# How does the spot size of light change the photocurrent?

I want to understand the photo electric effect. If I shine light at a surface then a photon can knock an electron off the surface if its energy $$h\nu$$ is larger than some minimum material specific energy $$W$$. So if $$\Delta E = h \nu - W > 0$$ the kinetic energy of this departed electron is now $$\Delta E.$$ Let's say my light source has some power $$P$$ and I spot all that light into an area $$A$$ on the surface, therefore the light intensity is $$P/A$$. I was told that a higher light intensity means more electrons take off the surface.

Here is where my confusion comes into play. Intensity is a function of both power and surface.

Case 1: I increase the intensity by increasing the power. This means more photons hit the surface per unit time and hence more electrons leave the surface.

Case 2: I increase the intensity by decreasing the surface. This shouldn't change anything because it's the same amount of photons hitting the surface per unit time. Perhaps even lower the electron count, since fewer atoms are in the spot.