Unit 13: Geometric Optics

Unit 13 covers how light behaves when it meets a boundary between materials and when it passes through lenses or reflects off mirrors. It includes the law of reflection, refraction and Snell's law, total internal reflection, dispersion, thin lenses, and mirrors, held together by the sign conventions that make every image calculation work.

13.1 Reflection

Light travels in straight lines until it meets a surface. The law of reflection says the angle of incidence equals the angle of reflection, written θ_i = θ_r. Both angles are measured from the normal, the line perpendicular to the surface at the point where the ray strikes. The incident ray, the reflected ray, and the normal all lie in the same plane.

A plane mirror forms an image with four dependable characteristics. The image is virtual, meaning no light actually passes through the image location. Your brain traces the reflected rays backward and concludes they came from behind the mirror. It is upright and the same size as the object, it sits the same distance behind the mirror as the object sits in front, and it is laterally reversed, so left and right appear swapped.

13.2 Refraction and Snell's Law

When light crosses into a material where it travels at a different speed, it bends. This bending is called refraction. The index of refraction of a material is n = c / v, where c = 3.00 × 10⁸ m/s is the speed of light in vacuum and v is the speed of light in the material. Because light slows down in any material, n is always greater than 1. Air is about 1.00, water is 1.33, and typical glass is around 1.52.

Snell's law relates the angles on the two sides of a boundary: n₁ sinθ₁ = n₂ sinθ₂. Here θ₁ is the angle in the first medium and θ₂ is the angle in the second medium, and both are measured from the normal. When light slows down entering a higher-n material, it bends toward the normal. When it speeds up entering a lower-n material, it bends away from the normal.

A worked example, recomputed fully. Light in air (n₁ = 1.00) strikes water (n₂ = 1.33) with an angle of incidence of 40° from the normal. From Snell's law, sinθ₂ = n₁ sinθ₁ / n₂ = (1.00)(sin 40°) / 1.33 = 0.6428 / 1.33 = 0.4833, so θ₂ = 28.9°. The ray bends toward the normal because it slowed down. The speed of light in the water is v = c / n = (3.00 × 10⁸ m/s) / 1.33 = 2.26 × 10⁸ m/s.

13.3 Total Internal Reflection

When light travels from a higher-n medium toward a lower-n medium, Snell's law makes the refracted angle larger than the incident angle. At one particular incident angle, the refracted ray would run exactly along the boundary, with θ₂ = 90°. That angle is the critical angle θ_c. For any incident angle larger than θ_c, there is no refracted ray at all. Every bit of the light reflects back into the first medium, which is total internal reflection.

Setting θ₂ = 90° in Snell's law gives n₁ sinθ_c = n₂ sin 90° = n₂, so sinθ_c = n₂ / n₁. Two conditions must both hold. First, the light must go from higher n to lower n, so n₁ > n₂. Second, the incident angle must exceed the critical angle, so θ₁ > θ_c.

A worked example. For light inside glass (n₁ = 1.52) heading toward air (n₂ = 1.00): sinθ_c = 1.00 / 1.52 = 0.6579, so θ_c = 41.1°. Any ray inside the glass that strikes the boundary at more than 41.1° from the normal reflects entirely back into the glass. For comparison, water (n = 1.33) to air gives sinθ_c = 1.00 / 1.33 = 0.7519, so θ_c = 48.8°.

13.4 Dispersion

The index of refraction of a material depends slightly on the wavelength of the light. In glass, shorter wavelengths experience a slightly larger n than longer wavelengths, so violet light bends more than red light when it refracts. White light passing through a prism spreads into a spectrum for this reason, with violet deviated the most and red the least. Dispersion is qualitative on the exam. You need the idea that n varies with color and that shorter wavelengths bend more, not a formula.

13.5 Thin Lenses

A converging lens is thicker in the middle than at the edges. It bends rays that arrive parallel to the axis so they meet at a focal point on the opposite side of the lens. A diverging lens is thinner in the middle. It spreads parallel rays apart so that they appear to come from a focal point on the same side as the incoming light. The focal length f is the distance from the lens to the focal point.

Two equations describe every thin lens image. The thin lens equation is 1 / f = 1 / d_o + 1 / d_i, where d_o is the object distance and d_i is the image distance. The magnification is m = h_i / h_o = −d_i / d_o, where h_o and h_i are the object and image heights. The minus sign in the magnification is part of the equation, not optional.

Sign conventions (AP convention: real is positive)

Learn this table before doing a single calculation. It is the same for lenses and mirrors, and it is where most points are lost.

Read an answer in this order. The sign of d_i tells you real or virtual. The sign of m tells you upright or inverted. The size of |m| tells you enlarged or reduced. A real image is formed where light rays actually converge, so it can be projected on a screen. A virtual image is formed where rays only appear to converge, so you see it by looking through the lens or into the mirror, but no screen will catch it.

