Suppose you are sitting outside in the dark and the video camera in the mobile is ON. If the clouds are lightning again and again, will this damage the video camera? As our eyes can get damaged by light (not thunder but maybe very intense light), I thought maybe the mobile camera can get damaged too.


2 Answers 2


Basically, no.

Of course there is some luminous intensity above which the cell sensor will die, but I don't think Lightning will get there.

  • Lightning only lasts about 30 micro-seconds so the absorbed energy per pixel is actually quite small
  • You observe lightning from a great distance (intensity drops off at the square of the distance)
  • The PRP (pulse repetition period) is extremely high compared to the pulse-width (e.g. low energy waveform)

Let's do some math...

According to the National Oceanic and Atmospheric Administration (NOAA) the average lightning bolt contains enough energy to light a 100W incandescent bulb for 3 months.

That's almost a billion (777,600,000) Joules of electrical energy

Only a small fraction of that energy is converted to optical energy (light), just like the incandescent (~3%).

16 lumen/Watt,electrical

1/683 lumen/Watt,optical

2.35% Watt,electrical --> Watt,optical

~18,200,000 Joules,optical per lightning strike (typical)

Assuming you are 1km from the lightning bolt (VERY CLOSE!), the emitted optical energy is spread over the surface of a sphere

100,000 cm in 1km

Area = 4 Pi r^2 = 125,660,000,000 cm^2

Irradiance = Energy / Area = 0.00015 Joules,optical/cm^2

Here is the human eye safety limit for collimated polarized light.

enter image description here

The average lightning bolt lasts 30 microseconds (same NOAA citation used above). Therefore the maximum safe irradiance from a lightning bolt is:

.035 * 30e-6 = 3e-7 J/cm^2

The cell phone sensor (let's use an iPhone1) has 2048 pixels in 0.358 cm:

1/(2048/0.358) = 1.75 micron width of a pixel

= 1.75^2 micron^2 area of a pixel

pixel area / illumination sphere = 2.43e-19

The amount of the original almost billion Joules of electrical energy that reaches an individual pixel in your imager is about 1/2.43e-19 times smaller!

4.43e-12 Joules,optical/pixel

Therefore you are below the human damage threshold by five orders of magnitude. Even if we assume the full electrical energy of the bolt was light you get:

2.43e-10 Joules,optical/pixel

Still three orders of magnitude too small.

Cell phone sensors can withstand greater intensity than the human eye without damage since they can handle internal temperatures in excess of 125C which the human eye can, obviously, not.

Cell phone sensors are also a lot less sensitive to light than the human retinal cells, furthering their withstanding ability.

You can recalculate for being even closer to the lightning bolt and for farther away by recomputing with the equations above.

Expanding the assumptions

Optics is tricky and in the analysis I made a few assumptions that could be expanded to clarify.

Intensity reduction

One commenter bizarrely states:

"Intensity doesn't drop with distance"

Nope. Physically impossible for non-collimated sources. You can prove this to yourself by observing the lightning bolt from different positions. If you can walk around the lightning bolt and still see it, it's an approximate isotropic radiator (emits light in all directions). Given that lightning is a highly chaotic process involving the production of plasma (high-energy random motion), this is expected.

The camera cannot collect the light (energy) emitted in directions that do not point towards the camera. Therefore, the intensity, just the energy available over the area in which it is captured, will decrease with (the square of) distance in a loss-less medium.

The energy impinging the area of the camera represents an ever smaller quantity of the total illumination sphere of the lightning bolt as you back away from it.

The effect of the lens (focusing)

I modelled the lightning bolt as a point source and disregarded the lens of the camera. The lens has an almost insignificant effect. Let's put the lens back in and assume it's perfect (no spherical or chromatic aberration, perfect installation, and loss-less).

First lets validate the point-source assumption. Typical lightning bolt strikes from about 6,000 feet (~1.8km) to the ground and is 1 inch (2.54 cm) in diameter.

Standing 1km away you would need a camera with at least a 61 deg field-of-view, a requirement which our example iPhone1 camera meets. The camera therefore catches all of the light from the bolt radiated in its direction.

To validate our point-source assumption, we need to show that the total light reaching the camera from a single on-bore source is equivalent to the total light of a set of weaker point-sources distributed along the path of the bolt. To clarify the math, I'll model the lightning bolt as a straight arcing (uniform radial distance to the camera) line and the lens in a 2D plane (but the extension to 3D -- e.g. 2D image plane -- is logically trivial).

$$ \mid I \mid = \frac{P \cdot A_{total}}{n \cdot A_{lens}} n $$

That is to say that if we divide the lightning bolt into n-many points of light, distribute them along its length, and reduce their intensity by n (evenly distribute the output power among the points), the total intensity of the light on the lens is the same as if radiated by an on-bore point-source (assuming the lens can see the whole extent -- arc length -- of the lightning bolt).

