Only a fraction of the solar radiation spectrum is visible to humans: the radiation with a wave length between 390 nm (violet) and 700 nm (red). That is the definition of light. Solar radiation received close to Earth but above atmosphere has a smaller fraction of energy shorter than 390 nm, with higher frequency than violet and called ultraviolet (UV), and a quite bigger fraction longer than 700 nm, with lower frequency than red called infrared (IR). UV and IR are just heat, not visible; UV is harmful for life. Average density of that radiation is 1361 W/m2.

At sea level, the solar radiation is filtered by the atmosphere. Some components are reduced. UV is filtered by the stratospheric ozone layer to a tiny portion. IR amounts to almost half the total solar power. The density of that radiation is about 1000 W/m2 when the sun is at the zenith (above your head). The corresponding illuminance (light flux per unit area) under clear sky is of more than 100,000 lm/m2 or 100,000 lux. It comprises direct solar radiation with its light fraction, and diffuse light from the visible sky and possibly other reflecting surfaces. 

This is a very short and simple introduction to the subject. Please pick more complete works like the classic Daylighting (Hopkinson, Longmore, Peterbridge) and Solar Engineering of Thermal Processes (Duffie & Beckman) and take good full two weeks to read and understand that material.

I agree with Alosio and add:

INSOLATION: it is the incident solar radiation and it is measures in watt / m^2 while the lux is the incident lumens per m^2. The lux is incident solar radiation scaled by the spectral response of the eye of the human being. The human eye responses only to a specific range of wavelengths of the electromagnetic radiation called Light. Then the light is only a small portion of the solar radiation.

The solar cells and photo detector has naturally a different spectral response different from the human eye. Radiometric is used with such detectors and converters of the solar radiation.

For more details please follow the link: http://depts.washington.edu/mictech/optics/me557/Radiometry.pdf

Best wishes

Hello

I agree with the previous answers of Aloisio and especially Abdelhalim which gave more details.

Hi, Dhanraj Chouhan! Do You found answer for Yours question?

I have the same problem.

I have cheap luxmeter UT383. It displays 64000 Lux for ClearSky conditions for noon and Sun height is near 66 degrees.

How to recalculate this Lux  to W/m2?

From numerous HandBooks on Physic I found this Magical numbers: 680 Lumen per 1 Watt; 4*Pi; 555nm. How to combine this numbers to obtain common relation E{W/m2}=E{Lux)*12.5e-3 ? What is the origin of Magical number 12.5e-3?

In may case this relation leads to 64000*12.5e-3=800W/m2.