# Light Intensity vs Visibility/Brightness?

I am curious about something that doesn't seem to be pointed out very much.

This is the double-slit intensity pattern as a graph:

This is what you're supposed to see in real-life:

If you look at the graph, the blue envelope has the middle maximum 24 times higher than the second maximum to the sides. However, in the picture we see the "second group of fringes" to the sides are as visible as the middle group.

At some point I actually had my hands on an actual single-slit setup and I measured the light at each point of the pattern with a photodetector and oscilloscope. I wasn't able to complete all of my measurements, but I thoroughly measured the intensity at a portion of the pattern and it fit the $\text{sinc}(x)$ curve fairly well, so I'm convinced that the graph above does give the right intensity distribution.

However, I'm still trying to understand why such small intensity light would look just as bright as something many times as intense. Is there an explanation accounting this? How does intensity relate to visibility?

Edit: Apparently, some people are saying the image above is likely to have been meddled with to show each group of fringes equally bright. This made me decide to get some of my own pictures of the single-slit pattern.

This is during the night:

This is during the day:

This is during the day, with more sunlight in the room:

This is during the day, but with a flash:

The picture with the flash completely eliminated all the other fringes in the picture, so it seems to support one of the answers involving exposure considerations of cameras.

• Might be of help, in addition to the nonlinear response of whatever medium recorded the picture at the bottom en.wikipedia.org/wiki/Weber%E2%80%93Fechner_law – user126422 May 10 '17 at 2:26
• Yeah, I'm not sure if the image is an accurate representation of the intensity pattern, either. Just sitting here at my Mac and using a little utility to measure image intensity at three points I get (R,G,B)=(232,114,55) for the highest intensity at the center, and (236,64,58) for the highest intensity at the "second maximum", and (15,19,23) for the black background. So, curiously, the intensity of red is higher at the secondary max than at the center. The center registers more 'green' than the secondary max, though. Clearly, this image cannot be used to directly measure the intensity profile. – Samuel Weir May 10 '17 at 2:37
• @SamuelWeir I took some pictures of my own, and they definitely look different compared to the original one that I posted. – Maximal Ideal May 10 '17 at 19:22
• @SpiralRain - Yeah, your pictures look better, but even so all of these pictures have central spots which are saturated (i.e., "over-exposed:). The image sensor just maxed-out at the center spot because there is so much light there. To actually get an image which accurately represents the image intensity, you would have to manually fiddle around with the camera's exposure settings so that the center spot is not overexposed. Also, would need a good lens so there is no lens flare or lens haze cast by the bright center spot that affects measurements of adjacent spots. – Samuel Weir May 10 '17 at 20:04

If you have ever tried to take a photograph of an interference pattern you will soon become aware of the difficulty of obtaining a picture of the pattern as "good" as that which you can see with your eyes.

You will find that if you get the exposure of the central part of the pattern right you cannot see higher order fringes.
Adjust things so that you get nice higher order fringes and the central part of the pattern becomes overexposed.

Then you can have a similar problem when displaying the image on a computer screen.

Here are two images which illustrate the problem.

The top picture shows the Sun correctly exposed but the building under exposed whereas the bottom picture shows the building correctly exposed whereas the Sun and the surrounding sky is overexposed.

The dynamic range (ability to process a range of light intensities) of a camera and a computer screen are far inferior to that of the eye.
So your picture of an interference pattern is not a true reproduction of the actual real life intensities.
The chances are that it was either computer generated or processed to show the salient features at the expense of getting the intensities wrong.

High dynamic range imaging has been made easier with the advent of digital cameras and computers where techniques are used to improve the quality of the final picture.

An example from the Wikipedia article shows how a number of images can be combined to produce the "desired" image although you must remember that the image you see might be degraded by the poor dynamic range of your computer screen.

Without any image processing you might get something like this:

.

So now take another picture with the central fringes correctly exposed and combine the centre of this image with the outer parts of the image above and you have a nice representation of the interference pattern that you can see with your eyes.

• I took pictures of single-slit patterns previously, and come to think of it I actually did notice that the middle spot was bright white as opposed to red. I'll post some of them. – Maximal Ideal May 10 '17 at 18:55

Strictly speaking, visibility/brightness is a perceptional construct i.e., it's subjective to the eye's perception of light. It's analogous to loudness in the case of sound. While the intensity remains constant at some distance and displays a trend of varying inversely with the square of the distance between the source and the observer, the visibility/brightness might not necessarily do the same. That's because it's not a physical construct. You cannot bring it down numerically unless you look into the perception biases in humans. Even then, the most you could do, for all practicality is bring it down qualitatively.

One notable bias in case of sound is the psycho-acoustic curve. How we perceive loudness is weighed by this curve. Certain frequencies are more predominant than others, even if the amplitude across all audible frequencies are evenly distributed.

The pressure level or Decibel gives you intensity, NOT loudness. The loudness can be found after weighing found intensity with an (mostly) experimentally-derived curve. Therefore, loudness ≠ intensity.

You can read more on Equal-loudness contour on Wikipedia.

My advice to you: Trust your instrument, it's most probably right every time. Use senses only as a back up to find obvious errors.