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If light travels as a wave rather than a straight line there should be a width limit for it to travel and penetrate matter - like a hight limit for automobils in roads due to bridges - My question is1.How to calculate this specific width ! 2. Can you use this formula to calculate the filtration percent of a sheet (on a beam of light passing through the sheet)? Tnx!

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    $\begingroup$ You need to be aware that the "wavy nature" of light is because the electric and magnetic fields of light are oscillating wrt time and space. The light doesn't follow a wavy path - it is the $E$- and $B$ field strength that has a e.g. $sin$ dependence. The direction in which light travels is still a straight line (and always perpendicular to the direction of $E$ and $B$ for a plane wave). $\endgroup$ – ahemmetter Dec 19 '18 at 10:26
  • $\begingroup$ Light is an EM wave $\endgroup$ – QuIcKmAtHs Dec 19 '18 at 11:00
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It seems you misinterpreted the wave nature of light.

It doesn't say that light follows a wavy path. Instead, light travels in a straight line, and it is the magnetic and electric fields that oscillate.

If you'd focus on one specific point in space, it would seem as if the magnetic and electric fields are getting stronger and weaker, and repeat in the other direction. Plotting the field strength and direction over time would result in a $sin$-curve for each field component. The same result you'd get if you froze the light wave and plotted the field strength and direction with respect to the position.

enter image description here

So, each part of the EM wave does in fact wave, but what about the whole light wave? For this we can calculate the Poynting vector, which tells us the energy flow: $$\vec S = \vec E \times \vec H$$

For a plane wave, we have the following expressions for $E$ and $H$ with propagation in $z$-direction:

$$\vec E(z,t) = \vec E_0 \sin(kz - \omega t)$$ $$\vec H(z,t) = \frac{\vec E_0}{Z_0} \sin(kz - \omega t)$$

where $Z_0$ is the impedance of free space ($377~\Omega$), which tells us the ratio of magnetic and electric field.

From the above formula we can see that the direction of the energy flow will be completely perpendicular to $\vec E$ and $\vec H$, since the cross product ($\times$) of two vectors returns a vector perpendicular to both. The remaining term gives the strength of the energy flow, which now oscillates at twice the frequency and has an average value of $$\langle S \rangle = \frac{1}{2Z_0} |E_0|^2$$


But in fact a light wave doesn't "fit" through arbitrarily small holes (or "apertures"). Light falling on an edge is diffracted and doesn't throw a neat, clear shadow line on a screen behind it. Instead, we see a diffraction pattern, which varies in intensity.

The same happens when light travels though a hole. Light falling on the edge of the hole is diffracted, changing direction and interfering with waves from other parts of the aperture. The smaller the aperture is, the more significant this effect becomes.

There is a limit to how small an aperture can be, given by the Fraunhofer diffraction limit: $$\alpha = \frac{1.22 \lambda}{d}$$ where $\alpha$ is the angle of the first dark ring (constructive interference), $\lambda$ is the wavelength and $d$ the diameter of the aperture.

enter image description here

For a given wavelength, a decrease in size of the aperture will quickly increase the angle of the first dark fringe and therefore increase the size of the spot of transmitted light. The light is spread out over an enormous area, and therefore very dim.

What if we have "more light"? An increase in intensity shows up as an increased amplitude of $E$ and $H$ fields. This in turn means that each spot on the screen is proportionally brighter if the amplitude of the incoming light wave is larger. However, the fraction of intensity on any given point to the total incoming intensity is determined only by the wavelength of the light and the aperture size - not the intensity/amplitude.

As you can see, the critical parameter that determines if light "fits through a hole" is the wavelength $\lambda$ and not the amplitude $E, H$ of the wave.

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  • $\begingroup$ Would be nice to add, that the meaning of the amplitude of light is occupied by the probility to find the particle somewhere on the observation screen (in the dark and bright fringes). The extent of the photons electric and magnetic field components is per definition infinite. $\endgroup$ – HolgerFiedler Dec 19 '18 at 11:48
  • $\begingroup$ @HolgerFiedler Thanks for the input! I added a short paragraph that I think picks up your suggestions, but leaves out the photon part. Since we are dealing with classical wave optics, it might be a bit confusing and not so fruitful to introduce photons into this answer. $\endgroup$ – ahemmetter Dec 19 '18 at 16:19
  • $\begingroup$ @ahemmetter I sincrerely thank you for your thourogh answer, I feel like I found the holes in my own knowledge of EM waves, I would start by following your suggestions of formulas to get a better grip of my question. $\endgroup$ – Amir Hoss Dec 22 '18 at 5:25

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