In sunlight a shadow looks darker relative to the sunlight, but is it actually brighter than when there is cloud cover?

Question

Considering the same point in a shadow on the ground, how does the incident light level vary when there's cloud cover diffusing the sunlight compared to no cloud cover?

In sunlight a shadow looks darker relative to the sunlight, but is it actually brighter than when there is cloud cover?

Example comparing the same scene in my garden, where the circled area is is interest.

(diffused sunlight f/1.8 1/385 4.44mm ISO62)

(sharp sunlight f/1.8, 1/1957 4.44mm ISO51)

I appreciate there are many factors that may influence this:

• Height of surrounding objects,
• Reflectivity of surrounding objects
• Time of day
• Thickness of cloud

Does anyone have a way of measuring this? i.e. long exposure of film or light sensor etc.

Update with experimental results

Following @Ruslans suggestion I have taken measurements using my phone Lux meter. It seemed sensitive on phone orientation so I have listed the readings from all orientations here and everaged them.

Day of March	time	Sunlight	Cloud cover	Lux <-	Lux /\	Lux ->	Lux /	Lux (Avg)
23	13:55	Sharp	thin 10%	4453	5737	7130	5788	5777
23	14:05	Diffuse	thin 10%	4589	5135	5318	4295	4834.25
23	14:10	Diffuse	thin 10%	10041	7969	9049	8495	8888.5
24	10:40	Diffuse	thin 100%	21060	19742	17105	19145	19263

By far the highest reading was with 100% light cloud cover providing diffusion from all directions. It doesn't take much cloud cover to push the diffuse light readings above the readings in the sharp shadow though.

Answer 1

I have a lux meter (MAX44009) on my window, the instrument's surface is perpendicular to the ground. The window direction is NNE, so direct sunlight only appears here in summer. This is not quite your setup (where your grass is horizontal), but I think it should be enlightening too.

Below are measurements from two days: one of them is cloudy, and the other is perfectly clear.

Vertical scale is labeled in lux, and horizontal scale's upper row denotes time in MSK time zone. As you can see, on the clear day maximum illuminance is $7500\,\mathrm{lx}$, while on the cloudy day it reaches $12\,000\,\mathrm{lx}$. But you can also see that at some points in time it drops considerably lower than the clear-sky illuminance: e.g. at 13:40 it's at $5800\,\mathrm{lx}$ on cloudy day while at clear day it's at $7300\,\mathrm{lx}$.

The conclusion is that clouds are very variable in their thickness and shape, and illuminance can get higher or lower than that from the clear sky.

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If you want to do your own measurements, you can easily do it with a smart phone: most of them have lux meters built-in, which are used to adjust screen brightness to match ambient lighting. Using a Lux Meter app (e.g. this one for Android) you can read your phone's measurements in real time.

Answer 2

As already mentioned by tom10, you could use two images with the same exposure (i.e. the same camera settings) to compare the brightness.

You do have two images, but their exposure settings are different. By increasing the exposure of the second (sunlit) image by about 2.7 stops¹ and decreasing the exposure of the first image by about one stop², we can emulate what the images would look like if the camera settings would have been the same:

At the top, I show (approximately) the same part of both images magnified. As you can see, the adjusted exposure is actually very similar $-$ the average brightness values are 31% (left) and 34% (right). So at least in your case, the brightness is (approximately) the same.

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This result is only approximate, but I think it is a good estimate. If you wanted a more precise measurement using a digital camera, you should set it to manual exposure settings and don't change them between taking the pictures. Also, the images should be taken one after the other as soon as possible, since the overall brightness could change.

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¹ I am not sure if you are familiar with the term "stop". In photography, it usually means doubling/halfing the amount of light that hits the sensor. You can read in more detail about this on Photography Stack Exchange: What does f-stop mean?, What is one “stop”?

² I could simply have increased the exposure of the right image by 2.7 stops, but I found this to reduce the quality too much, so I chose to adjust the brightness of both photos a bit.

Answer 3

In sunlight a shadow looks darker relative to the sunlight, but is it actually brighter than when there is cloud cover?

This is very hard to tell with the eye, and it's probably about the same. Shadows like this (where you can clearly see the objects within the shadows) are usually darker by a factor of 2-10, and cloudy days are usually darker by a factor of 2-10 (depending on the thickness of the cloud layer -- a well known number of solar panels). So there's no a priori way to know which is brighter.

For your picture, looking within an image from a single picture that was taken with a normal camera (ie, not post-processed to compress the dynamic range, etc), you can just compare pixel values (which are linear with intensity) and you can see that your shadow vs sunlight is different by about a factor of 3. You can't directly compare the pixel values to the above images because the ISO and shutter speed are different, but if they were the same, you could just read of the pixel values there too. (It's also easy to use shutter speed and ISO values to calculate a comparison, but it's easier and more accurate to take the picture with the same settings.)

Does anyone have a way of measuring this?

Almost any way of measuring the brightness should work, and it should be easy. The light intensity should be different by a factor of 2 to 10, so is well within an easy range of measurement. You could use either a light sensor with a sunlight range placed in the shadow, a camera where you can set the exposure entirely manually (shutter speed, aperture, ISO, etc) and take the same pictures as above, or a light meter.

An aside on relative vs absolute intensity perception:

That the shady spot looks darker on a sunny day is based on how human perception works in that human perception is usually making relative comparisons. This is particularly true in vision, which can work over many orders of magnitude in an absolute sense, but only about two orders of magnitude in a relative sense. For example, relative to our environment at the time, a black that has 1/100th of the light looks about the same to us as a black that has 1/1000th of the light, whereas if we were in an environment that only had 1/1000th of the light, we could adapt to that and still see. But there are many tricks that the visual system employs to gain this functionality, but, on the other hand, most sensors are absolute and not relative, so they make these measurements easily. Furthermore, perceived intensity (ie, brightness) is highly non-linear with intensity, direct perceptual comparisons are difficult.

There are several well known illusions which show that people are particularly bad at measuring absolute light levels, one of which is the checker shadow illusion, where in the image below, squares A and B have the same absolute brightness: