How can I estimate the optical power that a single-color LED generates?
So in some cases, you have it easy. As soon as the emitted light is out of range of the visible spectrum, the only measure for the manufacturer is the radiant flux which is the optical power, and exactly what we are after.

For example the NCSU033B is a UV-LED by Nichia and they give the following data:


Forward current: 500 mA
Forward voltage: 3.8 V
Radiant flux: 450 mW


The electrical power is 1.9 W, the light output is 450 mW. So it has an efficiency of 23.7%.

That's pretty simple right?



Now with visible light LED, you usually get a figure like lumen or candela. Which are based on the weighting of our eye. And things will get complicated.

So the next easy step is when the luminous flux in lumen is given. Under the assumption that we have a single wavelengh light source, which we of course never have.

We take again the weighting curve:

Public Domain, Link

In this diagram the weighting is given in a range from 0 to 1. What's missing is that at the peak you have  \$683~\frac{\text{lm}}{\text{W}}\$. And that factor is actually radiant flux to lumens. So we can write:

$$\Phi_e = \frac{\Phi_v}{683~\frac{\text{lm}}{\text{W}} V(\lambda)}$$

Where: \$\Phi_e\$ is the radiant flux in W, \$\Phi_v\$ is the luminous flux in lm and \$V(\lambda)\$ is the luminosity function.

The index e is usually there to denote energy, while v denotes visual, visual quantities are always weighted with the luminosity function.

Let's plug in an example: I've chosen the Osram Top LED and we can have a look at the different colors they are offering. The nice thing about this datasheet is, they give the luminous flux and luminous intensity, which we will visit in the next step. I wasn't able to find a datasheet which gives luminous flux and radiant flux - I guess that wouldn't be worth the trouble for the manufacturer.



Actually, and that's quite remarkable, Osram provides two wavelength figures. The one of the emitted wavelength and the dominant wavelength. As you can see the dominant wavelength is smaller, because of the form of the weighting function, our eye will "prefer" the wavelength nearer to 555 nm. We have to take the emitted wavelength for calculation, but I'm not sure how other manufacturers handle this. Well actually as long as we assume a single wavelength we always end up with some approximation.



And to do the calculation, we need the table for the luminosity function (sheet 1931 col observer column C).

Which gives us (linear interpolation, and datasheet values):


645 nm: 0.1382, Vf = 2.05 V, If = 20 mA, P = 41 mW, \$\Phi_v\$ = 2.2 lm
634 nm: 0.2266, Vf = 2.05 V, If = 20 mA, P = 41 mW, \$\Phi_v\$ = 2.7 lm
624 nm: 0.3330, Vf = 2.15 V, If = 20 mA, P = 43 mW, \$\Phi_v\$ = 3.5 lm
610 nm: 0.5030, Vf = 2.15 V, If = 20 mA, P = 43 mW, \$\Phi_v\$ = 3.7 lm
597 nm: 0.6693, Vf = 2.20 V, If = 20 mA, P = 44 mW, \$\Phi_v\$ = 2.7 lm


So we can calculate:


645 nm: \$\Phi_e = \frac{2.2~\text{lm}}{683~\frac{\text{lm}}{\text{W}} 0.1382}\$ = 23.3 mW
634 nm: \$\Phi_e\$ = 17.4 mW
624 nm: \$\Phi_e\$ = 15.4 mW
610 nm: \$\Phi_e\$ = 10.8 mW
597 nm: \$\Phi_e\$ = 5.9 mW


Which turns into these efficiencies:


645 nm: \$\eta = \frac{\Phi_e}{P_{el}} = \frac{23.3~\text{mW}}{41~\text{mW}}\$ = 56.8 %
634 nm: \$\eta\$ = 42.4 %
624 nm: \$\eta\$ = 35.8 %
610 nm: \$\eta\$ = 25.1 %
597 nm: \$\eta\$ = 13.4 %


So as the wavelength decreases the efficiency decreases, at least for this family of LED.



But what if my datasheet only supplies a value for luminous intesity which is measured in candela?

So we need a way to come from luminous intensity to luminous flux. Well for that we need something called the solid angle, which is measured in steradians. And a lumen is a candela multiplied by a steradian. A sphere has \$4\pi\$ steradians, so a light source emitting a candela uniformly in all directions would have 12.6 lumens. The solid angle is dimensionless like radian, but sr is often added to make clear why a candela suddenly turned into a lumen: \$1~\text{lm} = 1~\text{cd} * 1~\text{sr}\$

Now a LED has never a uniformly light distribution, at most it could cover a half sphere. Usually though it will have a smaller viewing angle. We take this as the apex angle of a cone. There is a neat little formula to calculate the solid angle of a cone with a given apex angle:

$$\Omega = 2 \pi (1 - \cos \theta)$$

Where: \$\Omega\$ is the solid angle and \$\theta\$ is half the apex angle of the cone.

With the Osram LED as before we get a range of candela in the datasheet and the apex angle of 120°:


645 nm: 0.355 ... 1.120 cd, mean: 0.7375 cd
634 nm: 0.450 ... 1.400 cd, mean: 0.9250 cd
624 nm: 0.560 ... 1.800 cd, mean: 1.1800 cd
610 nm: 0.710 ... 1.800 cd, mean: 1.2550 cd
597 nm: 0.450 ... 1.400 cd, mean: 0.9250 cd


So we first calculate our solid angle: \$\Omega = 2 \pi (1 - \cos 60°) = \pi ~\text{sr}\$ - that was easy for once.

And with that we get:


645 nm: \$\Phi_v = 0.7375~\text{cd} * \pi~\text{sr} = 2.32~\text{lm}\$ 
634 nm: \$\Phi_v\$ = 2.9 lm
624 nm: \$\Phi_v\$ = 3.7 lm
610 nm: \$\Phi_v\$ = 3.9 lm
597 nm: \$\Phi_v\$ = 2.9 lm


Which is more than what is given as typical value by Osram, but it's not very far off. I mean, the range given for the luminous intensity is huge and just taking the mean without knowing how their production varies is just a guess. Maybe keep this result in mind and take a bit less than the mean in future.

Actually, after thinking a bit more about it, I think the overestimation stems from not including the actual brightness per angle, as that will decrease from 0° to 60°.

With the lumen values, you can go and use the approach above again.



Now the above ways are an approximation based on simplifying:


the wavelength distribution of the LED to a single wavelength
the radiation characteristic to a simple cone


To do things right™ you will have to integrate over the actual distribution and radiation characteristic to get better values. Once you have mastered that, you can apply the same principle to any kind of light source. So you can go and do the same calculation for a white LED and see what happens there.

I'm actually inclined to do that. But it would probably explode this answer.