# Formula for secondary maxima in Fraunhofer Diffraction?

I was reading Fraunhofer diffraction and about the beautiful wave properties of light. Unlike Young's double slit experiment, I could not find a formula for the position of secondary maxima. The minima however are given by $$\sin(\theta_n)=\frac{n\lambda}{a}$$ where $$\lambda$$ is the wavelength of the light used and $$a$$ is the width of the opening. I could not find a formula for the maxima. Is it due to the fact that all the maxima other than the central are blurred out? One of my textbooks write that 'maxima are located nearly midway two consecutive minima'. The use of the word 'nearly' and the refusal of the writer to provide a formula for maxima made me wonder if there isn't a relation to find the exact positions of the secondary maxima.

Sorry if I sound naive and thank you.

The reason that most beginner's text-books don't give the exact locations of the secondary maxima, is simply because they are much harder to calculate. You need a lot more math to find them.

You start by calculating the intensity profile of single-slit diffraction. I skip the lengthy derivation here, because you can find it in any advanced text-book about wave optics. The final result is Fraunhofer's diffraction equation (see Diffraction - Single-slit diffraction):

$$I(\theta) = I_0 \left(\frac {\sin\left(\frac{d\pi}{\lambda}\sin\theta\right)} {\frac{d\pi}{\lambda}\sin\theta} \right)^2 \tag{1}$$ where $$d$$ is the slit width, $$\lambda$$ is the wavelength, and $$\theta$$ is the observed angle.

(image from Diffraction - single-slit diffraction)

By defining $$x=\frac{d\pi}{\lambda}\sin\theta$$, we can write (1) in a simpler way as $$I(\theta)=I_0\left(\frac{\sin x}{x}\right)^2 \tag{2}$$

Now it is straight-forward to find the $$x$$ values of the maxima and minima:

• The primary maximum is at $$x=0$$.
• The minima are at $$x=\pm\pi, \pm 2\pi, \pm 3\pi, \pm 4\pi, ...$$
• The secondary maxima are harder to find. Actually there is no analytical formula, and they can only be found by numerical methods. According to The unnormalized sinc function (Table 1) they are at $$x=\pm 1.429\ \pi, \pm 2.462\ \pi, \pm 3.470\ \pi, ...$$

Transforming from $$x$$ back to the original variables ($$d,\lambda,\theta$$) we have:

• The primary maximum is at $$\theta=0$$.
• The minima are at $$\sin\theta=\pm\frac\lambda d, \pm 2\frac\lambda d, \pm 3\frac \lambda d, \pm 4\frac\lambda d, ...$$
• The secondary maxima are at $$\sin\theta=\pm 1.429\frac\lambda d, \pm 2.462\frac\lambda d, \pm 3.470\frac\lambda d, ...$$

Now you can see: The secondary maxima are indeed approximately (but not exactly) half-way between the minima.

• Thank you. I read that the mathematical treatment of secondary maxima is beyond my current curriculum but one particular author in my country straight away gave the maxima at $\frac{(2n+1)\lambda}{d}$ which confused me because the result was seemingly compact and expressible. Your answer was enlightening. Feb 18, 2020 at 15:36