IFFT of Microphone Array Directivity Response
Your plot shows the magnitude of the array gain vs angle of arrival for the broadside beampattern. Before you can take meaning IFFT of this sequence, you need to follow the following steps.


The data in the plot is the magnitude of the array gain, which is stripped of all phase information. For IFFT to give desired result, we need to restore the phase information too. Fortunately, this can be easily accomplished by just changing the sign of the alternate lobes, assuming the underlying curve is of $\sin(Mx)/\sin(x)$ form.
The gain of a linear array looking at angle $\theta_0$ is proportional to $\dfrac{\sin\left( Ma(\cos(\theta)-\cos(\theta_0))\right) }{\sin\left( a(\cos(\theta)-\cos(\theta_0))\right) }$. So it makes more sense to take the IFFT of the array gain sequence when it is plotted against $\cos(\theta)$ as opposed to just $\theta$, or angle (as it is shown in your plot).


Also, assuming that the measurement has been made at the design frequency (i.e. sensor spacing $d = \lambda/2$), the result of the IFFT would provide the sequence of sensor weights. The number of FFT bins would only affect the zero-padding around the actual sensors.

I took the liberty of trying the above steps out on a set of data shown below. Note that it is very similar to your top plot.



The second subplot shown the result as we restore the sign by inverting the sign of the alternating lobes.

At this point, let theta and gain represent the x and y data in this plot (after sign correction).

Next, the data is resampled such that the x-axis is linearly spaced in $\cos(\theta)$ domain. Note that since $\theta$ ranges from 0 to 180, $\cos\theta$ would range from -1 to 1.

NFFT = 2^8;
xr = linspace(-1,1,NFFT);
yr = interp1(cosd(theta), gain, xr, 'linear', 'extrap');


The choice of NFFT is arbitrary, but it must be greater than the number of sensors.

The result of resampling is shown in the third subplot above. Notice how it looks more like a sinc function now.

Finally, we are ready for the IFFT.

wgt = fftshift(ifft(yr, NFFT)); % weight of sensors
% plot
figure;
plot((-NFFT/2:NFFT/2-1), abs(wgt), 'o-', 'linewidth', 2);
grid on;
xlim(20*[-1, 1]);
xlabel('Sensor location (\lambda/2)');
ylabel('Sensor weight');
title('IFFT result (after FFTSHIFT)');




Based on the above plot, it appears that there were 13 sensors in the experiment.