When dealing with double slit interference or interference of waves from numerous slits, the equation $nλ = d \sin θ$. However I do not understand why for calculating the angle for maxima and minima it is okay to consider the path difference of waves from each slit, where only one wave from each slit is considered. Surely, there are numerous waves from each slit, following Huygens' principle that each wave from a wave front is a source of a secondary wavelet. Therefore surely, there is interference of infinite waves from each slit. Therefore, surely it is unfair to assume that if one wave interferes in a certain way with another wave from another slit, then all the waves from each slit will interfere in the same way.


1 Answer 1


Actually, you're completely right. Most treatments of the double slit involve some kind of simplification. The usual idea is that we approximate each slit in the double slit to be a point source, so what we're really analyzing is the case of "two coherent point sources."

This approximation can be justified by the assumption that your slits are really tiny compared to the distance between them and the wavelengths. When each slit is tiny, there are "less amount of wavelets" coming out, and if there are less amount of wavelets coming out, there are less of them to be out of phase.

When you make your openings smaller and smaller, the two wavelets most out of phase are closer and closer, and they are more and more in phase.

Indeed, in the case of the single slit, when you make the slit opening smaller and smaller, the middle fringe on the target screen becomes wider. At some point, it becomes so wide that you might as rather approximate the slit as a point source.

Of course, you actually can do the full treatment of both the single slit and the double slit in detail (assuming far-field approximations). In fact, you can get the full blown function that describes the pattern. The derivation gets messy, but the idea is that you have infinitely many wavefronts coming out, and you want to "add up" the effects of each wavefront, each of which is infinitesimally small. It turns out this will require an integral to sum them up (much the same way people speak of integrals "adding up" infinitesimal area).

When you do this for a single-slit of width $a$, the intensity of the wave at the screen is $$ I(\theta) = I_{0}\frac{\sin^{2}(\tfrac{\pi a\sin\theta}{\lambda})}{(\tfrac{\pi a\sin\theta}{\lambda})^{2}}.$$ The double slit pattern (slit width $a$, distance between slits $d$) turns out to have a surprisingly similar function and the pattern is approximately given by

$$ I(\theta) = I_{0}\frac{\sin^{2}(\tfrac{\pi a\sin\theta}{\lambda})}{(\tfrac{\pi a\sin\theta}{\lambda})^{2}} \cos^{2}(\tfrac{\pi d\sin\theta}{\lambda}).$$

When we take $a\ll\lambda$, we have $\pi a\sin\theta /\lambda \approx 0$ so then $\sin^{2}(\tfrac{\pi a\sin\theta}{\lambda}) / (\tfrac{\pi a\sin\theta}{\lambda})^{2}\approx 1$ and the equation above becomes

$$ I(\theta)\approx I_{0}\cos^{2}(\tfrac{\pi d\sin\theta}{\lambda}). $$

This is the equation used to describe the double-slit pattern with "point sources" for "slits." Again, it's only a simplified idealization. You can see that in this equation, the maxima occur when $\tfrac{\pi d\sin\theta}{\lambda} = n\pi$ or rather $d\sin\theta = n\lambda$. But at the end of the day, this is nothing but a simplification/approximation.

Derivation Intro

I've been asked to provide a derivation of the formulas, which I thought would be an interesting exercise given that they're not easily found.

One rigorous way to do this is by using the Kirchhoff integral theorem applied to the wave equation with appropriate boundary conditions, then deriving Kirchhoff's diffraction formula, and applying appropriate limits/approximations. This approach can be found in various optics textbooks such as Eugene Hecht's Optics or Born and Wolf's Principles of Optics.

In my approach here, I will use Huygens's principle as a heuristic. A discussion of this approach and the equations are provided in the last chapter of Vibration and Waves by A.P. French. The derivation and results only apply to the case where the target screen is far away from the slits. This scenario is called Fraunhofer diffraction.

Case 1: Two Point Sources

Before we answer what happens with infinitely many wavefronts, it'd make sense to ask what if we have two point sources?

Two Point Sources

Given a point on the screen ahead, what is the straight path from one of the sources to that point? Since the screen is far away, we approximate these two paths to be parallel, as given by the two arrows in the image. The only difference between these two paths here is that one of them is longer by $d\sin\theta$.

