For waves, the intensity pattern for the double-slit experiment is given by the following equation (as of David Morin's notes on waves)

$I(x) = I(0) \frac{D}{\sqrt{x^{2} + D^{2}}}cos^{2} \left ( \frac{x \pi d}{\lambda \sqrt{x^{2} + D^{2}}} \right )$

Where $D$ is the distance between the measuring apparatus and the slits' wall, $d$ is the distance between slits, $\lambda$ is the wavelength of the incident light, $x$ is the distance along the apparatus' wall and I(0) is the intensity at x = 0.

This equation yelds a graphic like

This result is widely proved in standard textbooks.

On the other hand, although most introductory quantum mechanics books start with the double-slit experiment, the standard procedure if to only affirm that the same experiment done with particles (or by checking through which slit does each photon goes through) has a shape like

[2

Where the black graphics would refere to having one or the other slit open, while the net result of having both slits open would be simply the red graphic, which is a simple sum of the black graphics. This is usully presented as an empirical result, without further details on the math behind it. In fact, I haven't found any book or articles that breaks down the math of this result.

My personal guess is that each of the black graphics represents a Gaussian, centered in the position in the $x$-axis of its corresponding slit. Even then, I am not sure of how to proceed for the particular of case of incident waves whose photons behave like particles, specially because many question can be opened around it:

How do these Gaussians vary with the total intensity of the incident wave? Do they depend of wavelength of each photon? How is the intensity of each photon measured? Etc.

My main goal was checking if the total area below the intensity's graphic doesn't change after collapsing the wave function (i.e. there is no net energy difference between the experiments). This is an impossible task without further details on the latter graphics.

For waves, the intensity pattern for the double-slit experiment is given by the following equation (as of David Morin's notes on waves)

$$ I(x) = I(0) \frac{D}{\sqrt{x^{2} + D^{2}}}cos^{2} \left ( \frac{x \pi d}{\lambda \sqrt{x^{2} + D^{2}}} \right ) $$

Where $D$ is the distance between the measuring apparatus and the slits' wall, $d$ is the distance between slits, $\lambda$ is the wavelength of the incident light, $x$ is the distance along the apparatus' wall and I(0) is the intensity at x = 0.

This equation yelds a graphic like

This result is widely proved in standard textbooks.

On the other hand, although most introductory quantum mechanics books start with the double-slit experiment, the standard procedure if to only affirm that the same experiment done with particles (or by checking through which slit does each photon goes through) has a shape like

[2

Where the black graphics would refere to having one or the other slit open, while the net result of having both slits open would be simply the red graphic, which is a simple sum of the black graphics. This is usully presented as an empirical result, without further details on the math behind it. In fact, I haven't found any book or articles that breaks down the math of this result.

My personal guess is that each of the black graphics represents a Gaussian, centered in the position in the $x$-axis of its corresponding slit. Even then, I am not sure of how to proceed for the particular of case of incident waves whose photons behave like particles, specially because many question can be opened around it:

How do these Gaussians vary with the total intensity of the incident wave? Do they depend of wavelength of each photon? How is the intensity of each photon measured? Etc.

My main goal was checking if the total area below the intensity's graphic doesn't change after collapsing the wave function (i.e. there is no net energy difference between the experiments). This is an impossible task without further details on the latter graphics.