Calculating Luminous Flux: Solving for Total Light Output in a Room

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The discussion revolves around calculating the total luminous flux from an isotropic point source of 100 candela in a room. The initial calculation incorrectly used a solid angle of 2π steradians, leading to a result of 628 lumens. However, it was clarified that the walls and floor actually subtend a solid angle of π steradians, which is half of a hemisphere. Consequently, the correct luminous flux should be recalculated using this solid angle. The conversation concludes with an acknowledgment of the misunderstanding regarding the solid angle measurement.
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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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Amith2006 said:
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
 
Andrew Mason said:
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)? :confused:
 
Amith2006 said:
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)? :confused:
How many radians in half a circle?
 
Hootenanny said:
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
= 2(pi) steradians

I think radian is a 2 dimensional unit whereas steradian is a 3 dimensional unit of angle. I may be wrong.
 
Last edited:
Amith2006 said:
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
= 2(pi) steradians

I think radian is a 2 dimensional unit whereas steradian is a 3 dimensional unit of angle. I may be wrong.
No. You are right. A sphere subtends 4\pi steridians, so the hemisphere is 2\pi. Sorry about confusing you.

AM
 
That's ok. Thanks.
 
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