The effect of the lens becomes:

$$ \mid I \mid = \frac{A_{lens}}{A_{sphere}\cdot n_{pixels}} $$

The number of effected pixels is a function of the composition of the image (how much of the frame the lightening bolt covers), the size of the bolt (width and height), and the distance of the observer (how close are we?!).

Rather than roll-up all of these parameters, I took a sample set of lightning photographs (example) taken under approximately these conditions, threshold filtered them, and averaged the number of white pixels. Approximately 5% of the sensor pixels are covered by the lightning bolt.

For the iPhone1 camera with lens area approximately 0.1 cm^2, the capture ratio on an illuminated pixel is 7.48e-18 -- for a lens focusing gain of only about 31x. Given the magnitude of the margins, the focusing/lens just doesn't play an important-enough role.

On the origins of damage

[you assume] the damage caused by optical light would be by radiative heating

That's a perfectly fine assumption (a lot like assuming there is gravity). Heat transfer is really all energy transfer is. You can read more about CMOS sensors and how damage occurs here.

and that a device that can withstand a higher temperature than an eye is necessarily more resilient,

Given that damage comes from heat transfer, withstanding greater heat transfer is by definition more resilient.

also that being less sensitive would make a device more resilient

This deserves some nuance, but not really that important given the other factors. Being less sensitive requires that you are less reactive to or less absorbing of the phenomenon of interest. In the general case, that implies less energy transfer, because the assumption is that the phenomenon is taken in isolation (ceteris paribus). In practice, there are alternate parallel mechanisms, so it may not hold.

  • 2
    \$\begingroup\$ Nice analysis. If you are getting closer to a thunderbolt, I'd start to worry more about the EM-field (and my personal safety) than dead pixels in a camera. \$\endgroup\$
    – jippie
    Mar 17, 2013 at 7:03
  • \$\begingroup\$ i love this and a lot of work went into it, but it contains several nonsequiturs, that the damage caused by optical light would be by radiative heating, and that a device that can withstand a higher temperature than an eye is necessarily more resilient, also that being less sensitive would make a device more resilient... \$\endgroup\$ Mar 17, 2013 at 13:48
  • \$\begingroup\$ I agree with the conclusion, but one flaw in your analisys is that you ignored that the light is focused. That means the intensity doesn't drop with distance, only the area of the projected image. The intensity of the smaller bright spot remains the same. Still, the luminous diameter of a lightning bolt is probably small enough to be less than a pixel when projected, which would make your analisys correct again. In other words, you are probably right, but you should at least mention a obvious and possibly important effect like this. \$\endgroup\$ Mar 17, 2013 at 13:56
  • \$\begingroup\$ 1.75 um for pixel size not 175 um newer sensors are 1.4 um and even 1.2 um \$\endgroup\$ Mar 17, 2013 at 18:16
  • \$\begingroup\$ @rawbrawb -- doh! typo. The math is right though. \$\endgroup\$ Mar 17, 2013 at 19:08

No, the light from lightning will not damage the video sensor in your phone.

  • 1
    \$\begingroup\$ I'm not sure this is definitively true. There is video of concert light-show lasers damaging the CMOS sensor of Canon 5D2 cameras on the internet. Considering that this demonstrates excessive light can damage some sensors in certain situations, it's distinctly possible that lightning could replicate the event. Of course, it's probable that to get enough light to damage the sensor, you would have to be ridiculously, dangerously close to the lightning, bus it's a possibility. \$\endgroup\$ Mar 17, 2013 at 3:21
  • 1
    \$\begingroup\$ A laser and lightning are two very different things. There are special precautions that must be taken when using those light show lasers, and stupid things still happen. \$\endgroup\$
    – Matt Young
    Mar 17, 2013 at 4:20
  • \$\begingroup\$ The 5D2 also uses much larger and faster (read: collects more light) lenses and has bigger pixels (read: greater insolation per pixel) and laser light is collimated and polarized, whereas lightning is not. \$\endgroup\$ Mar 17, 2013 at 5:45
  • \$\begingroup\$ Remember that coherent (laser light) is not entirely the same. The coherent nature and focused beans makes it easy for a lens (or eye) to focus the beam precisely causing thermal damage to the sensor. A bright flash from lightening is probably not concentrated enough to cause the same level of damage. \$\endgroup\$
    – mfarver
    Mar 17, 2013 at 18:27

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