Geometrical Optics

Geometrical optics treats light as rays that travel in straight lines until they hit a boundary or curved surface. This holds whenever objects and apertures are much larger than the wavelength — true for essentially all lenses, mirrors, and eyes on the MCAT. The payoff: with one equation and a consistent sign convention you can predict where any image forms and whether it is real or virtual, upright or inverted.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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Reflection from Plane Surfaces

The Law of Reflection

Must know

The normal is the line perpendicular to the surface at the point of contact. The angle of incidence $\theta_i$ and angle of reflection $\theta_r$ are both measured from the normal (not the surface — a common error). The law of reflection:

$\theta_i = \theta_r$

For a plane mirror, the image is: as far behind the mirror as the object is in front; virtual; upright and same size; laterally inverted (left-right reversed).

Quick check: You stand 2 m in front of a plane mirror. How far is your image from you?

Answer: The image is 2 m behind the mirror, so you-to-image is $2 + 2 = 4$ m.

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Refraction and Snell's Law

Why Refraction Happens

Must know

Light slows in a medium. The index of refraction quantifies this:

$n = \frac{c}{v}$

where $v$ is the speed in the medium ($c \approx 3 \times 10^8$ m/s in vacuum). Since $v \leq c$, $n \geq 1$. Vacuum (and air, on the MCAT) $n = 1$; water $\approx 1.33$, glass $\approx 1.5$, diamond $\approx 2.4$.

Across an interface, frequency stays the same; speed and wavelength change, with $\lambda = \lambda_0/n$ (shorter wavelength in a denser medium). A ray bends because one side of the wavefront slows before the other:

• Low $n$ → high $n$ (air → glass): bends toward the normal
• High $n$ → low $n$ (glass → air): bends away from the normal

Snell's Law

Must know

$n_1 \sin\theta_1 = n_2 \sin\theta_2$

with both angles measured from the normal.

Worked example: A ray in air ($n_1 = 1.00$) hits glass ($n_2 = 1.50$) at $30°$. Find the angle of refraction.

$(1.00)(\sin 30°) = (1.50)(\sin\theta_2) \implies \sin\theta_2 = \frac{0.500}{1.50} = 0.333 \implies \theta_2 \approx 19.5°$

Low $n$ → high $n$, so it bends toward the normal: $19.5° < 30°$. ✓

Quick check: A ray travels from glass ($n = 1.5$) into water ($n = 1.33$). Does it bend toward or away from the normal?

Answer: Higher $n$ to lower $n$, so it bends away from the normal ($\theta_2 > \theta_1$).

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Dispersion

Must know

The index of refraction depends slightly on wavelength — this is dispersion. Shorter wavelengths (violet, blue) have a slightly higher $n$ than longer ones (red), so they refract more. A prism therefore spreads white light into a spectrum (violet bends most, red least); the same effect makes rainbows. Dispersion is also the cause of chromatic aberration in lenses.

Quick check: Which color is at the outer (top) arc of a rainbow?

Answer: Red. Red bends the least inside the droplet, exiting at a steeper angle that places it at the top of the arc.

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Total Internal Reflection

Must know

Going from higher $n$ to lower $n$, light bends away from the normal. As $\theta_1$ grows, the refracted ray reaches $\theta_2 = 90°$ at the critical angle $\theta_c$. Beyond $\theta_c$ there is no refracted ray — all the light reflects back. This is total internal reflection (TIR).

Conditions: (1) $n_1 > n_2$ (denser → less dense), and (2) $\theta_1 > \theta_c$.

Setting $\theta_2 = 90°$ in Snell's law:

$n_1 \sin\theta_c = n_2 \implies \sin\theta_c = \frac{n_2}{n_1}$

(e.g., glass→air: $\sin\theta_c = 1.0/1.5 \implies \theta_c \approx 41.8°$).

Passage-level

TIR underlies fiber optics (endoscopes) and the sparkle of diamond (small $\theta_c$ traps light).

Quick check: Can TIR occur when a ray goes from water ($n = 1.33$) into glass ($n = 1.5$)?

Answer: No. TIR needs $n_1 > n_2$ (denser to less dense); water to glass is the wrong direction.

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Spherical Mirrors

Concave vs. Convex

Must know

A spherical mirror is a section of a sphere whose center is the center of curvature $C$, a distance $R$ (radius of curvature) from the surface.

• Concave (converging): reflective inner surface; parallel rays converge.
• Convex (diverging): reflective outer surface; parallel rays diverge, appearing to come from behind the mirror.

Focal Length and the Mirror Equation

Must know

Parallel rays converge at (or appear to diverge from) the focal point $F$, halfway to $C$:

$f = \frac{R}{2}$

The image location comes from the mirror equation (same form as the thin-lens equation):

$\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$

($p$ = object distance, $q$ = image distance, $f$ = focal length.)

