# About 5% of the Power of a 100 W Light Bulb is Converted to Visible Radiation. What is the Average Intensity of Visible Radiation - Physics

Numerical

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'" = 5/100 xx 100 = 5 W

Hence, the power of visible radiation is 5 W.

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

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

"I" = ("P'")/(4pi"d"^2)

= 5/(4pi(1)^2)

= 0.398 W/m2

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

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

"I" = "P'"/(4pi("d"_1)^2)

= 5/(4pi(10)^2)

= 0.00398 W/m2

Concept: Electromagnetic Waves
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 8 Electromagnetic Waves
Exercise | Q 8.12 | Page 287
NCERT Class 12 Physics Textbook
Chapter 8 Electromagnetic Waves
Exercise | Q 12 | Page 287

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