Calculating the energy/power of incident light

  • Thread starter Thread starter PedroB
  • Start date Start date
  • Tags Tags
    Light
AI Thread Summary
The discussion focuses on calculating the energy input of incident light on a solar panel from a light bulb emitting at a single wavelength. To determine the power received by the solar panel, the fraction of the solid angle occupied by the panel as seen from the bulb must be calculated. The formula provided indicates that the power arriving at the panel is derived from the total power of the light bulb multiplied by this fraction divided by 4 Pi. For a circular solar panel, the solid angle is calculated using the cone formula, while non-circular panels complicate the calculations. Understanding solid angles is essential for accurate energy calculations in this context.
PedroB
Messages
16
Reaction score
0

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
 
Physics news on Phys.org
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?
 
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
 
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.
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top