Calculating the energy/power of incident light

  • Thread starter Thread starter PedroB
  • Start date Start date
  • Tags Tags
    Light
Click For Summary

Homework Help Overview

The original poster is investigating the optimum wavelength of light for solar panels, specifically seeking to determine the energy input of incident light from a 50W light bulb positioned at a certain distance from the solar panel. The focus is on calculating the power of the light that strikes the solar panel, considering the bulb emits light at a single wavelength.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to calculate the fraction of solid angle occupied by the solar panel as viewed from the light bulb. Questions arise regarding the definition and application of solid angles in this context, as well as requests for relevant formulas.

Discussion Status

Some participants have provided guidance on calculating the power received by the solar panel based on the solid angle it occupies. There is an ongoing exploration of the definitions and calculations involved, with no explicit consensus reached yet.

Contextual Notes

The original poster has indicated a lack of pertinent equations and has not made significant progress in their attempts to solve the problem. The discussion includes considerations for both circular and non-circular solar panels, with the latter being noted as potentially complex.

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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 19 ·
Replies
19
Views
9K
Replies
2
Views
3K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
7
Views
2K