Help calculating light intensity

[Mastagamer]

This is the question

Find the intensity of a light source that produces an illumination of 7.50 lux at a distance of 30m from the source.

This what I get but it's wrong
84823

This is the formula

E = Illumination
I = luminous intensity of the source ( in lumens)
r= distance between the source and the illuminated surface.

1 lux = 1lm/m^2

Formula E = I/(4*pi*(r^2)

So I did I = E*(4*pi*(r^2)

I =?
E=7.50
r = 30m

I = 7.50lux*(4*pi*(30m^2)


and got I = 84823

rounded in many different forms and I still get a wrong answer.. Any help would be appreciate it.

and is wrong because the computer practice test tells me but doesn't give me an answer.

 

[Andy Resnick]

Blech. Radiometry: that evil, dismal science. :)

First, you need to know more about the source- is it an isotropic point source? Extended Lambertian radiator? It matters. Second, check the units: 'lux' is a photometric unit, meaning the conversion to radiometric units (intensity) requires knowledge of the wavelength (or waveband!)

As written, the question is potentially meaningless, because of the (possible) mix of photometric and radiometric units. As it happens, there is a photometric term also called "luminous intensity" in units of lm/sr (candela [cd]). First, let's parse the question:

7.5 lux at a distance of 30 m: the incidance of light onto a surface 30 m away from the source is 7.5 lm/m^2. This implies (to me), that we have an isotropic point source freely radiating into 4*pi sr.

It's possible the question instead refers to an extended radiator, with an excitance of 7.5 lx. If the radiator is Lambertian, the radiated power does not depend on relative angle between the radiant surface and the detector, which simplifies things. The total radiated power is 7.5 *source area (lm), the received power is that fraction of solid angle (out of 2*pi) subtended by the detector.

Maybe the idea is to teach you how to convert areas into steradians? For small (solid) angles,$$\Omega = A/r^{2}$$

thus, 7.5 lx at 30 m converts to 8.3*10^-3 cd, if I did that correctly...

Like I said, radiometry and photometry can be incredibly confusing. Check out William Wolfe's "Introduction to Radiometry" (Tutorial Texts, SPIE press, vol TT29) for a really clear discussion.

 

[Moetasim]

Hello,

Thank you for reaching out for assistance with your light intensity calculation. It seems like you have the right formula and units, but there may be a mistake in your calculation. Let's break down the formula and see where the error may be occurring.

First, let's define the variables:
E = Illumination = 7.50 lux
I = Luminous intensity of the source (in lumens) = ?
r = Distance between the source and the illuminated surface = 30m

Next, let's plug in the values into the formula:
E = I / (4 * pi * (r^2))

Substituting the given values, we get:
7.50 lux = I / (4 * pi * (30m^2))

To solve for I, we need to isolate it on one side of the equation. This can be done by multiplying both sides by (4 * pi * (30m^2)):
7.50 lux * (4 * pi * (30m^2)) = I

Now, let's solve:
I = 7.50 lux * (4 * pi * (30m^2))
I = 2827.433388 lumen

So, the luminous intensity of the light source is approximately 2827.433388 lumen.

I hope this helps! Let me know if you have any further questions or if you need clarification on any part of the calculation. Keep up the good work in your scientific studies!