Help calculating light intensity

In summary, the conversation discusses finding the intensity of a light source that produces an illumination of 7.50 lux at a distance of 30m from the source. The formula used is E = I/(4*pi*(r^2)) and when plugged in the given values, the answer is 84823. However, there may be some confusion with units and the type of source being used, which could affect the final answer. The conversation also mentions the need to know the wavelength or waveband for accurate conversion from photometric to radiometric units. Reference to "Introduction to Radiometry" by William Wolfe is suggested for further clarification.
  • #1
Mastagamer
1
0
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.
 
Last edited:
Physics news on Phys.org
  • #2
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,[tex]\Omega = A/r^{2}[/tex]

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.
 
  • #3


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!
 

1. What is light intensity and why is it important to calculate it?

Light intensity is the amount of light energy that falls on a given surface area in a given time. It is important to calculate light intensity because it can help determine the brightness of a light source, the effectiveness of lighting in a space, and the amount of light needed for plant growth, among other things.

2. How is light intensity measured?

Light intensity is measured using a device called a light meter, which measures the amount of light in lux or foot-candles. Lux is the SI unit of illuminance, while foot-candles is the imperial unit. Both units measure the amount of light energy per unit area.

3. What factors can affect light intensity?

Several factors can affect light intensity, including the distance from the light source, the angle at which the light hits the surface, the transparency of the medium through which the light travels, and the reflectivity of the surface.

4. How can I calculate light intensity?

To calculate light intensity, you will need to know the luminous flux (measured in lumens) and the area over which the light is distributed. You can then use the formula: Light intensity = luminous flux/area. Make sure to use the same units for both values.

5. How can light intensity calculations be useful in everyday life?

Light intensity calculations can be useful in everyday life in a variety of ways. They can help with choosing the right lighting for different spaces, ensuring proper lighting for tasks that require precision, and determining the amount of light needed for plants to grow. They can also be useful in photography and other visual arts, as well as in understanding the impact of light pollution on the environment.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
725
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
910
  • Introductory Physics Homework Help
Replies
4
Views
972
  • Other Physics Topics
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
871
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
893
  • Introductory Physics Homework Help
Replies
5
Views
2K
Back
Top