$physics light diffraction$ $physics light diffraction$ $physics light diffraction$

These two minima limit the broad central maximum.Īn argument like the one applies if, in our imagination, we divide the slit into any even number of equal slices. On the other side of the axis of symmetry, sin θ = –λ/ a is also a minimum. Now, for every point in the top half of the slit, there is one in the bottom half a distance a/2 below and, at the angle that satisfies a sin θ = λ, If this distance is half a wavelength, i.e. The ray from the distance a/2 below has to travel an extra distance ( a sin θ/2). Typically in diffraction experiments, the slit is ~ 10 µm wide, while the distance to the screen might be ~ 1 m.) (Why parallel? Because the screen is distant. Consider parallel rays from both points, at angle θ to the axis of symmetry. a point at the very top of the lower half of the slit. Using the Huygens' construction, we consider a point at the very top of the slit, and another point a distance a/2 below it, i.e. We call the slit width a, and we imagine it divided into two equal halves. In effect, a pinhole in a wire grid polarizer will act like an ordinary pinhole for light whose E vector is parallel to the wires, and will act like a sheet of glass for light whose E vector is perpendicular to the wires.This animated sketch shows the angle of the first order minima: the first minimum on either side of the central maximum. The portion that has its E vector perpendicular to the wires will pass right through the wire grid polarizer. The portion of the scattered light that has its E vector parallel to the wires in the grid will be absorbed. Scattered light, though, will hit the material of the wire grid polarizer.

$physics light diffraction$

So, light that gets through the pinhole basically never touches the material that the pinhole has been punched into.Īs a result, if you make a pinhole in a wire grid polarizer and use the pinhole the way a pinhole is usually used, then most of the light, regardless of polarization, will go through the pinhole. The pinhole removes the scattered light, leaving a 'clean' beam. The only portions of the beam that hit the material outside the pinhole are portions due to light that has been scattered upstream by, e.g., dirt on the optics. Due to the geometry of the light beam and the lens, the beam waist is usually in the range from 5 to 40 microns in diameter, and the size of the pinhole is chosen to be slightly bigger than the beam waist. When a pinhole filter is used, it is typically at the focus of a lens, so a light beam is as small as it can be when it encounters the pinhole. Essentially none of the photons whose E vector is parallel to the wires can get through, even though there seems to be space between the wires. For all practical purposes, photons whose E vector is perpendicular to the wires just go through as if the wires were not there. However, only the photons whose E vector is aligned with the wires can induce current in the wires and be absorbed. That's not the way it works.Įach photon is effectively much wider than the distance between the wires, so every photon that hits the wire grid polarizer 'sees' the wires. You are imagining that some of the light hits the wires and some goes between the wires, and that the polarization filtering occurs only between the wires.

$physics light diffraction$

Physics light diffraction