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Nature and Propagation of Light (Index of refraction(n) (Light travels at…
Nature and Propagation of Light
Index of refraction(n)
Light travels at different speeds in a medium
\(n=\frac{c}{v}\)
c is the speed of light in a vacuum
When we enter a new medium frequency stays the same so if we're changing speeds it also means we're changing wavelengths
n=\(\frac{\lambda_0}{\lambda}\)
\(\lambda_0\) is the wavelength in a vacuum
This means the color red has a different wavelength in a medium that's not a vacuum, but it is still red
The smaller n is the faster v is
Polarization
Methods
By Polarizing Filters
Filters light coming through to only accept a certain orientation of electric fields
Malus' Law
Imagine an unpolarized light hits a polarizer, the only thing that passes through this polarizer is an Electric field parallel to the polarizing axis of the polarizer
When unpolarized light hits the first polarizer it was found that the Electric field becomes a quarter of what it was
This electric field comes continues moving forward to the second polarizer which we call the
analyzer
The only electric field allowed to go through must be parallel to the analyzer's axis
So the Electric field after the analyzer is \(Ecos\phi\)
\(I \alpha E^2\) so \(I=E^2=E_{max}^2cos^2(\phi)\)
\(I=I_{max}cos^2(\phi)\)
where \(\phi\) is the angle between the Polarizer's polarizing axis and the Analyzer's polarizing angle
Trend: What happens if you have 100 polarizers in a row?
1 more item...
The parallel component is doing the cos which we can imagine as mapping the E to its parallel component
This means I is halved
It doesn't matter which angle the field enters the polarizer
By Reflection
Imagine a side view of the place light hit reflection, this is the incident plance
It turns out that when polarized it is polarized perpendicular to this plane
Intensity
Going through the first polarizer halves the intensity
Going through the next polarizing our electric field only accepts what is parallel to the polarizing axis \(E=Ecos\phi\)
\(I\alpha E^2=E^2 cos^2\phi\)
\(I=I_{max}cos^2\phi\)
This is since the rays scattering in all directions have some component parallel to the polarizing axis
On average this means our total intensity from before is halved
Pattern
\(I_1=\frac{I_0 }{2}\)
\(I_2=I_1cos^2\phi_1=\frac{I_0}{2}cos^2\phi_1\)
\(I_3=I_2cos^2\phi_2=I_1cos^2\phi_1cos^2\phi_2=\frac{I_0}{2}cos^2\phi_1cos^2\phi_2\)
\(I_n=\frac{I_0}{2}cos^2\phi_1...cos^2\phi_{n-1}\)
n=number of polarizers-1
Remember that \(\phi\) is the angle between the current electric field orientation and the upcoming polarizers axis
Law of Reflection/Refraction
These come as a result of least time principle
When light hits a surface some of the light is reflected back while some light goes through the material(refraction)
Refraction
This is the idea that when light hits a surface some of it penetrates through the surface
Why doesn't light go straight when it enters at an angle?
Imagine light as a car,
\(n_{air}\)=smooth road
\(n_{b}\) =muddied road
When we're going at an angle one set of wheels would hit the muddied side first. The wheels that hit the mud start to slow down, but the wheels on the smooth road are still going fast
This causes our faster wheels to turn towards the slow wheels
Reflection
Drawing a normal line from the surface and the incident ray the
law of reflection
states that the angle from the incident ray to the normal line is the same as the angle from the normal line to the reflected ray
This is a
law
so it always applies even with surfaces that aren't plane
\(\theta_a=\theta_r\)
Fermat's Principle of Least Time
Nature is lazy and always wants to do things in the easiest ways possible
It turns out that the reason light reflects at the same angle it hits a surface is that this turns out to be the shortest time light needs to travel?
What's the fastest point from A to B? It's obviously just a straight line from one to the other.
This question becomes much harder if now we were asked what's the fastest route from A to B if A must hit the mirror first
We can solve this question by looking at an imaginary point \(B^\prime\) which is exactly across from B
Fastest way to get from A to \(B^\prime\) is just a straight line
WE can see that the distance from A to B is the same as A to \(B^\prime\) so this is the fastest time
How is light smart enough to know this?
Total Internal Reflection
When does this occur?
In most situations light will reflect some and refract some
But there are certain situations where everything is completely reflected. No refraction
This is when we're going from a slower medium to a faster one(light bends away from normal) at a certain angle so that the angle of reflection is 90 degrees
The critical angle means if any angle \(\ge \theta_{crit}\) then there is no refraction and all the rays are completely/totally reflected
We call the angle where the angle of refraction is 90 the
critical angle
\(n_asin\theta_{crit}=n_bsin\theta_b\)
\(n_asin\theta_{crit}=n_bsin90\)
\(sin\theta_{crit}=\frac{n_b}{n_a}\)
MUST HAVE \(n_b < n_a\)
Must be entering a medium where light travels faster since we HAVE to bend away from the normal
Applications
Changing Angles
If we enter at a 45 degree angle then if we're totally reflected all the rays are reflected at 45 degrees
Diamonds
Light enters the diamond and since diamond has a unusually high index of refraction the critical angle is small
This means light that enters the diamond is trapped for a long time, creating the shining effect we see
Fiber Optics
Dispersion
