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Question: About 6% of power of a 100 W light bulb is converted to visible radiation. The average in intensity ...

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

A

3.5×10−3Wm−23.5 \times 10^{- 3}Wm^{- 2}

B

5.1×10−3Wm−25.1 \times 10^{- 3}Wm^{- 2}

C

7.2×10−3Wm−27.2 \times 10^{- 3}Wm^{- 2}

D

2.3×10−3Wm−22.3 \times 10^{- 3}Wm^{- 2}

Answer

7.2×10−3Wm−27.2 \times 10^{- 3}Wm^{- 2}

Explanation

Solution

: Here, Power of bulb, = 100 W

As intensity, I=PowerofvisiblelightareaI = \frac{Powerofvisiblelight}{area}

=100×6/1004π(8)2=7.2×10−3Wm−2= \frac{100 \times 6/100}{4\pi(8)^{2}} = 7.2 \times 10^{- 3}Wm^{- 2}