10.4.1 N-slit Interference Pattern (Beyond the Syllabus)

We have had so much fun with two slits. Naturally, we want to add another slit to make it 3-slit interference, right?

Firstly, let me emphasize that we are talking about 3 slits which are equally spaced out. In other words, consecutive slits have the same slit separation d. Hence waves from consecutive slits arrive at (each point on) the screen with the same path difference d\sin \theta .

This has the important implication that when the 3 light rays arrive at a point on the screen, ray B will lag ray A by the same phase difference as ray C lags ray B.

Let’s consider what happens when \displaystyle d\sin \theta =\lambda .

Ray C will arrive exactly one cycle after ray B, which arrives exactly one cycle after ray A. So all 3 sinusoidal waves arrive in phase at P, superposing constructively to form a resultant wave of amplitude 3A (assuming the amplitude of each wave is A).

Let’s now consider what happens when \displaystyle d\sin \theta =\frac{\lambda }{3}.

Ray C will arrive one-third of a cycle after ray B, which arrives one-third of a cycle after ray A. It is like we have 3 sinusoidal waves spaced apart with 120° between (A and B), (B and C) and (C and A). Do you realize that these 3 sinusoids are going to sum up to zero? The 3 rays interfere destructively to cancel one another out!

Amazingly, D.I. also occurs among these 3 waves when \displaystyle d\sin \theta =\frac{{2\lambda }}{3}.

With some thought, we can deduce other values of d\sin \theta at which C.I. and D.I. will occur.

C.I. when d\sin \theta =D.I. when d\sin \theta =
0\displaystyle \frac{1}{3}\lambda , \displaystyle \frac{2}{3}\lambda
λ\displaystyle (1+\frac{1}{3})\lambda , \displaystyle (1+\frac{2}{3})\lambda
2λ\displaystyle (2+\frac{1}{3})\lambda , \displaystyle (2+\frac{2}{3})\lambda

Alright. I guess we are now confident enough to add even more slits!

If we have 4 slits (and 4 rays), the first D.I. occurs when the 4 waves are spaced apart by 360{}^\circ \div 4=90{}^\circ . This happens when the path difference is  \displaystyle d\sin \theta =\frac{\lambda }{4}.

If we have 10 slits (and 10 rays), the first D.I. occurs when the 10 waves are spaced apart by 360{}^\circ \div 10=36{}^\circ . This happens when the path difference is  \displaystyle d\sin \theta =\frac{\lambda }{{10}}.

By extension, we can deduce that for N slits:

Bright fringes are formed at angles θ where \displaystyle d\sin \theta =n\lambda , n=0,1,2...

Dark fringes are formed at angles θ where \displaystyle d\sin \theta =(n+\frac{m}{N})\lambda , n=0,1,2... and m=1\text{ }..\text{ }N-1

The graphs below show how the interference pattern’s intensity profile changes as we add more and more slits (while maintaining the same slit separation d).

Note that as N increases the number of bright fringes do not change. Neither do their positions. However, they do become brighter, since they have amplitude NA and intensity {{N}^{2}}{{\mathrm{I}}_{0}} (assuming each ray arrives on the screen with amplitude A and intensity I0). In the graphs above, the intensity has been “equalised” across the different N values because we have chosen to plot the normalized intensity \displaystyle \frac{\mathrm{I}}{{{{\mathrm{I}}_{{\max }}}}}, where {{\mathrm{I}}_{{\max }}} is {{N}^{2}}{{\mathrm{I}}_{0}}.

Note also that as N increases, the number of dark fringes increases. In fact, there are N-1 dark fringes between two consecutive bright fringes. When N is a large number, we see dark fringes everywhere, except at those locations where there are very narrow and bright fringes.

Video Explanation   

Making Sense of the Diffraction Grating Pattern

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