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# About 5% of the Power of a 100 W Light Bulb is Converted to Visible Radiation. What is the Average Intensity of Visible Radiation - CBSE (Science) Class 12 - Physics

#### Question

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 1 m from the bulb?

(b) at a distance of 10 m?

Assume that the radiation is emitted isotropically and neglect reflection.

#### Solution

Power rating of bulb, P = 100 W

It is given that about 5% of its power is converted into visible radiation.

P'=5/100xx100=5 W

Hence, the power of visible radiation is 5W.

(a) Distance of a point from the bulb, d = 1 m

Hence, intensity of radiation at that point is given as:

I=(P')/(4pid^2)

=5/(4pi(1)^2)=0.398 W/m^2

(b) Distance of a point from the bulb, d1 = 10 m

Hence, intensity of radiation at that point is given as:

I=(P')/(4pid_1^2)

=5/(4pi(10)^2)=0.00398 W/m^2

Is there an error in this question or solution?

#### APPEARS IN

NCERT Solution for Physics Textbook for Class 12 (2018 to Current)
Chapter 8: Electromagnetic Waves
Q: 12 | Page no. 287

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Solution About 5% of the Power of a 100 W Light Bulb is Converted to Visible Radiation. What is the Average Intensity of Visible Radiation Concept: Electromagnetic Waves.
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