Sign conventions (real-is-positive): $q > 0$ = real image in front of the mirror; $q < 0$ = virtual image behind. $f > 0$ = concave (converging); $f < 0$ = convex (diverging). Object in front: $p > 0$.

Magnification:
$m = -\frac{q}{p}$

$|m| > 1$ magnified, $|m| < 1$ diminished; $m > 0$ upright, $m < 0$ inverted.

Real vs. Virtual Images

Must know

A real image forms where rays actually intersect (projectable onto a screen); a virtual image forms where rays only appear to diverge from (not projectable). For mirrors: real is in front ($q > 0$), virtual is behind ($q < 0$).

Ray Diagrams for Concave Mirrors

Must know

Three canonical rays (any two suffice): (1) parallel ray reflects through $F$; (2) ray through $F$ reflects parallel; (3) ray through $C$ reflects back on itself.

Worked example: Object 30 cm in front of a concave mirror, $f = 10$ cm. Find $q$, $m$, and describe the image.

$\frac{1}{30} + \frac{1}{q} = \frac{1}{10} \implies \frac{1}{q} = \frac{2}{30} \implies q = +15 \text{ cm}, \quad m = -\frac{15}{30} = -0.5$

$q > 0$ → real, in front; $m < 0$ → inverted; $|m| = 0.5$ → diminished. Consistent with an object beyond $C$. ✓

Quick check: Where do you place an object to get a virtual, upright, magnified image in a concave mirror?

Answer: Between $F$ and the mirror ($p < f$). This is how a makeup/shaving mirror works.

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Thin Lenses

Converging and Diverging Lenses

Must know

A thin lens refracts entering and exiting, but we treat refraction as happening at one plane.

• Converging (convex) lens: thicker at center; parallel rays converge to a real focal point on the far side. $f > 0$.
• Diverging (concave) lens: thinner at center; parallel rays appear to diverge from a virtual focal point on the near side. $f < 0$.

Memory aid: convex = converging = $f > 0$; concave = diverging = $f < 0$.

The Thin-Lens Equation and Sign Conventions

Must know

$\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$

Sign conventions (real-is-positive): real object on incoming side $p > 0$; real image on outgoing (far) side $q > 0$, virtual image on incoming side $q < 0$; converging lens $f > 0$, diverging $f < 0$. Magnification is again $m = -q/p$.

Ray Diagrams for Thin Lenses

Must know

Converging lens, three canonical rays: (1) parallel ray refracts through far focal point $F_2$; (2) ray through near focal point $F_1$ refracts parallel; (3) ray through the optical center passes straight through. (For a diverging lens, ray 1 refracts as if from the near focal point.)

Worked example: Object 20 cm from a converging lens, $f = 30$ cm (object inside $f$). Find $q$.

$\frac{1}{20} + \frac{1}{q} = \frac{1}{30} \implies \frac{1}{q} = -\frac{1}{60} \implies q = -60 \text{ cm}, \quad m = -\frac{-60}{20} = +3$

$q < 0$ → virtual, same side as object; $m = +3$ → upright, magnified 3×. This is exactly how a magnifying glass works. ✓

Quick check: A diverging lens always produces what type of image for a real object?

Answer: Always virtual, upright, diminished — regardless of object position. ($f < 0$, $p > 0$ always give $q < 0$, $0 < m < 1$.)

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Lens Strength and Diopters

Must know

Lens power $P$ is measured in diopters (D):

$P = \frac{1}{f}$

with $f$ in meters. Converging $P > 0$; diverging $P < 0$. (E.g., $+2$ D → $f = +0.5$ m.)

Quick check: A lens has $f = 25$ cm. What is its power in diopters?

Answer: $f = 0.25$ m, so $P = 1/0.25 = +4.0$ D.

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Combination of Lenses

Must know

For thin lenses in contact, powers add:

$P_{total} = P_1 + P_2 \qquad \frac{1}{f_{total}} = \frac{1}{f_1} + \frac{1}{f_2}$

Optional

If lenses are separated, treat them sequentially — the image from lens 1 is the object for lens 2 (the separated-system formula is not required).

Quick check: A $+3$ D lens is placed in contact with a $-1$ D lens. What is the combined focal length?

Answer: $P_{total} = 3 + (-1) = +2$ D, so $f_{total} = 0.50$ m = 50 cm.

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Lens Aberration

Know the logic

Spherical aberration: rays through the edges of a spherical lens/mirror focus at a slightly different point than rays through the center, blurring the image. Corrected with parabolic shapes, a smaller aperture, or compound designs.

Chromatic aberration: because $n$ depends on wavelength (dispersion), colors focus at slightly different distances, giving colored fringes. Corrected with an achromatic doublet.

