Photometry / radiation question

[ginnerpip]

Hey guys (and girls)

I've got a rather annoying question.

If you were given the radiant power (45 mW) and luminous flux (22 l), and then was told it was spread uniformly spread over a hemisphere. (a) how do find the luminous intensity??, (b) the illuminance 1.8m away, and (c) illumance 5.0 cm straing out, and 3.0cm to the side.

The first questioj is really annoying me, any help??

Thanks

 

[HallsofIvy]

Hey guys (and girls)

What about the rest of us?

 

[Galileo]

Phew, all those funky optical jargon units are pretty annoying sometimes.
Let's see:
Radiant energy is Q_e, so radiant power is probably \frac{dQ_e}{dt}, measured in Joules and J/s (Watts).
While luminous energy is measured in lm-s (talbot)
and the luminous flux is \frac{dQ_v}{dt} where Q_v is the luminous energy. The unit of flux is lm (lumen).
the Luminous Intensity (candlepower) I_v is lm/sr (lumen per steradian) or cd (candela).

You have to have some formula for switching from photometric units to radiometric units. According to my opicts book it's:
photometric unit = K(\lambda)\times radiometric unit
where K(\lambda) is the luminous efficacy.

The first question is not so tough. Since the luminous flux is 22 lm and the luminous intensity is lm/sr and a hemisphere has 2\pi steradian, the luminous intensity is 11\pi lm/sr.

You really don't need the radiant power after all.