# How to derive the intensity formula of a diffraction grating?

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?

(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 $$I(\theta) = |E(\theta)|^2$$