Worked Lens Examples

Example 1: converging lens, real image. A converging lens has f = +10 cm. An object is placed at d_o = +30 cm, which is beyond twice the focal length. Solve the thin lens equation for d_i: 1 / d_i = 1 / f − 1 / d_o = 1 / 10 − 1 / 30 = 3 / 30 − 1 / 30 = 2 / 30 = 1 / 15, so d_i = +15 cm. The positive sign means a real image, 15 cm from the lens on the opposite side from the object.

Now the magnification: m = −d_i / d_o = −15 / 30 = −0.5. The negative sign means inverted, and |m| = 0.5 < 1 means reduced. If the object is 4.0 cm tall, the image height is h_i = m · h_o = (−0.5)(4.0 cm) = −2.0 cm. The minus sign records the inversion, and the image is 2.0 cm tall. Summary: real, inverted, reduced.

Example 2: converging lens, virtual image. The same lens, f = +10 cm, but now the object is at d_o = +5 cm, inside the focal length. Then 1 / d_i = 1 / 10 − 1 / 5 = 1 / 10 − 2 / 10 = −1 / 10, so d_i = −10 cm. The negative sign means a virtual image, 10 cm from the lens on the same side as the object. You see it by looking through the lens, the way a magnifying glass works.

The magnification is m = −(−10) / 5 = +2. The positive sign means upright, and |m| = 2 means enlarged. A 4.0 cm object gives an 8.0 cm tall upright image. Summary: virtual, upright, enlarged.

13.6 Mirrors

A concave mirror curves inward like the inside of a spoon. It converges reflected rays to a real focal point in front of the mirror, so f is positive. A convex mirror curves outward. It diverges reflected rays so they appear to come from a focal point behind the mirror, so f is negative.

Mirrors use the same two equations as lenses: 1 / f = 1 / d_o + 1 / d_i and m = −d_i / d_o. The sign conventions carry over with one change of scenery. A positive d_i means a real image located in front of the mirror, on the same side as the object, where the reflected light actually travels. A negative d_i means a virtual image behind the mirror surface, where no reflected light goes.

Worked example: concave mirror. A concave mirror has f = +15 cm. An object is placed at d_o = +45 cm in front of it. Then 1 / d_i = 1 / 15 − 1 / 45 = 3 / 45 − 1 / 45 = 2 / 45, so d_i = +22.5 cm. The positive sign means a real image, 22.5 cm in front of the mirror. The magnification is m = −22.5 / 45 = −0.5, so the image is inverted and reduced. A 6.0 cm tall object gives an image 3.0 cm tall and upside down. Summary: real, inverted, reduced, in front of the mirror.

13.7 Ray Diagrams

A ray diagram traces a few special rays to locate an image. Learn the path of each ray in words, because exam questions describe diagrams without showing them. In every diagram, solid lines are real light paths and dashed lines are backward extensions that your brain follows. An image forms where rays actually meet (real) or where their extensions meet (virtual).

Converging lens, object outside the focal length. Three principal rays: (1) a ray arriving parallel to the principal axis refracts through the lens and passes through the focal point on the far side; (2) a ray through the center of the lens continues straight without bending; (3) a ray through the near focal point refracts through the lens and emerges parallel to the axis. The refracted rays intersect on the far side, marking a real, inverted image.

Converging lens, object inside the focal length. The same three rays, but after refraction they spread apart on the far side. Extend them backward with dashed lines to the object's side, where the extensions meet. That meeting point is a virtual, upright, enlarged image.

Diverging lens. Three principal rays: (1) a ray arriving parallel to the axis refracts so that its backward extension passes through the near focal point, while the ray itself bends away from the axis; (2) a ray aimed at the center of the lens continues straight; (3) a ray aimed at the far focal point refracts and emerges parallel to the axis. The backward extensions meet on the object's side, always giving a virtual, upright, reduced image.

Concave mirror. Three principal rays: (1) a ray arriving parallel to the axis reflects through the focal point in front of the mirror; (2) a ray through the focal point reflects parallel to the axis; (3) a ray striking the vertex, the center point of the mirror surface, reflects with equal angles, θ_i = θ_r. The reflected rays meet in front of the mirror for a real image.

Convex mirror. Two useful rays: (1) a ray arriving parallel to the axis reflects so that its backward extension comes from the focal point behind the mirror; (2) a ray aimed at the focal point behind the mirror reflects parallel to the axis. The extensions meet behind the mirror, always giving a virtual, upright, reduced image.

Comparing the Four Devices

Confusions That Cost Points

Practice Questions

Original questions written for this guide in the style of the AP exam. Answers and explanations are on the next pages, so complete the questions before checking them.

Answer Key

1. A. From Snell's law, sinθ₂ = (n₁ / n₂) sinθ₁ = (1.00 / 1.50)(sin 30°) = (0.6667)(0.5) = 0.3333, so θ₂ = 19.5° from the normal. B assumes the light does not bend at all. C inverts the index ratio, computing (1.50 / 1.00)(sin 30°) = 0.75, which gives 48.6°; that ratio would bend the ray away from the normal even though the light slowed down. D confuses the reference line: the computed 19.5° is measured from the normal, and the angle from the surface would be 90° − 19.5° = 70.5°.

