Question

About $5\%$ of the power of a $100W$ light bulb is converted to visible radiation. What is the average intensity of visible radiation
A) At a distance of 1 m from the bulb?

About $5\%$ of the power of a $100W$ light bulb is converted to visible radiation. What is the average intensity of visible radiation
B) At a distance of $10m$?

Answer 1

A) Given, Power of the bulb, $P=100W$
$5\%$ of its power is converted into visible radiation. Therefore, power of visible radiation,
$P_{rad}=\frac{5}{100}\times 100=5W$
Distance of a point from the bulb, d =1 m
Hence, intensity of radiation at that point is given as:
$I=\frac{P_{rad}}{4\pi d^2}$
Substituting the values, we get
$I=\frac{5}{4\pi (1{)}^2}=0.398W/m^2$
Final Answer : $0.398W/m^2$

B) Given,
Power of the bulb, $P=100W$
$5\%$ of its power is converted into visible radiation. Therefore, power of visible radiation,
$P_{rad}=\frac{5}{100}\times 100=5W$
Distance of a point from the bulb, $d_1=10m$
Hence, intensity of radiation at that point is given as:
$I=\frac{P_{rad}}{4\pi d_1^2}=\frac{5}{4\pi (10{)}^2}=0.00398\frac{W}{m^2}$
Final Answer: $0.00398W/m^2$