http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/gratint.html
Could anyone in simple terms explain to me how the peaks are narrower as the number of slits increases in a diffraction grating? Note i've only come across double slit diffraction and was just introduced to this recently, and the explanation by hyperphysics is a bit confusing with formula.
If we start with the double slit with all the usual assumptions.
As a result of the incident plane wave two coherent sources ($A$ and $B$) which produce waves with the same wavelength $\lambda$, are in phase with one another and of the same intensity.
These waves overlap to produce an interference pattern.
Consider two parallel rays coming from the two sources at an angle $\theta$ to the normal of the plane containing the two sources.
Somewhere a long way away or by the use of a lens these rays will superimpose and because they have travelled different distance ($BC)$ there phase relative to one another would have changed by an amount $\delta = \dfrac {d \sin \theta}{\lambda} \, 2 \pi$ where $d$ is the separation of the slits.
If $\theta =0$ then $\delta =0$ and the waves from the two sources arrive in phase.
Assuming that the individual "displacements" $y$ of the waves at the same position are given by $y_1 = A \cos (2\pi f t)$ and $y_2 = A \cos (2\pi f t+\delta)$, where $A$ is the amplitude and $f$ is the frequency, then the resultant displacement is the sum of the individual displacements $y = A \cos (2\pi f t) + A \cos (2\pi f t+\delta)$
Rather than use trig identities lets have a look at particular values of the phase difference $\delta$.
If $\theta =0$ then $\delta =0$ and the waves from the two sources arrive in phase.
The resultant displacement is $2A \cos (2\pi f t)$.
This means that the amplitude of the combined waves is $2A$ and the intensity which is proportional to the amplitude squared is proportional to $(2A)^2 = 4A^2$.
If the path difference $AB = \frac \lambda 4$ which corresponds to a phase difference $\delta = \frac \pi 2$ you can show that the resultant amplitude is $\sqrt 2 A$ corresponding to an intensity (proportional to) $2A^2$.
If we assume that the slit separation $d = 2 \lambda$ then $\delta = 4 \pi \sin \theta$ and the angle $\theta = \sin^{-1} \left ( \frac 1 4 \right ) \approx 14.5^\circ $
So your zero order two slit fringe has a variable intensity dropping from the centre value of $4A^2$ when $\theta = 0^\circ$ to $2A^2$ when $\theta = \pm 14.5^\circ$.
If the path difference is $\frac \lambda 2$, $\delta = \pi$, with $\theta \approx 30^\circ$, the resultant displacement is zero.
So you could say that the width of that central fringe is $\pm 30^\circ$.
The addition of a third slit means that the resultant disparagement is now the sum of three waves $y_1 = A \cos (2\pi f t), \, y_2 = A \cos (2\pi f t+\delta)$ and $ y_3 = A \cos (2\pi f t+2\delta)$ which gives
$y = A \cos (2\pi f t) + y_2 = A \cos (2\pi f t+\delta)+ y_3 = A \cos (2\pi f t+2\delta)$
When $\delta = 0$ and $\theta = 0^\circ$ the resultant amplitude is $3A$ and so the intensity is $9A^2$.
The first minimum (zero) occurs when $\delta = \frac {\pi}{3}$ ie a path difference of $\frac \lambda 3$ with $\theta = \sin^{-1} \left ( \frac 1 6 \right ) \approx 9.6^\circ $.
So as a result of adding a third slit the intensity increases from $4A^2$ to $9A^2$ but the width of the central fringe decreases from $\pm 14.5^\circ$ to $\pm 9.6^\circ$.
The fact that the intensity of the central fringe for three slits is to be expected as more light is emitted by three slits than two slits and the fringe is "compressed" into a smaller angle.
Repeating for four slits one gets, $\delta = \frac \pi 2$, a central intensity of $16A^2$ and angular width of the central fringe of $\pm \sin^{-1} \left ( \frac 1 8 \right ) \approx \pm 7.2 ^\circ$
It might not escaped your notice that if $N$ is the number of slits the angular width of the central fringe is $\pm \sin^{-1} \left ( \frac {1}{2 N} \right )$
So a diffraction grating with a $5000$ slits in total with produce a central fringe of width $\pm \sin^{-1} \left ( \frac {1}{10 000} \right ) = \pm 0.0057^\circ$ and a central intensity of $25\,000\,000 A^2$.
The analysis which I have done above can be greatly simplified with the use of phasors in that the addition of sinusoidal functions of the same frequency can be likened to the addition of vectors.
So if the addition $A \cos (2\pi f t) + A \cos (2\pi f t+\delta)$ needs to be done the following digram can be drawn with the sum shown as the result of a "vector" addition.
Here are some of these phasor diagrams relating to two slit interference.
And for more slits.
I think that the 10 slit examples shown quite nicely that you do not need to great a phase difference between adjacent slits to get a resultant of zero.
Consider there is a huge amount of slits. Then, the waves travelling through each slit will all have different phases, and these phases will be equally distributed (if the slits are equally spaced). However, when you sum a huge number of sinus whose phases are equally distributed, you get something almost null: thus, the more slits there are, the nuller the diffraction figure will be. Nevertheless, at null incidence, there is no phase difference between each slit, so the are constructive interferences, and you see an intensity peak. Why is this maximum intenser as the number of slits increase ? You can understand this using energy conservation. There is always the same amount of energy that comes through the slits, but as the number of slits increases, it is more and more focused, so the maximum intensity increases as well.
This is no substitute for Spirine's answer but, a very short summary is this…
Increasing the number of slits increases the number of angles at which destructive (or almost destructive) interference can occur, while the condition for constructive interference remains just the same ($n \lambda=d\ \text{sin} \theta$ for a transmission grating at normal incidence).
