In a double slit experiment, does each and every photon leave a dot on the screen in the bright area?

The photon is a quantum mechanical entity.

Number 1) cannot be right, because the experiments with single photons at a time give single dots as photon footprints not bright regions .

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

The photons arrive one at a time at the left, the interference appearing slowly with the accumulations on the right, a probability distribution.

The boundary condition problem "photon + two slits of given width a distance apart" defines the wave function of the system,$$Ψ$$ . The $$Ψ^*Ψ$$ is the probability distribution for the accumulation of photons.

It will depend on the experiment if every photon that leaves the source and hits the double slits leaves a footprint, the efficiency of the screen. There will be experimental errors. In principle every photon that passes the double slit should end on the screen.

Which one is right?

I think that neither of the two options that you presented is completely right, but I think that the true answer contains elements of both choices.

Without getting into the real physics (i.e., the math):

• The geometry of the experiment (two slits, and the screen) defines a wave function.
• Any photon that gets past the slits will make at most one mark on the detector, but the detector probably is less than 100% efficient, and some photons may fail to leave a mark.
• Considering only those photons that leave a mark, the wave function predicts the spatial distribution of the marks that they leave.

Edit: Oops! I forgot to say, "...and the wavelength of the light." You can't know the wave function if you don't know the wavelength of the photons that you're sending through.