Calculating the energy/power of incident light

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In summary, the conversation discusses a method for determining the optimum wavelength of light for solar panels through calculating the energy input from a light bulb at a specific distance from the panel. The formula for calculating the fraction of solid angle occupied by the panel is provided and it is noted that this method may become more complicated for non-circular panels.
  • #1
PedroB
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Homework Statement



Simply put, I am conducting an investigation that determines the "optimum" wavelength of light for solar panels (ie the wavelength which produces the highest conversion efficiency).

To do this however, I need to know the energy input of incident light on a particular solar panel (dimensions irrelevant), if say a 50W light bulb (that emits light at only one wavelength) were used. Despite looking all over the internet I have not yet found the solution to my problem. Simply put, If a lightbulb is "x" cm away from the solar panel, and light (of 1 wavelength) shines on the panel, what is the energy (or power which fundamentally is energy per second) of the incident light that strikes ONLY the solar panel.

Homework Equations



I have found no pertinent equations that may aid my situation

The Attempt at a Solution



Nothing worth mentioning
 
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  • #2
You have to calculate the fraction of solid angle which is occupied by the solar panel, as seen from the light bulb. The power the solar panel gets is then the total power of the light bulb times this fraction divided by 4 Pi, which is the solid angle of a sphere. Is that what you wanted?
 
  • #3
Yes, it is something along those lines. If you don't mind (I really appreciate the effort, don't want to seem disrespectful) can you please post the formula (or link to said formula)? Also, from what I gathered from my research, solid angles are and I quote "two-dimensional angle in three-dimensional space". Is this the case here? Thanks in advance
 
  • #4
You got that right with the definition of a solid angle.
Since the energy is radiated by a bulb in all directions simultaneously and with the same intensity, the fraction of the energy taken by the solar panel is equal to the fraction of "sky" it occupies as seen from the light bulb. For the calculation of solid angles of common objects, have a look at Wikipedia: http://en.wikipedia.org/wiki/Solid_angle
So for example for a circular solar panel with radius r at a distance d from the center of the bulb, the solid angle it occupies is the one of a cone, which Wikipedia has as 2 Pi (1 - cos theta), where tan theta = r/(2 d). So since the full sphere ("all sky") has solid angle 4 Pi, the power arriving is P = P_0 1/2 (1 - cos (arctan(r/(2 d)))), where P_0 is the power of the light bulb.
If you have a non-circular solar panel, the calculations get very messy, but they may be doable.
 
  • #5


I would recommend using the equation E = hc/λ, where E is the energy of a photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the incident light. This equation can be used to calculate the energy of a single photon at a specific wavelength. Then, you can use the equation P = E/t, where P is power, E is energy, and t is time, to calculate the power of the incident light on the solar panel. By knowing the power of the incident light, you can then determine the efficiency of the solar panel at that specific wavelength. Additionally, you may also need to take into account factors such as the angle of incidence and the reflectivity of the solar panel in your calculations. I would also suggest consulting with other experts in the field of solar energy to ensure the accuracy and validity of your investigation.
 

1. How do you calculate the energy of incident light?

The energy of incident light can be calculated using the equation E = hc/λ, where E is the energy in joules, h is Planck's constant (6.626 x 10^-34 J*s), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength in meters.

2. What is the relationship between energy and power of incident light?

Energy and power of incident light are directly proportional. This means that as the energy of the light increases, so does the power. The equation for power is P = E/t, where P is power in watts, E is energy in joules, and t is time in seconds.

3. How is the power of incident light measured?

The power of incident light can be measured using a device called a photometer or a power meter. These instruments measure the amount of light energy per unit time, typically in watts. The power can also be calculated by measuring the voltage and current of the incident light and using the equation P = IV.

4. What factors affect the energy and power of incident light?

The energy and power of incident light can be affected by several factors, including the intensity or brightness of the light source, the distance between the light source and the object, and the properties of the medium through which the light is traveling (such as absorption or scattering).

5. Can the energy or power of incident light be altered?

Yes, the energy and power of incident light can be altered by changing the properties of the light source, such as its intensity or wavelength. The energy and power can also be altered by manipulating the medium through which the light is traveling, such as using lenses or filters to focus or block certain wavelengths of light.

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