Regimes of optics:
📐 Geometrical optic:
- All components considered are very large compared to the wavelength
- Interference and diffraction effects are ignored.
🍎 Physical optics:
- Apparatus used is comparable to the wavelength of light
- Interference and diffraction are noticeable
⚡ Electromagnetism:
- Full solution of Maxwell’s equations
- No approximations
🍇 Quantum optics:
- Discrete nature of photons is important
- Ex: light interacting with matter and laser emissions
path length
$$ \begin{aligned} l=\sqrt{x^2+y^2}+\sqrt{(h-y)^2+x^2} \end{aligned} $$
Applying Fermat’s principle
$$ \begin{aligned} \frac{\text d l}{\text dy}&=\frac y{ \sqrt{x^2+y^2}}-\frac{h-y}{\sqrt{(h-y)^2+x^2}} \\ &=0 \end{aligned} $$
Thus we get
$$ \frac{y}{\sqrt{x^2+y^2}}=\frac{h-y}{\sqrt{x^2+(h-y)^2}} $$
Which is equivalent to $\sin(\theta_i)=\sin(\theta_r)$
$$ \boxed{\theta_i=\theta_r} $$
We draw the wavefronts perpendicular to the beam
We see that
$$ \begin{aligned} p n_1&=q n_2 \\ p=d\sin(\theta_i) \; &\quad\; \; q=d\sin(\theta_r) \end{aligned} $$
Putting it together we get
$$ n_1d\sin(\theta_1)=n_2d\sin(\theta_r) $$
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/42e0ea0e-95d5-4c19-95fc-d81186497349/Snells_law.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/42e0ea0e-95d5-4c19-95fc-d81186497349/Snells_law.png" width="40px" /> Snell’s law:
$$ n_1\sin(\theta_i)=n_2\sin(\theta_r) $$
</aside>
If the refraction angle $\theta_r=90\degree$ then $\theta_i=\theta_c$ known as the critical angle
Light when $\theta_i>\theta_c$ will be totally internally reflection (not pass through the less dense medium)
🗒️ Note: We will use Paraxial approximation $\theta \simeq \sin(\theta) \simeq \tan(\theta)$
We can write
$$ \begin{aligned} \alpha_v&=\alpha _R+\theta \\ \alpha _R&=\alpha_u + \theta \end{aligned} $$
Thus
$$ \begin{aligned} \alpha_v+\alpha_u&=2 \alpha_R \\ \tan(\alpha_v)+\tan(\alpha_u)&=2\tan(\alpha_R) \\ \frac{y}{v}+\frac{y}{u}&=\frac{2y}{R}
\end{aligned} $$
$$ \boxed{\frac 1u+ \frac 1v =\frac 2R} $$
