# Intensity of Two Slits of Finite Width Pattern

I am a college sophomore in an introductory E&M course. Young & Freedman presents the resulting diffraction/interference pattern when shining monochromatic light through two slits of finite width (finite as opposed to infinitesimally small such that only interference would occur). The mathematical expression for the resulting pattern is explained to be the intensities of the double slit interference and of the finite width single slit diffraction multiplied together.

However, the text does not explain why we multiply the intensity for the double slit interference pattern with the intensity of the finite width single slit pattern. Why do we multiply the two intensities together to get the resulting pattern?

Why do we not analyze the resulting pattern by superimposing the waves of interference and the waves of diffraction?

• 1st year physics keeps it simple to make the theory work. There are problems if you dig deeper ... for ex. water waves thru a single slit only diffract but do not interfere ... contrary to Huygens (1600s). In quantum optics you learn a more advanced theory (Feynman/Dirac ... each photon finds its own path (historically stated as " interferes with itself"). In the dark areas of interference patterns there are no photons, bright areas have all. Photons take predetermined paths ... probably determined by virtual fields of exited electrons of atoms. Apr 27, 2022 at 12:29
• Back when I taught diffraction, I didn't drop this result as-is on my students' laps but simply used the Huygens-Fresnel principle to compute the diffraction by each point of each slit, and then added them up (with a few subtleties) to compute the interference pattern. The result you mentioned simply emerged as a consequence. May 6, 2022 at 8:21
• Every single edge has a diminishing Diffraction pattern (see straight edge diffraction) and they are all the same depending on the light source. A double slit has four edges and they all overlap on the detection screen to create the double slit pattern. The patterns are created by photons that diffract around the edges or scatter off the edges. See “Single Edge Certainty” at billalsept.com May 10, 2022 at 0:10

As already mentioned in @Miyase's comment you can derive this result by applying the Huygens-Fresnel principle. Consider two slits (each having width $$a$$, at a distance $$g$$) illuminated by light of wavelength $$\lambda$$ (and hence wave number $$k=\frac{2\pi}{\lambda}$$).

The following calculation is not essentially difficult, but requires some familiarity with complex analysis. Therefore it is usually not taught in an introductory physics course.

The light amplitude $$E(\theta)$$ into direction $$\theta$$ can be calculated straight-forward by summing the contributions

• by the first slit ($$x$$ from $$-\frac g2-\frac a2$$ to $$-\frac g2+\frac a2$$)
• and by the second slit ($$x$$ from $$+\frac g2-\frac a2$$ to $$+\frac g2+\frac a2$$)

The path difference of each contributing ray (compared to the path length of the ray originating from the center of the double slit at $$x=0$$) is $$x\sin\theta$$. And hence its phase is $$kx\sin\theta$$.

Summing these contributions you get \begin{align} E(\theta) &=E_0\left( \int_{-g/2-a/2}^{-g/2+a/2}e^{ikx\sin\theta}dx +\int_{+g/2-a/2}^{+g/2+a/2}e^{ikx\sin\theta}dx \right) \\ &=E_0\left(e^{-\frac i2 kg\sin\theta}+e^{+\frac i2 kg\sin\theta}\right) \int_{-a/2}^{+a/2}e^{ikx\sin\theta}dx \\ &=2E_0\cos\left(\frac 12 kg\sin\theta\right) \frac{\sin\left(\frac 12 ka\sin\theta\right)}{\frac 12 k\sin\theta} \end{align}

From this you get the intensity $$I(\theta)=|E(\theta)|^2$$ (using $$I_0=4E_0^2a^2$$ as an abbreviation) $$I(\theta)=I_0 \cos^2\left(\frac 12 kg\sin\theta\right) \left(\frac{\sin\left(\frac 12 ka\sin\theta\right)}{\frac 12 ka\sin\theta}\right)^2$$

You see, you got a product of two factors. The first factor is the interference pattern of 2 infinitesimally narrow slits separated by distance $$g$$. And the second factor is the diffraction pattern of a single slit of width $$a$$.