About 5% of the power of a 100 W light bulb is converted to visible radiation. The average intensity of visible radiation at a distance of 1 m from the bulb:

Options

A. `0.4Wm^(-2)`
B. `0.5Wm^(-2)`
C. `0.6Wm^(-2)`
D. `0.8Wm^(-2)`

Explanation

To solve the problem, we need to calculate the average intensity of visible radiation from a 100 W light bulb at a distance of 1 meter. Here are the steps to arrive at the solution:

### Step 1: Determine the power converted to visible radiation
The power of the light bulb is given as 100 W, and it converts 5% of its power to visible radiation.

\[ \text{Visible Power} = 5\% \text{ of } 100 \text{ W} = \frac{5}{100} \times 100 = 5 \text{ W} \]

### Step 2: Use the formula for average intensity
The average intensity \( I \) of radiation from a point source can be calculated using the formula:

\[ I = \frac{P}{A} \]

where \( P \) is the power and \( A \) is the area over which the power is distributed. For a point source, the area \( A \) at a distance \( R \) is given by the surface area of a sphere:

\[ A = 4\pi R^2 \]

In this case, \( R = 1 \text{ m} \).

### Step 3: Calculate the area at 1 meter
Substituting \( R = 1 \text{ m} \) into the area formula:

\[ A = 4\pi (1)^2 = 4\pi \text{ m}^2 \]

### Step 4: Calculate the average intensity
Now, substitute the values of \( P \) and \( A \) into the intensity formula:

\[ I = \frac{P}{A} = \frac{5 \text{ W}}{4\pi \text{ m}^2} \]

### Step 5: Simplify the expression
Now we can compute the average intensity:

\[ I = \frac{5}{4\pi} \text{ W/m}^2 \]

### Final Calculation
Using the approximate value of \( \pi \approx 3.14 \):

\[ I \approx \frac{5}{4 \times 3.14} \approx \frac{5}{12.56} \approx 0.398 \text{ W/m}^2 \]

### Conclusion
The average intensity of visible radiation at a distance of 1 meter from the bulb is approximately \( 0.398 \text{ W/m}^2 \).

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