2. C. Total internal reflection needs light going from higher n to lower n with an incident angle above the critical angle. For water to air, sinθ_c = 1.00 / 1.33 = 0.7519, so θ_c = 48.8°, and 60° exceeds it. D fails the direction condition: going from air to water, the refracted ray always exists. A has the right direction but the wrong angle: for glass to water, sinθ_c = 1.33 / 1.52 = 0.875, so θ_c = 61.0°, and 30° is below it. B has the right direction but 30° is below the water-to-air critical angle of 48.8°.

3. D. From 1 / d_i = 1 / f − 1 / d_o = 1 / 20 − 1 / 30 = 3 / 60 − 2 / 60 = 1 / 60, d_i = +60 cm. The positive sign means a real image on the opposite side from the object. The magnification is m = −60 / 30 = −2, so the image is inverted and enlarged. A adds 1 / d_o instead of subtracting it, giving 1 / 20 + 1 / 30 = 5 / 60 and d_i = 12 cm. B misreads the positive image distance as a virtual image on the object's side. C forgets the minus sign in m = −d_i / d_o and calls the image upright.

4. B. For a mirror, a negative image distance means a virtual image behind the mirror surface, 12 cm back. C treats the minus sign as meaningless and describes a real image. D confuses the meaning of the signs: the sign of d_i tells you real or virtual, while orientation comes from the sign of m (here m = −(−12) / d_o is positive, so the image is upright). A is the reflex to "fix" a negative answer; a negative d_i is a valid result that describes a virtual image.

5. C. The focal length is negative for a diverging lens: 1 / d_i = 1 / f − 1 / d_o = −1 / 15 − 1 / 30 = −2 / 60 − 2 / 60 = −4 / 60, so d_i = −15 cm. Then m = −(−15) / 30 = +0.5, giving a virtual, upright, reduced image. D forgets that f is negative and uses +15 cm, which gives d_i = +30 cm and a real image that a diverging lens can never form. A gets d_i right but drops the minus sign in the magnification formula, calling the image inverted. B substitutes the focal length in place of the object distance, computing −1 / 15 − 1 / 15 = −2 / 15.

6. B. A convex mirror has a negative focal length, so with a positive object distance the image distance always comes out negative: the image is virtual, behind the mirror. The magnification m = −d_i / d_o is then positive (upright) and less than 1 (reduced). A is impossible; a convex mirror never forms a real image. C contradicts the positive magnification. D contradicts the sign convention: d_i is negative for every convex mirror image.

7. D. A ray parallel to the principal axis is one of the principal rays for a converging lens, and it refracts through the focal point on the far side. A describes no bending at all. B describes a different principal ray, the one through the center of the lens. C confuses lenses with mirrors; lenses refract, they do not reflect the ray back.

8. A. Dispersion happens because the index of refraction depends on wavelength: shorter wavelengths see a slightly larger n in glass, so violet bends the most. B gets the frequency order backward; red has the lowest frequency of the visible colors. C ignores the wavelength dependence of n that causes dispersion in the first place. D reverses the physics: the larger index for violet means it travels slower through the glass, not faster.

One-Page Recall Check

Say each answer out loud before you look back, and mark the ones you cannot finish. Anything you cannot say out loud yet belongs in your flashcard deck. In Rycal, add those items to the Geometric Optics deck and let spaced review bring them back over the next few days.

• State the law of reflection and explain how the angles are measured.
• Describe the image formed by a plane mirror: type, orientation, size, and location.
• Write Snell's law and define the index of refraction.
• Work the air-to-water refraction example from scratch: n₁ = 1.00, n₂ = 1.33, θ₁ = 40°, and find θ₂.
• State both conditions for total internal reflection and write the critical angle formula.
• Compute the critical angle for light going from glass (n = 1.52) to air.
• Explain dispersion and say which visible color bends the most in glass.
• Write the thin lens equation and the magnification equation, including the minus sign.
• State the full AP sign convention for f, d_o, d_i, and m.
• Work the converging lens example from scratch: f = +10 cm, d_o = 30 cm. Find d_i and m and describe the image.
• Explain what changes when the object is placed inside the focal length of a converging lens.
• Describe the image a diverging lens always forms and the image a convex mirror always forms.
• Work the concave mirror example from scratch: f = +15 cm, d_o = 45 cm.
• List the three principal rays for a converging lens and the three for a concave mirror.

Key terms for this unit

Law of reflection, Angle of incidence, Normal, Plane mirror, Virtual image, Real image, Index of refraction, Snell's law, Total internal reflection, Critical angle, Dispersion, Converging lens, Diverging lens, Focal point, Focal length, Thin lens equation, Magnification, Sign convention, Concave mirror, Convex mirror, Principal ray.

About this guide. Written for Rycal and aligned to the College Board AP Physics 2 course framework, Unit 13. All questions and explanations are original Rycal writing. Rycal is independent and is not affiliated with or endorsed by the College Board.