# Interference pattern of a fan of plane waves

• baseballfan_ny
In summary, the angle the nth plane wave makes with respect to the z-axis is the angle it makes with respect to the n =1 plane wave, ##n \Delta \theta##, minus the 1/2 the total angle between the n = N and n = 1 plane waves, ##\Delta \theta \frac {(N - 1)} {2} ##, giving me ##\theta_n = \Delta \theta (n - \frac {N - 1} {2} ) ##. Angles below the z-axis are negative and above are positive. Therefore, ## \theta_n = -\theta_{n - \frac {N - 1} {2

#### baseballfan_ny

Homework Statement
A fan of ##N## plane waves are propagating symmetrically with respect to the z axis, as shown in figure below. The angular spacing between successive members of the fan is fixed and equal to ##\Delta \theta##. Describe the interference pattern observed on a plane perpendicular to the z axis.
Relevant Equations
Intensity of interference of waves ## \alpha | A_1e^{i\phi_1} + A_2e^{i\phi_2}|^2##

So I've kind of made the assumption that there will be an odd number of plane waves and the same amount above and below the z-axis. Then, using the diagram below, I determined the angle the nth plane wave makes with respect to the z-axis to be the angle it makes with respect to the n =1 plane wave, ##n \Delta \theta##, minus the 1/2 the total angle between the n = N and n = 1 plane waves, ##\Delta \theta \frac {(N - 1)} {2} ##, giving me ##\theta_n = \Delta \theta (n - \frac {N - 1} {2} ) ##. Angles below the z-axis are negative and above are positive. Therefore, ## \theta_n = -\theta_{n - \frac {N - 1} {2}}## for ## n > \frac {N - 1} {2} ##.

Making the assumption that they all have the same wavelength (and hence wavenumber ##k## and amplitude, the nth plane wave would be of the form (also ##\Delta z## is distance along z-axis to perpendicular plane we are interested in) ...$$E_n = E_0 e^{ik(x\sin(\theta_n) + \Delta z \cos(\theta_n))}$$

Then to get the intensity I would need the magnitude squared of the sum of this term over all n...

$$I \propto | \sum_{n = 1}^N E_0 e^{ik(x\sin(\theta_n) + \Delta z \cos(\theta_n))} | | \sum_{n = 1}^N E_0 e^{-ik(x\sin(\theta_n) + \Delta z \cos(\theta_n))} |$$

From here, I might have messed up. I used the idea that ##\theta_n = -\theta_{n - \frac {N - 1} {2}}##. I then used Euler's formula and this condition to rewrite the sum as a sum of cosines and one complex exponential corresponding to the plane wave centered on the z-axis (the x = 0 one), which gave me an algebraic mess. I'm going to skip showing that because as I was writing this post I thought of another way that maybe gave me a reasonable answer?

That other way:
Maybe these sums should be added using the geometric sequence formula: ##S_n = \frac {a(1 - r^{n+1})} {1-r} ##. Calculating the r factor...
$$r = \frac { E_0 e^{ ik(x\sin( \theta_{n+1} ) + \Delta z \cos( \theta_{n+1} ) ) } } { E_0 e^{ ik(x\sin(\theta_n) + \Delta z \cos(\theta_n) ) } } = e^{ ik( x\sin(\theta_{n+1}) + \Delta z \cos(\theta_{n+1} ) ) - ik( x\sin(\theta_n) + \Delta z \cos(\theta_n) ) } = e^{ik \gamma}$$

So then I make my sum from 0 to N-1 instead of 1 to N to apply the summation formula...

$$I \propto \frac {E_0 [1 - e^{ik\gamma N}]} {1- e^{ik\gamma} } * \frac {E_0 [1 - e^{-ik\gamma N}]} {1- e^{-ik\gamma} } = \frac {E_0^2 [1 - e^{ik\gamma N}] - e^{-ik\gamma N}] + 1} {1 - e^{ik\gamma} - e^{-ik\gamma} + 1} = \frac {E_0^2 [2 - 2\cos{k\gamma N} ]} {2 - 2\cos{k\gamma} } = \frac {E_0^2 [1 - \cos{k\gamma N} ]} {1 - \cos{k\gamma} }$$

So there would be maxima when ## k\gamma N = m\pi##, where m is odd? I'm still not sure this is right because I'm not sure how to explain what happens when ##cos{k\gamma}## in the denominator equals 1 and the whole thing blows up...

Last edited:
Delta2
baseballfan_ny said:
That other way:
Maybe these sums should be added using the geometric sequence formula: ##S_n = \frac {a(1 - r^{n+1})} {1-r} ##. Calculating the r factor...
$$r = \frac { E_0 e^{ ik(x\sin( \theta_{n+1} ) + \Delta z \cos( \theta_{n+1} ) ) } } { E_0 e^{ ik(x\sin(\theta_n) + \Delta z \cos(\theta_n) ) } } = e^{ ik( x\sin(\theta_{n+1}) + \Delta z \cos(\theta_{n+1} ) ) - ik( x\sin(\theta_n) + \Delta z \cos(\theta_n) ) } = e^{ik \gamma}$$
In order to have a geometric series, the ##r## factor needs to be a constant (i.e., independent of the term index ##n##). This is not the case here. It does look like a mess.

But if you can assume that all of the ##\theta_n## are small so that ##\sin(\theta_n) \approx \theta_n## and ##\cos(\theta_n) \approx 1##, you will have a geometric series.

baseballfan_ny
TSny said:
In order to have a geometric series, the ##r## factor needs to be a constant (i.e., independent of the term index ##n##). This is not the case here. It does look like a mess.

But if you can assume that all of the ##\theta_n## are small so that ##\sin(\theta_n) \approx \theta_n## and ##\cos(\theta_n) \approx 1##, you will have a geometric series.
This is exactly the hint I needed and it worked perfectly. I apologize for the late acknowledgment. Thanks for the help!

TSny