An intense light source radiates uniformly in all directions. At a distance $5 m$ from the source, the radiation pressure on absorbing surface is $9 \times 10^{- 6} P a$ . Find the total average power output of source.

8.5 \times 10^{5} W

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6.5 \times 10^{5} W

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8.5 \times 10^{3} W

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6.5 \times 10^{3} W

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Solution

The correct option is A $8.5 \times 10^{5} W$
The radiation pressure $p = I / c$ where $I =$ intensity of radiation and $c =$ velocity of light.
Total average power $= I \times a r e a = I \times 4 π r^{2} = p c \times 4 π r^{2} = 9 \times 10^{- 6} \times 3 \times 10^{8} \times 4 π \times 5^{2} = 8.5 \times 10^{5} W$

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An intense light source radiates uniformly in all directions. At a distance 5m from the source, the radiation pressure on absorbing surface is 9×10−6Pa. Find the total average power output of source.

The correct option is A 8.5×105WThe radiation pressure p=I/c where I= intensity of radiation and c= velocity of light. Total average power =I×area=I×4πr2=pc×4πr2=9×10−6×3×108×4π×52=8.5×105W

An intense light source radiates uniformly in all directions. At a distance $5 m$ from the source, the radiation pressure on absorbing surface is $9 \times 10^{- 6} P a$ . Find the total average power output of source.