We're dealing with 2D waves where they spread out in a circle from each source, but along each line we essentially have a 1D wave. In the picture above, if $x$ is the distance along one of the arrows, the wave is given by $A\sin(kx-\omega t)$. If the right arrow has total distance $x=D$, the left arrow has total distance $x=D+d\sin\theta$ (all that matters is that if you draw a really long triangle, the two longest sides would differ by about $d\sin\theta$).

At the screen, the wave due to the right source is $A\sin\big(kD-\omega t\big)$. The wave due to the left source is $A\sin\big( k(D+d\sin\theta)-\omega t \big)$. Therefore, our total wave is

$$ A\sin\big(kD-\omega t\big) + A\sin\big( k(D+d\sin\theta)-\omega t \big). $$

To clean this up a little, let $\phi = kD-\omega t$ and let $\Delta\phi = kd\sin\theta$. This gives the expression

$$ A\sin(\phi)+A\sin(\phi+\Delta\phi). $$

This sum of sines is actually just a single sine wave in disguise! We have to use some non-trivial trig identities. This one is given on Wikipedia in the section called "Linear combinations". Then

$$ A\sin(\phi)+A\sin(\phi+\Delta\phi) = \underbrace{\sqrt{A^{2}+A^{2}+2A^{2}\cos(\Delta\phi)}}_{\text{Amplitude}}\sin(\cdots). $$

The amplitude (using the half-angle formula) is given as

\begin{align*} \sqrt{2A^{2}+2A^{2}\cos(\Delta\phi)} &= A\sqrt{2}\sqrt{1+\cos(\Delta\phi)} \\ &= 2A\sqrt{ \frac{1+\cos(\Delta\phi)}{2} } \\ &= 2A\cos(\tfrac{\Delta\phi}{2}). \end{align*}

Ooookay. One more thing. If we back up, we need to clarify that $k$ is called the wave number. It is related to wavelength by $k = 2\pi/\lambda$. With this, we know $\Delta\phi = kd\sin\theta = 2\pi d\sin\theta/\lambda$. Therefore, the wave is $2A\cos\left(\frac{\pi d\sin\theta}{\lambda}\right)$. The intensity is given by the square of the amplitude, so $$ I(\theta) = 4A^{2}\cos^{2}\left(\frac{\pi d\sin\theta}{\lambda}\right). $$

Case 2: Single-Slit Pattern

The above was given just so that we have the general approach to looking at these problems, so you can attack the problem where you have three point sources or $N$ many, and this will be useful here.

We have a single slit of width $a$. The strategy is to split the wave into $N$ waves in the spirit of Huygens. The wave sources are equally spaced out by $\Delta s = a/N$. Each wave source contributes $1/N$ of the original wave.


This will mirror the reasoning employed for the case of two-point sources. Let $D$ be the distance between the exact center of our slit to the target. The distance from each point to the target is then $x = D + s_{n}\sin\theta$ where $s_{n}$ is the position of the source relative to the center (if the source is farther to the right, $s_{n}$ shall be greater).

The wave due to each source is now $(A/N)\cos\big(k(D + s_{n}\sin\theta) - \omega t \big)$. Since we already noted $\Delta s = a/N$, we have $\Delta s / a = 1/N$. Summing up all the waves, we have

$$ \sum_{n} A\cos\big(k(D+s_{n}\sin\theta) - \omega t \big)\frac{\Delta s}{a}. $$

We are now going to use a trick. The trick is to take advantage of Euler's identity $e^{iu} = \cos(u)+i\sin(u)$. This will seem cheap, but we will simply replace all of the $\cos u$ terms by $e^{iu}$, and we will understand that we're dealing only with the real part when necessary.

The key to this is that the "real part" is additive, so $\text{Re}\; (e^{iu}+e^{iu'}) = \text{Re}\; (e^{iu}) + \text{Re}\; (e^{iu'})$. This kind of property does not hold if we are multiplying complex numbers, and this only works because we are adding things. Also, what we're doing respects integration, so $\int_{a}^{b} \text{Re}\;e^{iu}\;du = \text{Re}\;\int_{a}^{b}e^{iu}\;du$. The magic of this trick is that it makes all of the trigonometry significantly easier to deal with.

