Often in my book they have referenced the fact that as $N$ (the number of slits) becomes very large, the angular width of the first maxima in a diffraction grating can be approximated as follows: $$\Delta \sin\theta \sim \frac{\lambda}{Nd} $$

I am wondering how to proof this statement, but cant make much progress.


2 Answers 2


This question is very much related to Help understanding resolution between two light beams which perhaps should be read first?

The first thing to remember is that in general the number of slits in a diffraction grating $N$ is very large so $N\gg 1$.

In the diagram below there are $N+1 \approx N$ slits of which only a few are shown.

enter image description here

For the $m^{\rm th}$ order the basic diffraction grating equation is $m\lambda = d \sin \theta_{\rm m}$ where $d$ is the distance between adjacent slits and the path difference between adjacent slits is $m\lambda$.

In my diagram the path difference between the slit labelled $A$ and that labelled $I$ is $Nm\lambda = IJ$ where $N$ is the number of slits in the grating.
The light from all slits in the grating constructively interfere and this results in a principal maximum in direction $\theta_{\rm m}$.

Now consider a change of direction by $\Delta \theta_{\rm m}$ such that the path difference between the light from slit $A$ and slit $I$ is $Nm\lambda + \lambda$. This would result in the slits in the bottom half of the grating and the corresponding slits in the top half of the grating, eg slit $E$ and slit $I$, producing waves which are half a wavelength out of phase and so interfere destructively.
This results in the first subsidiary minimum. The path difference between slit $A$ and slit $I$ could also be could also have been $Nm\lambda - \lambda$ and that would produce the first subsidiary minimum on the other side of the principal maximum.

From the diagram $AJ = D \cos \theta_{\rm m} = Nd \cos \theta_{\rm m}$.

For the triangle with angle $\Delta \theta_{\rm m}$ you have $\lambda = Nd \cos \theta_{\rm m} \:\Delta \theta_{\rm m}$ which then gives the result

$$\cos \theta_{\rm m} \:\Delta \theta_{\rm m} = \Delta (\sin \theta_{\rm m}) = \frac{\lambda}{Nd}$$

Note that this relationship gives the angular half width of a principal maximum.


Using the grating equation (I'm replacing the $d$ variable with the $a$ variable for clarity)

$$a(sin(\theta^{'})-sin(\theta))=m \lambda$$

setting $m=1$,differentiating, and noting that the $\theta$ - the incident angle - is constant

$$ cos(\theta{'})d\theta{'}=d\lambda/a.$$

Dropping the prime on $\theta$ since there is only 1 angle of interest, and approximating the derivative with $\Delta$ $$\Delta sin(\theta) \approx \Delta\lambda/a.$$

Using the definitions of resolving power to find $\Delta\lambda$

$$R=\lambda/\Delta\lambda$$ $$\Delta\lambda=\lambda/R$$

and $$ R=Nm$$

$$\Delta\lambda=\lambda/Nm$$ (noting that $m=1$) yields $$\Delta sin(\theta) \approx\lambda/Na.$$ Finally, replacing $a$ with $d$ $$\Delta sin(\theta) \approx\lambda/Nd$$


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.