Here, the sun is way off to the left and the rays of the sun are parallel and we have a surface that transects this beam of light. The beam's power can be denoted as phi = EA, so the irradiance of the beam can be represented as the power of the beam divided by the area A (on the surface). If we had a real surface that was tilted, we now have a different area but the same power hitting it, so this means we have less power per unit surface area hitting this space.

Here, the sun is way off to the left and the rays of the sun are parallel and we have a surface that transects this beam of light. The beam's power can be denoted as phi = EA, so the irradiance of the beam can be represented as the power of the beam divided by the area A (on the surface). If we had a real surface that was tilted, we now have a different area but the same power hitting it, so this means we have less power per unit surface area hitting this space.