Reflection

Laws of reflection

• the incident ray, reflected ray and the normal, at the point of incidence, all lie in the same plane
• $\angle i = \angle r \rightarrow$ angle of reflection and angle of reflection are equal

Plane mirror

Refraction

• following formulas assume the incident ray, refracted ray and normal all lie in the same plane

Refractive index

$${}_1n_2 = \frac{n_2}{n_1} = \frac{v_1}{v_2} = \frac{\lambda_1}{\lambda_2} = \frac{\sin(\theta_1)}{\sin(\theta_2)}$$

${}_1n_2 = \frac{n_{2}}{n_{1}} =\frac{1}{{}_2n_1}$

Absolute refractive index

the refractive index of a vacuum/air is $n_\text{air} = 1$.

• the absolute refractive index of an optical medium is ${}_\text{air} n_\text{medium} = n_\text{medium} / 1$
• every other physical medium is more optically dense than air
• e.g. $n_{\text{water}} = 1.33$

$$\begin{aligned} n_1 \sin(\theta_1) &= n_2 \sin(\theta_2) && \text{(General Snell's Law)} \\ \frac{n_2}{n_1} &= \frac{\sin(\theta_1)}{\sin(\theta_2)} && \text{(Rearrange for relative index)} \\ \frac{n}{1} &= \frac{\sin(i)}{\sin(r)} && \text{(Set } n_1 = 1 \text{ for vacuum/air)} \\ n &= \frac{\sin(i)}{\sin(r)} && \text{(Absolute Refractive Index)} \end{aligned}$$

Relative refractive index

• for a pair of optical media, the ratio $\sin \theta_1 / \sin \theta_2$ is constant — this is the relative refractive index of medium 2 w.r.t. medium 1

$$\begin{aligned} {}_1n_2 &= \frac{v_1}{v_2} && \text{(Definition of relative index)} \\ n_1 &= \frac{c}{v_1} \implies v_1 = \frac{c}{n_1} && \text{(Absolute index of medium 1)} \\ n_2 &= \frac{c}{v_2} \implies v_2 = \frac{c}{n_2} && \text{(Absolute index of medium 2)} \\ {}_1n_2 &= \frac{c/n_1}{c/n_2} && \text{(Substitute velocities)} \\ {}_1n_2 &= \frac{n_2}{n_1} && \text{(Final relative index formula)} \end{aligned}$$

Snell's law

$n_{1}\sin(\Theta_{1})=n_{2}\sin(\Theta_{2})$

chaining

$$\begin{aligned} n_{13} &= n_{12} \times n_{23} && \text{(The Chaining Law)} \\ \frac{n_3}{n_1} &= \left(\frac{n_2}{n_1}\right) \times \left(\frac{n_3}{n_2}\right) && \text{(Explanation with absolute refractive indices)} \end{aligned}$$

Critical Angle

• the critical angle c of a pair of media is the angle of incidence in the optically denser medium when the angle of refraction in the optically less dense medium is 90°.

$$\begin{aligned} \sin(c) &= \frac{n_{\text{rare}}}{n_{\text{dense}}} && \text{(General Critical Angle)} \\ \sin(c) &= \frac{1}{n_{\text{dense}}} && \text{(If rare medium is air/vacuum, } n=1 \text{)} \\ n_{\text{dense}} &= \frac{1}{\sin(c)} && \text{(Rearranged to find index of the dense medium)} \end{aligned}$$

• if moving from medium 1 (dense) to medium 2 (less dense)

$$\begin{aligned} \sin(c) &= {}_1n_2 && \text{(If Medium 1 is Dense and Medium 2 is Rare)} \\ \end{aligned}$$

other useful formulas

Real vs apparent depth

when standing in air look at a pool(water), glass block, etc, the object is shallower appears shallower than it really is

$$\begin{aligned} n_{\text{dense}} &= \frac{\text{Real Depth}}{\text{Apparent Depth}} && \text{(Viewing from rare to dense)} \end{aligned}$$

Parallel glass slab

when light passes through a rectangular block of glass, it bends towards the normal and then away from the normal when it emerges from the glass. The exiting ray is parallel to the original entering ray, but it has shifted to the side. This shift is the Lateral displacement, or Lateral shift ($d$ or $x$)

$$\begin{aligned} d &= \frac{t \sin(i - r)}{\cos(r)} \hspace{1em} \text{(Lateral Shift Formula)} \\ \\ \text{Where:} \\ d &= \text{Lateral displacement (distance between the parallel rays)} \\ t &= \text{Thickness of the glass slab} \\ i &= \text{Angle of incidence} \\ r &= \text{Angle of refraction at the first boundary} \end{aligned}$$

Apparent Shift

when placing a glass slab directly on a piece of paper, it appears to have moved upwards. The length of this upward shift ($s$) is given by

$$s = t \left(1 - \frac{1}{n}\right)$$

(Where $t$ is the thickness of the slab and $n$ is the refractive index of the slab).

Lenses

other cases

The Thin Lens & Magnification Equations

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$ $m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$

Alternative Forms

To find Image Distance ($d_i$):

$d_i = \frac{d_o f}{d_o - f}$

To find Object Distance ($d_o$):

$d_o = \frac{d_i f}{d_i - f}$

To find Focal Length ($f$): $f = \frac{d_o d_i}{d_o + d_i}$

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Example

• $f$ is (+) because it is a convex lens.
• $d_o$ is (+) because the object is sitting on the left side.
• $d_i$ is (+) because the image formed on the right side.
• $h_o$ is (+) because your starting arrow points up.
• $h_i$ is (-) because your final image arrow points down.
• $m$ is (-) because the image is upside down.
  • m = -1 as both object and image are at $2f$