Hi everyone. I am using a coherent source to illuminate the sample and collect the intensity of Fourier plane via CCD. The formula is |F{x}|.^2, where F denotes the Fourier transform.

And I hope to collect the blurred (defocus) Fourier plane in order to reduce the light intensity. So I hope to put the camera out of the focal plane (Fourier plane). After reading the references [1-2], I am confused since I have two different ideas of the formulation.

1. |F{x*h}|.^2, where * denotes the convolution kernel and h is brought by the lens before the camera. According to the references [1], the defocus in the Fourier plane is equivalent to multiplication of the Fourier transform of the kernel.

2. |F{x}*h'|.^2. If I treat the F{x} as the object plane that I want, so putting out the focal plane is equivalent to convoluting with the kernel h'.

I am stuck in these two ideas. So one of them should be wrong. Any suggestions? Thanks in advance!

Reference:

[1] https://ocw.mit.edu/courses/mechanical-engineering/2-71-optics-spring-2009/video-lectures/lecture-25-resolution-defocused-optical-systems/MIT2_71S09_lec25.pdf

[2] https://ocw.mit.edu/courses/mechanical-engineering/2-71-optics-spring-2009/video-lectures/lecture-26-depth-of-focus-and-field-polarization-wave-plates/MIT2_71S09_lec26.pdf