Question

A \(1\) kW bulb radiates light uniformly in all directions. The intensity at a point on the surface of the surrounding sphere of area \(200\,\text{m}^2\) is (in \(\text{W m}^{-2}\))

• A. \(2\)
• B. \(4\)
• C. \(5\)
• D. \(6\)
• E. \(8\)

Hint:

For isotropic sources (uniform in all directions), always use \(I=\frac{P}{4\pi r^2}\) or directly \(P/A\) if area is given.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Intensity of radiation is defined as the power transmitted per unit area, where the area is measured perpendicular to the direction of propagation of the energy. The problem states that a bulb radiates uniformly in all directions, so the energy is spread out over the surface of a sphere.
Step 2: Key Formula or Approach:
The formula for intensity (I) is:
\[ I = \frac{P}{A} \] where P is the power of the source and A is the area over which the power is distributed.
Step 3: Detailed Explanation:
We are given the following values:
- Power of the bulb, \( P = 1 \text{ kW} \). We must convert this to Watts:
\( P = 1 \times 1000 \text{ W} = 1000 \text{ W} \).
- The area of the surrounding sphere, \( A = 200 \text{ m}^2 \).
Now, we can calculate the intensity using the formula:
\[ I = \frac{P}{A} = \frac{1000 \text{ W}}{200 \text{ m}^2} \] \[ I = 5 \text{ W m}^{-2} \] Step 4: Final Answer:
The intensity at a point on the surface of the sphere is 5 Wm\(^{-2}\). This corresponds to option (C).