# Image of Sun

Image of Sun by a magnifying glass

Assume that you are located at 33º North in Phoenix, Arizona , having a 50 mm wide magnifying glass of $f = 75 mm$. The intensity of the rays from the sun is 980 $W/m^2$ perpendicular to the Earth. The size of sun in the sky is 8.0 mrad. Please find the intensity of the image of the sun made on the ground by the magnifying glass. (Earth is tilted 23º towards the sun at this time).

The sun rays make with the normal to the surface of the Earth an angle of $=10 deg$ . The intensity normal to the surface of the earth is thus

$I=I_0*cos⁡ ∝$  ,where $I_0=980 W/m^2$

The Sun is seen on an angle $β=8 mrad$ . The image of the Sun given by the lens is has the same angular diameter which means

$d/f=tan⁡(β)=β$ so that $d=f*β$  (for this relation to work $β$ is in rad)

The total power on the surface of the lens is

$P=I*S(lens)=I_0*cos⁡∝*S(lens)$

The total intensity of the bright spot produced by the lens (image of the sun) is

$I(spot)=P/S(image) =I_0 cos⁡∝*S(lens)/S(image) =I_0*cos⁡ ∝*D^2/d^2 =$

$=I_0*cos⁡ α*D^2/(f^2 β^2 )$

$I(spot)=980*cos⁡(10 deg)*(0.05/0.075)^2*1/0.008^2 =6.70*10^6 W/m^2$