# Triple slit interference

1. Nov 8, 2008

### kasse

1. The problem statement, all variables and given/known data

Three slits of width a and d (separation of adjacent slits). d << the distance to the screen. Show that the intensity we observe is

$$I (\vartheta) = I_{1} \left[1 + 2 cos \left(\frac{2 \pi d sin\vartheta}{\lambda} ) \right] ^{2}$$

2. The attempt at a solution

For an angle $$\vartheta$$ we see that the waves from slit 1 and 3 must travel a distance dsin$$\vartheta$$ shorter and longer than the waves from slit 2 to get to a point on the screen.

Total wave amplitude far from the slits:

A = A1 + A2 + A3 = $$A _{0}sin(kx - \omega t - \varphi )+ A _{0}sin(kx - \omega t) + A _{0}sin(kx - \omega t + \varphi )$$

Am I on the right track? How do I find an easier expression for A?

Last edited: Nov 8, 2008
2. Nov 9, 2008

### kasse

I think I should use complex notation here. How do I represent A in complex form? My book (and other sources on the internet that I've found) is not very clear here. My best guess is

$$A = A_{2}[exp(ikd sin\vartheta ) + 1 + exp(-ikd sin\vartheta )]$$

Can I then write $$exp(ikd sin\vartheta ) + exp(-ikd sin\vartheta ) = 2cos(kd sin \vartheta)$$? Why?

Last edited: Nov 9, 2008
3. Nov 12, 2008

### weejee

1. You were actually right in your first attempt, except that you should square the amplitude and obtain the time average.

2. Yet, as in your second attempt, using complex notation is far easier.

3. exp(ix) = cos(x) + isin(x) can be used to prove the identity you want to use.
Above identity can be proved using taylor's expansion.