# Luminous flux problem

by Amith2006
Tags: flux, luminous
 P: 424 Please help me with this problem. # An isotropic point source of 100 candela is fastened to the ceiling of a room. What is the total luminous flux falling on all the walls and floor? I solved it in the following way: Here Luminous intensity(I) = 100 cd, Total solid angle(w) = 2 (pi) steradian Total luminous flux falling on all the walls and floor = I x w = 100 x 2(pi) = 100 x 2 x 3.14 = 628 lumens Is it right?
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 Quote by Amith2006 Please help me with this problem. # An isotropic point source of 100 candela is fastened to the ceiling of a room. What is the total luminous flux falling on all the walls and floor? I solved it in the following way: Here Luminous intensity(I) = 100 cd, Total solid angle(w) = 2 (pi) steradian Total luminous flux falling on all the walls and floor = I x w = 100 x 2(pi) = 100 x 2 x 3.14 = 628 lumens Is it right?
The walls and floor subtend a solid angle of $\pi$, not $2\pi$.

AM
P: 424
 Quote by Andrew Mason The walls and floor subtend a solid angle of $\pi$, not $2\pi$. AM
I thought the walls and the floor subtended a hemispherical solid angle at the ceiling. Could you please explain how the solid angle is (pi)?

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## Luminous flux problem

 Quote by Amith2006 I thought the walls and the floor subtended a hemispherical solid angle at the ceiling. Could you please explain how the solid angle is (pi)?
How many radians in half a circle?
P: 424
 Quote by Hootenanny How many radians in half a circle?
There are (pi) radians in a half circle. But I think there are 4(pi) steradians in a sphere. So there are 2(pi) steradians in a hemisphere.
Solid angle(w) = Area of hemisphere/R^2
= [2(pi)R^2]/R^2
No. You are right. A sphere subtends $4\pi$ steridians, so the hemisphere is $2\pi$. Sorry about confusing you.