In the notes I have, they have a diffraction grating with $2N + 1$ slits, a slit width of $2a$ and a slit spacing of $d$. They then say that the equation for the diffraction intensity pattern is given by:

$$I = I_0 \left( \frac{\sin((N+\frac{1}{2})kd\sin\theta)}{\sin(\frac{1}{2}kd\sin\theta)} \right)^2 \left( \frac{\sin(ka\sin\theta)}{ka\sin\theta} \right)^2 $$

They don't, however, give any proof or reason why this is the formula. I have been looking online for a way to justify this formula but I cant find anything. Anybody have a nice proof for this?


1 Answer 1


diffraction grating
(image from Antonine education)

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

  • of all the slits ($n$ from $-N$ to $+N$)
  • and of the parts of each individual slit ($x$ from $-a$ to $+a$)

The path difference of each contributing ray (compared to the path length of the ray originating from the center of the grating) is $(nd+x)\sin\theta$. And hence its phase is $k(nd+x)\sin\theta$.

Summing these contributions you get $$ \begin{align} E(\theta) &= E_0 \sum_{n=-N}^{+N} \int_{-a}^{+a} e^{ik(nd+x)\sin\theta} \text{d}x \\ &= E_0 \left( \sum_{n=-N}^{+N} e^{iknd\sin\theta}\right) \left( \int_{-a}^{+a} e^{ikx\sin\theta} \text{d}x \right) \\ &= E_0 \left( \frac{\sin((N+\frac{1}{2})kd\sin\theta)}{\sin(\frac{1}{2}kd\sin\theta)} \right) \left( 2a\frac{\sin(ka\sin\theta)}{ka\sin\theta} \right) \end{align} $$

And finally you get the intensity by taking the absolute square of the amplitude $$I(\theta) = |E(\theta)|^2$$


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