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Physics Question on Electromagnetic waves

About 6%6\% of the power of a 100 W100\, W light bulb is converted to visible radiation. The average intensity of visible radiation at a distance of 8 m8\, m is (Assume that the radiation is emitted isotropically and neglect reflection.

A

3.5×10−3 W m−23.5 \times 10^{-3}\, W\,m^{-2}

B

5.1×10−3 W m−25.1 \times 10^{-3}\, W\,m^{-2}

C

7.2×10−3 W m−27.2 \times 10^{-3}\, W\,m^{-2}

D

2.3×10−3 W m−22.3 \times 10^{-3}\, W\,m^{-2}

Answer

7.2×10−3 W m−27.2 \times 10^{-3}\, W\,m^{-2}

Explanation

Solution

Here, power of bulb =100 W= 100\, W As intensity, I=Power of visible lightareaI = \frac{Power\,of \,visible \,light}{area} =100×6/1004π(8)2 \frac{100 \times6/100}{4\pi\left(8\right)^{2}} =7.2×10−3 W m−2= 7.2 \times 10^{-3}\,W\,m^{-2}