## Saturday, March 17, 2012

### Light Concentration by a Lens

We are now in a position to do something that I wanted to do at the beginning of February and that is to determine the light concentrating power of a lens. First we need to redo the plot of the focal region to get a slightly different view. There seems to be some discrepancy in the vertical scale of this plot from the previous one but we will use these values and treat them as nominal.

The rays from the lens come to a focus over a range of 6 μm in the depth of field. The best focus appears to be at just under 5.1 cm from the center of the lens where the radius of the focal area is 15 μm. We can now use the flux tube method to calculate the irradiance of the focal area. If the light passing through the lens has an irradiance of 1 kW/m2* then the flux is the irradiance, E1, times the transmittance of the lens times its cross-sectional area, A1. This is equal to the flux passing through the focal cross-sectional area which enables us to determine the irradiance there. This comes to 0.4 gigawatts per square meter which is quite large.

If this energy was absorbed by matter at the focus its temperature would have to be raised considerably so that the radiation emitted would equal the radiation absorbed. We can calculate a concentration factor for the lens, FC, by dividing the two irradiances. If the focus is too sharp one risks damaging the components of an optical system. This includes the human eye so caution is definitely advised.

These focus computations that involved ray tracing for the spherical, parabolic and hyperbolic lenses used idealized models and Snell's law of refraction. The actual situation is a little more complicated since light is scattered by imperfections in the lens and can be internally reflected at the boundaries of the media. One would expect the sharpness of the focus to deviate somewhat from the idealized model.

*Edit (Mar 21): Deleted the line mentioning the Sun since there was an error for the concentration of light from it. The Sun and its image both have the same angular size but the cross-sectional radius of the image, rimg, is determined by its position which is at the focal length of the lens, f. A corrected calculation for the irradiance of the Sun's image is given below. ES = 1360 W/m2 is the solar constant but the transmission factor T was left at its nominal value of 1.

The result can be expressed in terms of the F-number of the lens. The concentration of sunlight is not as high as was originally indicated but the result is still much larger than the solar constant.