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Question: Answered & Verified by Expert
About 5% of the power of 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation at a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection.
PhysicsElectromagnetic WavesNEET
Options:
  • A 4×10-3 W m-2
     
  • B 5×10-3 W m-2
  • C 6×10-3 W m-2
     
  • D 7×10-3 W m-2
     
Solution:
2438 Upvotes Verified Answer
The correct answer is: 4×10-3 W m-2
 
Total power = 100 W

Visible radiation power = 5% of total power

=5100×100=5 W

At a distance of 10 m, the energy distributed in the form of sphere Area of sphere = 4π (radius)2

The intensity of visible radiation

= Power Area =54×3·14102=4×10-3 W m-2

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