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

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Homework Help Overview

The discussion revolves around calculating the total luminous flux from an isotropic point source of light in a room, specifically focusing on the luminous intensity and the solid angle subtended by the walls and floor.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between luminous intensity and solid angle to determine luminous flux. There is debate over whether the walls and floor subtend a solid angle of 2π or π steradians.

Discussion Status

Some participants have provided calculations and reasoning regarding the solid angle, while others have questioned these assumptions. Clarifications about the definitions of radians and steradians are also being discussed.

Contextual Notes

There is a focus on the geometric interpretation of solid angles in relation to hemispherical and spherical configurations, with some confusion about the correct values being addressed.

Amith2006
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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 [itex]\pi[/itex], not [itex]2\pi[/itex].

AM
 
Andrew Mason said:
The walls and floor subtend a solid angle of [itex]\pi[/itex], not [itex]2\pi[/itex].

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 [itex]4\pi[/itex] steridians, so the hemisphere is [itex]2\pi[/itex]. Sorry about confusing you.

AM
 
That's ok. Thanks.
 

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