Why is it that there are two different equations for the minima of a single slit diffraction pattern and a double slit interference pattern?

In the double slit experiment, the two slits act as two point sources of coherent light, thus maxima occur when the two rays have a path difference that is dsinθ=n λ and minima occur when dsinθ=(n λ)/2 where d is the separation of the two slits and n is an integer. In other words, if their path difference is an integer multiple of wavelength then there will be constructive interference and if there is half an integer multiple of wavelength then there will be destructive interference.

However, a different equation is used for finding the minima of a single slit experiment; (b/2)sinθ=(n λ)/2, where b is the slit width of the single slit. The reasoning behind this derivation for the minima of single slit diffraction is that according to Hyugen's principle, there are an infinite number of light source in a single slit. Thereby, by taking a light source at the top of the slit and a light source at the middle of the slit, their path difference will hence be (b/2)sinθ. Additionally, two coherent sources of light interfere destructively if their path difference is half an integer multiple of wavelength, and hence (b/2)sinθ=(n λ)/2 for minima. From that equation, it can then be simplified to bsinθ=n λ.

Comparing the maxima equation for the double slit and minima equation for the single slit, we get that they are the same. Additionally, the minima equation for double slit and single slit are opposite of each other, despite both are built on a similar concept.

Another question is that, in the single slit experiment, why must the light source at the top and middle of the slit be used to derive where the minima occurs? Why can't it be the light source at the top and bottom of the slit? By taking the first point and the last point, their path difference will be bsinθ, and for destructive interference to occur, bsinθ=(n+1/2)λ, which would hence correspond with the equation for minima in a double slit experiment.

Last question is that, why isn't there an equation for maxima in a single slit?


2 Answers 2


About "(...) the minima equation for double slit and single slit are opposite of each other, despite both are built on a similar concept." no one has answered yet.

The answer to this conundrum lies in the fact that two-slit interference we consider only the very edges of a slit to be wavelet sources separated by $D$ and in one-slit interference we consider infinite number of sources in between the edges of the slit (including those extreme points). Their average distance to a point on the screen is equal to the distance between the slit centre and that point on the screen.

If you want the reasoning behind this, read on. What you'll find below is heavily based on Crawford's "Waves".

So in the far field limit, the pattern for two sources is defined by their relative phase difference (optical path difference) $$\Delta\varphi=k(r_2-r_1)=kD\sin(\theta)=\frac{2\pi}{\lambda}D\sin(\theta)$$ Two electric fields superpose giving their sum: $$E(r,\theta,t)=A(r)\cos(\omega t+\varphi_1)+A(r)\cos(\omega t+\varphi_2)\\=A(r,\theta)\cos(\omega t+\varphi_{avg}-kr)$$ where $A(r,\theta)=2A(r)\cos(\omega t+\frac{1}{2}(\varphi_2-\varphi_1)+\frac{1}{2}\Delta\varphi)$.

Having that irradiance is $I(t)=n\varepsilon_0c|E(t)|^2$, and that the electric field amplitude averaged over time $$\left<|E(t)|^2\right>=\frac{1}{2}A^2(r,\theta)$$ this gives averaged intensity $$I(\theta)=I_{max}\cos^2\left(\frac{1}{2}(\varphi_2-\varphi_1)+\frac{1}{2}\Delta\varphi\right)$$ When the two sources are initially in phase this simplifies to $$I(\theta)=I_{max}\cos^2\left(\frac{1}{2}\Delta\varphi\right)=I_{max}\cos^2\left(\frac{\pi}{\lambda}D\sin(\theta)\right)$$ From periodicity of $\cos^2$ function we get bright fringes at $\frac{m\pi}{\lambda}D\sin(\theta)=m\pi$ or $D\sin(\theta)=m\lambda$ for $m\in(0,\pm1,\pm2,...)$.

Now for diffraction on single slit. A slit of size $D$ can be divided into $N$ point sources (separated by $d$) of wavelets every one superposing with another to give average electric field (as in interference between two point sources) - so this will become generalized case of the previous one. $$E=\sum_{j=1}^{N} A(r)\cos(kr_j-\omega t)$$ or in terms of complex amplitude: $$E_c=A(r)e^{-i\omega t}(\sum_{j=1}^{N} e^{ikr_j})$$ Having all $N$ elements emitting rays at an angle $\theta$ (which is approximately valid in the far field limit), the $N$-th element travels distance $r_N=r_1+(N-1)d\sin(\theta)$. And so the complex amplitude becomes $$E_c=A(r)e^{-i\omega t}e^{ikr_1}S$$ where $S=\sum_{j=1}^{N}e^{i\Delta\varphi(j-1)}$. Because $$S(e^{i\Delta\varphi}-1)=e^{i\Delta\varphi N}-1$$ dividing each side by $e^{i\Delta\varphi}-1$ leads to $$S=e^{i\frac{1}{2}(N-1)\Delta\varphi}\frac{\sin(\frac{1}{2}N\Delta\varphi)}{\sin(\frac{1}{2}\Delta\varphi)}$$ And to complex amplitude $$E_c=A(r)e^{-i\omega t}e^{ikr}\frac{\sin(\frac{1}{2}N\Delta\varphi)}{\sin(\frac{1}{2}\Delta\varphi)}$$ where $r = r_1+\frac{1}{2}(N-1)d\sin(\theta) = r_1+\frac{1}{2}D\sin(\theta)$ is the distance between the centre of the slit and a point on the screen at an angle $\theta$ created between the ray coming from the slit's centre and the line perpendicular to the screen.

