Geometric Optics

Reflection

• Law of reflection: $\theta_i = \theta_r$ (angles measured from the normal).
• Plane mirrors form virtual, upright, same-size images.

Refraction and Snell's Law

$n_1 \sin\theta_1 = n_2 \sin\theta_2$ where $n$ is the index of refraction ($n = c/v$).

• Light bends toward the normal when entering a denser medium.
• Total internal reflection occurs when $\theta_1 > \theta_c$, where: $\sin\theta_c = \frac{n_2}{n_1} \quad (n_1 > n_2)$

Mirrors

Concave (converging) mirror:

• Center of curvature $C$, focal point $f = R/2$.
• Can form real or virtual images depending on object position.

Convex (diverging) mirror:

• Virtual focal point behind the mirror.
• Always forms virtual, upright, smaller images.

Lenses

Converging (convex) lens: focal length $f > 0$. Diverging (concave) lens: focal length $f < 0$.

Thin-Lens / Mirror Equation

$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$

Magnification

$M = -\frac{d_i}{d_o} = \frac{h_i}{h_o}$

• $|M| > 1$: enlarged; $|M| < 1$: reduced.
• $M > 0$: upright; $M < 0$: inverted.

Sign Conventions

• Real images: $d_i > 0$ (same side as outgoing light for lenses).
• Virtual images: $d_i < 0$.

Problem: An object is placed $30\ \text{cm}$ from a converging lens with $f = 20\ \text{cm}$. Where is the image and what is the magnification?

Solution:

$\frac{1}{30} + \frac{1}{d_i} = \frac{1}{20}$ $\frac{1}{d_i} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$ $d_i = 60\ \text{cm}$

$M = -\frac{d_i}{d_o} = -\frac{60}{30} = -2$

The image is real ($d_i > 0$), inverted ($M < 0$), and magnified ($|M| = 2$), located $60\ \text{cm}$ from the lens.

• Reflection: $\theta_i = \theta_r$. Refraction: $n_1 \sin\theta_1 = n_2 \sin\theta_2$.
• Total internal reflection occurs beyond the critical angle.
• Thin-lens equation: $1/d_o + 1/d_i = 1/f$.
• Magnification $M = -d_i/d_o$ tells you size and orientation of the image.