Why does color vision improve in color-blind persons using these filter glasses from Enchroma? Will a color blind person be able to see the same colors on a television? I'm asking, because the colors for every point on a TV screen are combinations of red, blue and green.
1 Answer
Short answer
Optical filters can optimize color contrast, but never restore color blindness. Filters remove colors, they can never add something. However, glasses like this can prove helpful for certain sub-populations of people that are partially color blind.
Background
First, the term color blindness is deceptive. It is better to talk about colour vision deficiency (CVD). This, because many people labeled as color blind in fact have all the red, green and blue cones like the normally-sighted. These people have protanomaly (abnormal, but still functional red cone photopigment), or deuteranomaly (abnormal green cone photopigment). These two conditions are mild and don't affect daily life too much (source: NEI / NIH).
The other two most common forms of CVD are protanopia and deuteranopia. Protanopes have no working red cone cells and deuteranopes have no green cones. These conditions are more severe and do affect daily life as color perception is compromised. These conditions can be coined color-blindness, as these folks miss one type of cone altogether (source: NEI / NIH).
The glasses of Enchroma target the first two conditions, namely the anomalies and not protanopia or deuteronopia. The Technology page of Enchroma explains why. Basically, protanomaly and deuteranomaly are characterized by the red and green pigments having a shifted spectral maxima. The end result is the same, namely more overlap between the spectral absorbance curves (shown for deuteranopia in Fig. 1). In normal vision, the overlap is already pretty extensive, meaning that green light activates a large portion of red cones and vice versa. In protanopia or deuteronopia this effect is further increased, resulting in a degraded capability to separate reds and grees. For example, protanopes perceive reds, oranges, and yellows as greener, and colors are not as bright. Deuteranopes perceive yellows and greens as more red and they have difficulty telling violet from blue (source: NEI / NIH).
Fig. 1. Left: normal cone spectral distribution vision, right: cone distribution in Deuteranopes. source: Enchroma
What Enchroma has come up with is actually really clever (!) - they've put a spectral notch filter in place and wedge it in-between the red and green spectral optima, thereby removing the large area of overlap. This reduces brightness, but it will enhance contrast when the glasses are worn (Fig. 2).
Fig. 2. Spectral notch filter to enhance green-red color contrast. source: Enchroma
And yes, above considerations hold for any colored object, including monitor pixels.
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1$\begingroup$ I am colourblind and own a pair of the enchroma glasses. For me (experiances seem to vary) the glasses certainly make colours look different however there is only limited improvement between difficult red-green differentiation. I've yet to try a colourblindness test wearing them! $\endgroup$– TroysephJun 14, 2016 at 13:34
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$\begingroup$ I've been reading around a lot and I still don't understand how this is meant to work. Surely it just makes the wearer blind to wavelengths of light filtered out by the glasses? The wavelengths on either side of the notch (not being blocked) are going to excite the red and green cones about the same amount, just as they were before, and so will still be difficult to distinguish. I haven't found an explanation that feels sufficient yet. $\endgroup$ Sep 14, 2017 at 22:18
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$\begingroup$ The wavelengths on the side will excite as they did before, but the regions of high overlap are excluded by the notch. Imagine that the area on the left excites G 1 unit and R 0.2 units, while the are covered by the notch is 5 units G and 5 units R. With the notch, you have 1:0.2 or a contrast of 5. Without the notch you have (1+5):(5.2) or a contrast of 1.15. Contrast would be enhanced almost 5-fold. $\endgroup$ Jan 11, 2020 at 4:14