# Optics

In electromagnetic theory, the intensity of light is proportional to the square of the oscillating electric field which exists at any point in space. The Fourier transform of this signal is the equivalent of breaking the light into it's component parts of the spectrum, a mathematical spectrometer.

One simple example application of the fourier transform in optics is the diffraction of light when it passes through narrow slits. The ideas represented here can be equally applied to acoustic, x-ray, and microwave diffraction, or any other form of wave diffraction.

Suppose we have the experiment shown in the picture below where we have a slit in an opaque screen. If the gap is of width W then the transmission function will be This square transmission function is an equivalent to the to a filter response such as an ideal band-pass filter. Light is only transmitted between the edges of the slit.

The Fourier transform of this gives us it's frequency content, which in turn is used to calculate the diffraction intensity at different angles from the normal. The resultant intensity is given by the square of the fourier transform: In other words, on the screen there is a bright region based around the slit which rapidly falls off to either side of the slit. Roughly what you would expect if you shone a light through a hole in a wall.

If the screen is altered to give us two parallel slits, with width W and distance D apart, then the transmission function is: If we the use this, and the Fourier transform to calculate the diffracted intensity we get This represents closely spaced maxima and minima (light and dark bands) that gradually decrease the further from the center of the two slits you go.

This is the Fourier transform used to calculate the result of a Young's double slit experiment.

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