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Luminous flux problem

by Amith2006
Tags: flux, luminous
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Amith2006
#1
Jun25-06, 02:14 PM
P: 423
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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Andrew Mason
#2
Jun25-06, 07:11 PM
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Quote 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 [itex]\pi[/itex], not [itex]2\pi[/itex].

AM
Amith2006
#3
Jun26-06, 02:38 AM
P: 423
Quote Quote by Andrew Mason
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)?

Hootenanny
#4
Jun26-06, 05:27 AM
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Luminous flux problem

Quote 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?
Amith2006
#5
Jun26-06, 11:44 AM
P: 423
Quote 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
= 2(pi) steradians

I think radian is a 2 dimensional unit whereas steradian is a 3 dimensional unit of angle. I may be wrong.
Andrew Mason
#6
Jun26-06, 07:22 PM
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Quote Quote by Amith2006
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
Amith2006
#7
Jun27-06, 01:13 AM
P: 423
That's ok. Thanks.


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