gives:
\[ I_2 = \frac{1000}{100} = 10 \, \text{W/m}^2 \]

### Final Answer:
The intensity at 100 m from the point source is \( I_2 = 10 \, \text{W/m}^2 \).

---