A solar panel is tilted at an angle $\backslash$beta from the horizontal, show that the intensity of the direct solar radiation on the solar panel, I$\backslash$beta $($W$/$m$2),$ pointing towards the Equator, is given by the following equation:

I$\backslash$beta $=$ IHZ $[$sin$(\backslash$alpha $+\backslash$beta $)/$sin$(\backslash$alpha $)]$

Where $\backslash$alpha is the altitude ofthe Sun and IHZ is the intensity of the direct solar radiation on a horizontal surface.

$($b$)$ For Curepipe $($latitude $20\backslash$deg S$)$ on June $21$st$,$ IHZ $=400$ W$/$m$2,$ sketch I$\backslash$beta verses $\backslash$beta $(0\backslash$deg $90\backslash$deg $).$Which angle gives the maximum direct solar radiation intensity onto the panel?

Discuss the shape of your curve.

$($c$)$ For Sydney $($latitude $34\backslash$deg S$)$ on December $21$st$,$ IHZ $=950$ W$/$m$2,$ sketch I$\backslash$beta verses $\backslash$beta $(0\backslash$deg $90\backslash$deg $).$Which angle gives the maximum direct solar radiation intensity onto the panel? Discuss the shape of your curve.

$($d$)$ Use the formula for I$\backslash$beta to show or explain that for some values of $\backslash$alpha $,$ the direct component of solar radiation on a tilted surface can be higher than the direct component on a flat surface.