We are going to go back to the sum above and write

$$ \sum_{n} Ae^{i\left[ k(D+s_{n}\sin\theta)-\omega t\right]} \frac{\Delta s}{a}. $$

Huygens's principle deals with infinitely many wavefronts. This means we are meant to send $N\rightarrow \infty$. The above sum then transforms into

\begin{align*} \int_{-a/2}^{a/2} \frac{A}{a}e^{iks \sin\theta} e^{ikD-i\omega t} \;ds &= \frac{Ae^{ikD-i\omega t}}{a}\left( \frac{e^{iks\sin\theta}}{ik\sin\theta} \right) \Big|_{s=-a/2}^{s=a/2} \\[1em] &= \frac{Ae^{ikD-i\omega t}}{aik\sin\theta}\left( e^{ika\sin\theta/2}-e^{-ika\sin\theta/2} \right) \\[1em] &= \frac{Ae^{ikD-i\omega t}}{aik\sin\theta}\left(2i\sin\left(\frac{ka\sin\theta}{2}\right)\right)\\[1em] &= A\frac{\sin\left( \tfrac{ka\sin\theta}{2} \right)}{\left(\tfrac{ka\sin\theta}{2}\right) }e^{ikD-i\omega t} \\[1em] &= \underbrace{A\frac{\sin\left( \tfrac{\pi a\sin\theta}{\lambda} \right)}{\left( \tfrac{\pi a\sin\theta}{\lambda} \right)}}_{\text{Amplitude}}e^{ikD-i\omega t}. \end{align*}

Finally, the intensity is the square of the amplitude so

$$ I(\theta) = A^{2}\frac{\sin^{2}\left( \tfrac{\pi a\sin\theta}{\lambda} \right)}{\left(\tfrac{\pi a\sin\theta}{\lambda} \right)^{2}}. $$

Case 3: Double-Slit Pattern

Each slit has width $a$. The distance between the center of each slit is $d$.


Just as before, we split the wave in each opening into $N$ waves, respectively. For each opening, $\Delta s = a / N$. As we send $N\rightarrow \infty$ for both slits, we obtain the integrals

$$ \int_{-\frac{d}{2}-\frac{a}{2}}^{-\frac{d}{2}+\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds + \int_{\frac{d}{2}-\frac{a}{2}}^{\frac{d}{2}+\frac{a}{2}}\frac{A}{a}e^{iks\sin\theta}e^{ikD-i\omega t}\;ds. $$

After evaluating the integrals, performing some serious trig acrobatics, and simplifying, we obtain

$$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\left[ \sin\left(k(\tfrac{d}{2}+\tfrac{a}{2})\sin\theta\right)-\sin\left(k(\tfrac{d}{2}-\tfrac{a}{2})\sin\theta\right) \right].$$

We can hatch open this expression by applying sine addition formulas and getting cancellation. This yields

$$ \frac{Ae^{ikD-i\omega t}}{ak\sin\theta/2}\cdot 2\cos\left(k\left(\tfrac{d}{2}\right)\sin\theta\right)\sin\left(k\left(\tfrac{a}{2}\right)\sin\theta\right). $$

As before, the wavenumber is $k = 2\pi/\lambda$, so now the expression equal to

$$ \frac{Ae^{ikD-i\omega t}}{(\tfrac{\pi a\sin\theta}{\lambda})} \cdot 2\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right) $$

and this is

$$ \underbrace{2A\frac{\sin\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}\cos\left(\tfrac{\pi d\sin\theta}{\lambda}\right)}_{\text{Amplitude}}e^{ikD-i\omega t}. $$

By squaring the amplitude, we obtain

$$ I(\theta) = 4A^{2}\frac{\sin^{2}\left(\tfrac{\pi a\sin\theta}{\lambda}\right)}{\left(\tfrac{\pi a\sin\theta}{\lambda}\right)^{2}}\cos^{2}\left(\tfrac{\pi d\sin\theta}{\lambda}\right). $$

  • $\begingroup$ If ever an answer deserved to be accepted by the OP, it would be this. $\endgroup$
    – garyp
    Commented Jun 17, 2023 at 17:51

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