Part 5 - Light Waves | rezarezvan.com

In this part we’ll cover light waves and their properties.

Light is actually electromagnetic radiation. The visible light we can see is usually between 380 - 740 nm.

General Conventions

As we remember from mechanical waves $v = \lambda f$ , this also applies for light waves, but our $v$ is instead $c$ .

c = \lambda f = 3 \cdot 10^8 \ \mathrm{m/s}.

If we look at light like usual mechanical waves, it quickly becomes very cluttered in our sketch. Therefore we usually draw light as a plane wave. Which usually just means a singular line, instead of many.

Refractive Index

We’ve seen that if a wave enters a new medium it gains a new velocity, transmission and reflection. The same goes for light.

The refractive index tells us how much slower the light becomes in the new medium, compared to the speed in vacuum (or air, as we usually just set it as the same).

n = \frac{c}{v} \Leftrightarrow v = \frac{c}{n},

where $v$ is the new speed in the new medium.

Snell’s Law

Snell’s law is a formula which describes the relationship between the input angle which a light wave has and after, along with the different refractive indexes. I’m not going to prove it, it’s quite simple trigonometry but,

Theorem 1 (Snell’s law)

n_1 \sin(\theta_i) = n_2 \sin(\theta_r)

Reflection

When a wave reflects against a more dense medium (which means a lower propagation speed) - the reflected wave is ‘shifted’ by $\pi$ rad.

Meaning the reflected wave/pulse/light is the ‘inverse’ of the input wave, the inputs high becomes the reflections low etc.

When a wave reflects against a less dense medium (which means a higher propagation speed) - the reflected wave is not shifted. Thus the input wave and the reflection wave is the ‘same’ wave.

Interference with light

Imagine we have two glass panes - we put a small object in between, which is almost negligible. We can imagine that the light inside these two glass panes travels a distance of $d$ . Therefore the total geometric travel is then $2d$ .

The optical way is then $2d + \frac{\lambda}{2}$ .

And the phase difference is $\frac{2d}{\lambda} \cdot 2\pi + \pi$ .

We achieve constructive interference if,

2d + \frac{\lambda}{2} = m\lambda \Leftrightarrow 2d = m\lambda - \frac{\lambda}{2},

where $m = 1, 2, 3, \dots$

Young’s interference experiment

This is quite a visual proof/experiment and I’m too lazy currently to fix pictures for this sooo,

d \sin(\theta) = m\lambda.

This results in a ‘maxima’ or when it shines light - then we have 0th maxima, 1st maxima and so on.

And for minima,

d \sin(\theta) = (2m + 1) \frac{\lambda}{2}.

And just a general formula for the wave,

y = 2A \cos\left(\frac{\phi}{2}\right) \sin\left(kx - \omega t + \frac{\phi}{2}\right).

Also just to dump some formulas here,

\tan(\theta) = \frac{y}{L}.

And for ‘small’ angles (typically $\theta \lesssim 10^\circ$ ),

\tan(\theta) \approx \sin(\theta) = \frac{y}{L}.

Reflection

Another way to measure reflection is by using a special kind of mirror and doing some stuff :)

We can calculate the reflection like,

R = \left(\frac{n_1 - n_2}{n_1 + n_2}\right)^2.

Total Reflection

Total reflection is the phenomenon that occurs when the transmission wave becomes parallel with the ‘medium switch’. This is quite an easy equation, because in practice this just means that the transmission wave has an angle of $90^\circ$ ,

\begin{align*} n_1 \sin(\theta_1) & = n_2 \sin(\theta_2), \newline n_1 \sin(\theta_1) & = n_2. \end{align*}

Interference in thin layers

For interference of light in thin layers - we have to consider what medium we have before, inside and after.

If we have for example, air on both sides (which results in 1 phase shift), this will result in interference for maximum,

\begin{align*} 2dn_2 + \frac{\lambda}{2} & = m\lambda, & m & = 1, 2, 3, \dots \newline 2dn_2 - \frac{\lambda}{2} & = m\lambda, & m & = 0, 1, 2, 3, \dots \end{align*}

So generally,

2dn_2 + \text{possible phase shift} = m\lambda.

For the minima,

2dn_2 = m\lambda, \quad m = 0, 1, 2, 3, \dots

So generally,

2dn_2 + \text{possible phase shift} = m\lambda + \frac{\lambda}{2}.

Diffraction

Diffraction is the phenomenon that occurs when we have a width of our openings. What happens is that, since we now have an opening with a width, there will be many light ‘points’ that will radiate. This means that these points, most likely, will have destructive interference with each other more often, than if we have just 2 light ‘points’.

This leads to,

b \sin(\theta) = m\lambda.

This is also why the intensity will wear off with distance.

Optical resolution

So with this in mind, we can now understand why, if we have a big opening, for example a big lens, why the pictures come out better.

Because if we increase $d$ , then $\sin(\theta)$ must become larger.

We can also see this from,

\frac{\lambda}{\Delta\lambda} = m N,

where $N$ is the amount of openings.

Conclusion

This was it for light waves - yes a very short part since a lot of the intuition and proofs are visual and I can’t be bothered to put pictures of these at the moment :)

So take the formulas and go! :)