Calculating Energy Flux Density and Candle Power

AI Thread Summary
The discussion centers on calculating energy flux density and candle power, specifically using the standard candle's radiant intensity of 1/683 W/sr. To find the energy flux density at 1.06 meters from the candle, the participants note the need to convert watts per steradian to watts per square meter using the solid angle of a sphere. For the candle power of a 75-watt light bulb, it is suggested to assume uniform emission into 4π steradians. The conversation also touches on the historical context of measurement units in photonics, highlighting the complexities of converting between various units like lumens and watts. Overall, the thread emphasizes the importance of understanding the relationships between these units for accurate calculations.
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Homework Statement



A standard candle has radiant intensity equal to 1/683 W/sr.

[The unit of measurement is watts per steradian. The steradian (sr) is the standard unit for solid angles: If a patch on the surface of a sphere of radius r has area A, then the solid angle subtended by the patch at the center of the sphere is A/r^2. The solid angle of an entire sphere is 4π because the area is 4πr^2.]

a) What is the energy flux density (power per unit area) at a distance of 1.06 meters from the candle?

b)What is the candle power of a 75 watt light bulb?

Homework Equations



I figured out the power to be 1.84E-2 W

The Attempt at a Solution



I'm not sure how to solve this one.
 
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There are 4pi sterads in a sphere.
How many m^2 are there in a sphere of radius 1.06m
So you can work out how many m^2 / sterad.

a, You know the watts / sterad so getting w/m^2 is easy.

b, just assume the bulb emits uniformaly into 4pi sterads.
 
As it says in the querstion the part of the surface area of a sphere that is subtended by a steradian is r^2.

As it says in the question the light energy passing through this area is 1/683 (weird!) Watts and r is 1.03 m.

The answer required is Watts per square meter. We have the Watts and we have the square meters (so it's not necessary to calculate the total power of the candle) ...
 
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Thanks again mgb and catkin as well, I got them both.
 
catkin said:
As it says in the question the light energy passing through this area is 1/683 (weird!)...

No stranger than 1 horsepower = 746 Watts: many widely used physical units substantially predate the introduction of the metric system and, later, SI. (There's a ton [ = 907 kg.] of such units in fields like engineering or astronomy...) The metrical congresses standardized the relationship of these units to the metric system, but often failed in efforts to replace them with something that's a nice power of ten. (There's still debate on questions like whether astronomers should keep using light-years or parsecs instead of something that's just 10^13 m long...)
 
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The 683 is to match the eye's response and is based on an early unit.
Defined "as the light produced by a pure spermaceti (sperm whale oil) candle weighing one sixth of a pound and burning at a rate of 120 grains per hour."
The new metric Candela is even worse = "luminous intensity of a blackbody at the freezing point of liquid platinum which was to be 58.9 international candles per square centimeter."

Add in lumens/lux/EV and a bunch of other units and photonics is a pain. Usually end up working either in photons/sec or watts/sec.
 
mgb_phys said:
... the light produced by a pure spermaceti (sperm whale oil) candle weighing one sixth of a pound and burning at a rate of 120 grains per hour ...
Wonderful!
 
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