About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation (a) at a distance of 1m from the bulb? (b) at a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection.
Given: The power rating of bulb is 100 W, 5 % of the power is converted into visible radiation.
a)
The power of visible radiation is given as,
P v = 5 100 × P
Where, P v is the power of visible radiation and P is the power rating of bulb.
By substituting the values in the above expression, we get
P v = 5 100 × 100 W = 5 W
The intensity of radiation at d = 1 m is given as,
I = P v 4 π d 2
Where, I is the intensity of radiation and d is the distance at which intensity is measured.
By substituting the values in the above expression, we get
I = 5 4 π ( 1 ) 2 = 0.398 W / m 2
Thus, the intensity of radiation at distance of 1 m from the bulb is 0.398 W / m 2 .
b)
The intensity of radiation at d = 10 m is given as,
I = P v 4 π d 2
Where, I is the intensity of radiation and d is the distance at which intensity is measured.
By substituting the values in the above expression, we get
I = 5 4 π ( 10 ) 2 = 0.00398 W / m 2
Thus, the intensity of radiation at distance of 10 m from the bulb is 0.00398 W / m 2 .
Given: The power rating of bulb is 100 W, 5 % of the power is converted into visible radiation.
a)
The power of visible radiation is given as,
Where,
By substituting the values in the above expression, we get
The intensity of radiation at
Where,
By substituting the values in the above expression, we get
Thus, the intensity of radiation at distance of 1 m from the bulb is
b)
The intensity of radiation at
Where,
By substituting the values in the above expression, we get
Thus, the intensity of radiation at distance of 10 m from the bulb is
