Question

$5$% of the power of $200W$ bulb is converted into visible radiation. The average intensity of visible radiation at a distance of $1m$ from the bulb is:

• A. 0.5 $W/m^2$
• B. 0.8 $W/m^2$
• C. 0.4 $W/m^2$
• D. 2 $W/m^2$

Answer 1

The correct option is B 0.8 $W/m^2$

Power of the bulb = $200W$

Power of the bulb, emitted effectively = Power$\times$ Efficiency

$=200W\times 0.05=10W$

Thus, intensity at a distance $r=\frac{P}{4\pi r^2}=\frac{10W}{4\times \pi \times 1^2{m}^2}$

$=0.8W/m^2$