Question

If 11% of the power of a 200 W bulb is converted to visible radiation, then the intensity of the light at a distance of 100 cm from the bulb is

• A. \( 10.5 \ \text{W m}^{-2} \)
• B. \( 5.25 \ \text{W m}^{-2} \)
• C. \( 3.5 \ \text{W m}^{-2} \)
• D. \( 1.75 \ \text{W m}^{-2} \)

Hint:

To find intensity from a point source, use \( I = \frac{P}{4\pi r^2} \), where \( P \) is the power radiated and \( r \) is the distance.

Answer 1

The Correct Option is D

Step 1: Calculate useful power for visible radiation
\[ P = 11% \text{ of } 200 \ \text{W} = \frac{11}{100} \times 200 = 22 \ \text{W} \] Step 2: Use formula for intensity
\[ I = \frac{P}{4\pi r^2} \] Given \( r = 100 \ \text{cm} = 1 \ \text{m} \) \[ I = \frac{22}{4\pi (1)^2} = \frac{22}{4\pi} = \frac{11}{2\pi} \approx 1.75 \ \text{W m}^{-2} \]