I have been wondering about this question for quite some time. Assuming an ideal case, the energy from photons hitting solar cells is converted into electric energy as described by the equation:

$RI^2t=W\equiv E=\hbar\nu$

where $\nu$ is the frequency of photons. Using a lens won't increase the frequency of photons, thus no extra electricity is generated.

Am I correct in thinking that no extra electricity will be generated by solar cells when a lens is used to focus light onto them?

  • $\begingroup$ If this was true, we could use a lens to spread out the photons, and generate the same energy from less light... and then by using more solar cells, more energy from the same amount of light! $\endgroup$
    – user253751
    May 6, 2015 at 7:53

4 Answers 4


Your equation is partly correct. You've calculated the energy per photon ($\hbar \nu$), but you've neglected the number of photons. That's why the units don't match (power is energy per unit time, while you've only got energy for each photon).

The ideal power (energy per unit time) depends on the area of the solar panel, $A_p$, the number of photons striking it per unit time ($\Phi$) and the energy of each photon, $E$, such that $W_{Ideal} =A_p \cdot \Phi \cdot E$.

A lens or mirror can focus light (a flux of photons) onto a small area. Under really ideal conditions, the area of the lens ($A_L$), would replace $A_p$ in the formula above. So, if the lens is larger than the solar panel, it can capture a larger flux of photons and direct them to the panel, increasing the power.

  • $\begingroup$ So if lens can increase electricity production, are they currently in used? $\endgroup$
    – TBBT
    May 5, 2015 at 23:56
  • $\begingroup$ Indeed $\endgroup$
    – Dan
    May 6, 2015 at 0:35
  • $\begingroup$ I might be wrong, but sun rays angle changes through out the day. Isn't this make using lens a bit non-practical? $\endgroup$
    – TBBT
    May 6, 2015 at 1:56
  • 2
    $\begingroup$ Lenses are used in a few cases, because they do allow more solar energy gathered per square inch of solar cell (by concentrating more square inches of incident light into one area). In some cases, this actually does provide benefits (the highest efficiency solar cells require substantial flux of photons to do their job). However, it is tricky to use because that also means there is an increased demand for heat dissipation, which is not easy. About two years ago, i did hear of a solar cell whose optimum efficiency was achieved at 100+ suns of lux, requiring lensing to reach that. $\endgroup$
    – Cort Ammon
    May 6, 2015 at 5:14
  • 1
    $\begingroup$ @TBBT Lenses can increase production by increasing the surface area, but so can mirrors which reflect light onto whatever is generating power. Take a look at these: en.wikipedia.org/wiki/Concentrated_solar_power $\endgroup$
    – Patrick M
    Nov 22, 2015 at 6:22

Yes, increasing the illumination on a solar cell by using lenses or mirrors increases the electric power output.

However, there are limiting factors. The efficiency of a solar cell goes down with temperature. The current stays roughly proportional to the photon flux, but the open circuit voltage goes down as the semiconductor junction is heated. Still, more flux yields more power, although not quite linearly.

Keep going, and the solar cell gets so hot that the semiconductor it's made from no longer acts like a semiconductor. That's around 150°C for silicon. If you can keep the cell cool, you can hit it with higher photon flux. However, other non-linear effects start to get in the way and you start getting diminishing returns at high flux levels.

  • $\begingroup$ Would, then, a small amount of magnification, say 1.5 or 2x, be worthwhile on currently available solar systems? Not factoring the difficulty of achieving magnification, focus change, etc. Just from the point of view of your typical consumer 100 watt solar panel. $\endgroup$
    – user5650
    Mar 19, 2016 at 18:28
  • $\begingroup$ @Frank: It all depends on how much temperature rise the panel can tolerate. On a hot day in August in Phoenix, probably not, since the panel is probably already close to its max temperature. In other cases, 1.5x or 2x could be fine. Keep in mind that 15% efficient means 85% inefficient, so 85% of the sunlight goes to heating the panel. At roughly 1.2 kW per square meter worst case, that's 1 kW per square meter of heat. Can it take 1.5 or 2 kW per square meter? Probably sometimes, but not likely in the general case if it's not designed for that up front. $\endgroup$ Mar 20, 2016 at 14:34

The greater the photon density at the frequency of interest, the more the power as the photons excite the semiconductor's electrons to higher energy orbitals to the band-gap and beyond. That being said,as Olin stated, the increase is not linear. Eventually as temperature increases, the increased photon intensity results in ever decreasing increments in power.

My suggestion would be to use lens filters and other methods to reject the photon wavelengths that are not of benefit. We only want the photons of wavelength that are tuned to excite the electrons across the band-gap for that particular semi-conductor.

Any photons that don't do this only cause increase in temperature. So you want to increase the incident photon density of only the photons of relevance.

You can actually cool the solar arrays by aluminum heat sinks beneath them that run water through them for water heater purposes. I saw such a setup at a trade show. It was by a Spanish company but I don't recall the name. The setup combined solar power for both electricity and convective water heating.


A new way to convert solar power has been discovered at the University of Michigan. Please check out : https://phys.org/news/2011-04-solar-power-cells-hidden-magnetic.html

It uses the magnetic component of light which manifested itself when high intensity light goes through a transparent but non-electrically conductive material, glass for example. The light must be focused to an intensity of 10 million watts per square centimeter. Sunlight isn’t this intense on its own, but new materials are being sought that would work at lower intensities.

Your concentration of light using lenses and mirrors has limited potential for extracting more energy from the conventional solar cell, but it would certainly increase electrical energy with this new method.

  • $\begingroup$ As you mentioned, this doesn't answer the question. $\endgroup$
    – hazzey
    May 8, 2017 at 20:09

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