Key MCAT point: Mirrors do not show chromatic aberration (reflection doesn't depend on $n$) — hence large telescopes use mirrors.

Quick check: Why don't mirrors suffer from chromatic aberration?

Answer: The law of reflection doesn't involve $n$, so all wavelengths reflect identically and focus at the same point.

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Optical Instruments, Including the Human Eye

The Human Eye

Must know

The eye is a converging lens system projecting a real, inverted image on the retina. The cornea provides most of the fixed converging power; the crystalline lens provides variable power. Accommodation is the ciliary muscles changing the lens curvature (more power) to focus on near objects. Near point ≈ 25 cm (young adult); far point = infinity (normal eye).

Myopia, Hyperopia, Presbyopia, Astigmatism

Must know

The corrections:

• Myopia (nearsighted): eye too long/strong; image forms in front of retina; distance blurry; far point < ∞. Corrected with a diverging (–) lens.
• Hyperopia (farsighted): eye too short/weak; near objects blurry; near point farther than normal. Corrected with a converging (+) lens.
• Presbyopia: lens stiffens with age, reducing accommodation; near vision worsens. Corrected with converging (+) reading glasses.
• Astigmatism: irregular corneal curvature blurs at all distances. Corrected with cylindrical lenses.

Passage-level

For a myopic far point of 0.50 m, the corrective power is $P = -1/0.50 = -2.0$ D (negative of the far-point distance).

Condition	Problem	Blurry at	Corrective Lens
Myopia	Eye too long / strong	Distance	Diverging (–)
Hyperopia	Eye too short / weak	Near	Converging (+)
Presbyopia	Lens stiffens with age	Near	Converging (+)
Astigmatism	Irregular cornea	All	Cylindrical

Other Optical Instruments

Passage-level

• Magnifying glass: single converging lens, object inside $f$ → virtual, upright, magnified image.
• Compound microscope: two converging lenses; the objective forms a real, magnified intermediate image that the eyepiece magnifies again. Total $M$ = (objective)(eyepiece).
• Refracting telescope: large objective + eyepiece; $M = f_{obj}/f_{eye}$. Large telescopes use mirrors to avoid chromatic aberration.
• Camera: single converging lens forms a real, inverted, reduced image; focus by changing lens-to-sensor distance.

Quick check: The objective produces a 40× intermediate image; the eyepiece acts as a 10× magnifier. What is the total magnification?

Answer: $M_{total} = 40 \times 10 = 400\times$ (the "40× objective + 10× eyepiece" labeling).

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Common Confusions & Tricks

1. Always measure angles from the normal, not the surface — otherwise you get the complement of the right angle.

2. The sign of $f$ is everything. Converging lenses and concave mirrors → $f > 0$; diverging lenses and convex mirrors → $f < 0$.

3. "Concave/convex" flips meaning for mirrors vs. lenses. Mirror: concave = converging, convex = diverging. Lens: convex = converging, concave = diverging.

4. Real image is in front of a mirror but behind a lens — wherever the rays actually converge. The sign conventions encode this ($q > 0$ = real on both, but "real side" differs).

5. A virtual image is always upright; a real image is always inverted (single lens/mirror, real object). Reliable answer check.

6. Myopia = minus; hyperopia = plus. Myopic eyes over-converge, so diverge first.

7. Diopters require $f$ in meters. $f = 25$ cm = 0.25 m → $P = 4$ D.

8. TIR requires the right direction — only high $n$ → low $n$; no critical angle the other way.

9. $f = R/2$ is for spherical mirrors only, not lenses. (The lensmaker's equation is not tested.)

10. A convex mirror or diverging lens → always virtual, upright, diminished. Guaranteed shortcut.

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Key Equations

Equation	Variables & When to Use
$\theta_i = \theta_r$	Reflection; both angles measured from the normal
$n = c/v$	Index of refraction: $v$ = speed in medium
$n_1 \sin\theta_1 = n_2 \sin\theta_2$	Snell's law for refraction at a flat interface
$\sin\theta_c = n_2/n_1$	Critical angle for TIR; requires $n_1 > n_2$
$f = R/2$	Focal length of a spherical mirror
$\dfrac{1}{p} + \dfrac{1}{q} = \dfrac{1}{f}$	Mirror/lens equation; $p$ = object dist, $q$ = image dist
$m = -\dfrac{q}{p}$	Magnification; $m > 0$ upright, $m < 0$ inverted, $
$P = \dfrac{1}{f(\text{in meters})}$	Lens power in diopters; converging $> 0$, diverging $< 0$
$P_{total} = P_1 + P_2$	Powers add for thin lenses in contact
$M_{telescope} = \dfrac{f_{obj}}{f_{eye}}$	Angular magnification of a refracting telescope