Taking now only the real part of the complex amplitude we obtain

$$E(r,\theta,t)=\left[A(r)\frac{\sin(\frac{1}{2}N\Delta\varphi)}{\sin(\frac{1}{2}\Delta\varphi)}\right]\cos(kr-\omega t)=A(r,\theta)\cos(kr-\omega t)$$

For $N=2$ as in the two-source problem of Young's experiment, this becomes $$E(r,\theta,t)=\left[A(r)\frac{\sin(\Delta\varphi)}{\sin(\frac{1}{2}\Delta\varphi)}\right]\cos(kr-\omega t)=\left[2A(r)\cos\left(\frac{1}{2}\Delta\varphi\right)\right]\cos(kr-\omega t)$$ This is on par with the previous result where we assumed initial in-phase relation of the source wavelets.

Furthermore, from consideration of $N\rightarrow\infty$ we then have that the total phase difference across the slit is $\Phi=kD\sin(\theta)=(N-1)\Delta\varphi\approx N\Delta\varphi$.

Taylor-serializing $\frac{\sin(\Delta\varphi)}{\sin(\frac{1}{2}\Delta\varphi)}$ in the $N\rightarrow\infty$ limit and having that $\Phi\rightarrow 0$ when $\theta\rightarrow 0$, electric field becomes: $$E(r,\theta,t)=\left[A(r,0)\frac{\sin(\frac{1}{2}\Phi)}{\frac{1}{2}\Phi}\right]\cos(kr-\omega t)$$

And so the intensity profile on a screen: $$I(\theta)=I_{max}\frac{\sin^2(\frac{1}{2}\Phi)}{(\frac{1}{2}\Phi)^2}$$

The dark fringes appear every time $\frac{1}{2}\Phi=m\pi=\frac{1}{2}\frac{2\pi}{\lambda}D\sin(\theta)$ or $m\lambda=D\sin(\theta)$, where $m\in(\pm1,\pm2,\pm3...)$.

So, the result is expected. The only difference between these two experiments is in fact spatial distribution. In reality these two phenomena coexist and there is no pure diffraction-free interference pattern.


For single/double slit experiments interference/Huygen's principle is employed which is a classical/historical explanation for observations. The math used provided an explanation for the observations and there have been discussions on the conflicts. Classical is still taught but a more advanced explanation is available that better explains slit diffraction:

In the modern understanding, whether it's photon or electron, each one has a wave function to describe it's propagation. As a single event a wave/particle is traveling towards the slits and chooses one based on probability, the probability is based on the varying EM field of the wave/particle interacting with the varying EM field of the slit(s) (lots of electrons moving in the slit material), the varying causes the randomness of which slit is chosen. The wave propagation function also causes the banding pattern seen, it's a solution of the wave propagation equations with the EM properties and dimensions of the slit(s). Maxwell's laws for EM are used.

The classical model could not explain diffraction when photons/electrons were used fired one at a time, no inference between photons was possible yet the diffraction pattern still emerged. The single photon is said to be "interfering with itself" going through the slit(s) but using field theory (the photon is a wave in the EM field) it would be better explained as an interaction with the slit(s).

The term "interference" can not explain what happens in a double slit experiment and leads to a lot of debate, it's outdated. The classical theory math was great as a first step but as you note it has inconsistencies. Huygen's principle explained water waves interacting with objects, although I'm not sure how accurately.

  • $\begingroup$ Yes that makes very good sense and is very logical. I am reading more of your posts and they are very informative. It is expensive to make good slits, it would be good to use sharp ones, dull ones and insulators vs conductors to see if patterns changes are measurable. $\endgroup$ Aug 22, 2018 at 3:21
  • $\begingroup$ you are able to perform these setups? Let me know about the results. $\endgroup$ Aug 22, 2018 at 4:20
  • $\begingroup$ No, unfortunately. $\endgroup$ Aug 22, 2018 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.