What is Geometric Optics?
Geometric optics treats light as traveling in perfectly straight lines called rays. This approximation works when objects and apertures are much larger than the wavelength of light. Chapter 24 uses this model to predict where images form — from mirrors and lenses — using a single master equation:
1/xo + 1/xi = 1/f
x_o = object distance · x_i = image distance · f = focal length
This one equation handles mirrors (reflection) and thin lenses (refraction) with only sign convention changes. The magnification is always:
m = hi / ho = −xi / xo
Roadmap of this deck
- Slide 2: Sign conventions — the most critical foundation
- Slides 3–8: Concave mirrors — all cases with worked examples
- Slide 9: Convex mirrors
- Slides 10–13: Thin lenses — converging and diverging
- Slide 14: Magnification deep dive
- Slides 15–16: What-if scenarios
- Slide 17: Common exam traps
- Slide 18: Summary cheat sheet
Light is a wave with wavelength ~400–700 nm. Why can we pretend it travels in straight rays? Because the objects in this chapter (mirrors, lenses, the objects themselves) are centimeters in size — thousands of times larger than the wavelength. In that regime, diffraction effects are negligible and straight-line propagation is an excellent model. This is exactly the regime of everyday vision, photography, telescopes, and eyeglasses.
Physical optics (Chapter 25) handles the wave nature — interference, diffraction, thin-film colors. For Chapter 24, the ray model is all you need.
The Sign Convention: Get This Wrong, Get Everything Wrong
The mirror and lens equation look identical, but the sign conventions differ in one key place. Memorize these. Every problem in Chapter 24 requires them.
For Mirrors
- xo: Always positive (object is in front of mirror)
- xi > 0: Real image — forms in front of the mirror (same side as the object)
- xi < 0: Virtual image — appears to be behind the mirror
- f > 0: Concave mirror (converging)
- f < 0: Convex mirror (diverging)
- f = R/2 (always! — this is a common trap)
For Lenses
- xo: Always positive
- xi > 0: Real image — forms on the far side of the lens (opposite the object)
- xi < 0: Virtual image — forms on the same side as the object
- f > 0: Converging (convex) lens
- f < 0: Diverging (concave) lens
Magnification — same for both mirrors and lenses
m = hi / ho = −xi / xo
m < 0 → inverted image | m > 0 → upright image
|m| > 1 → magnified | |m| < 1 → reduced | |m| = 1 → same size
For mirrors, the convention is that distances measured on the same side as the incoming light (in front of the mirror) are positive. A real image forms where reflected rays converge — in front of the mirror. A virtual image seems to come from behind the mirror (where light doesn't actually reach), so it gets negative xᵢ.
For lenses, refracted rays cross to the other side. A real image forms where the transmitted rays converge — on the far side. That's positive xᵢ. A virtual image (diverging lens, or object inside F of converging lens) appears on the same side as the object — negative xᵢ. Same logic, just mirrored geometry.
Drawing Ray Diagrams for Concave Mirrors
You don't need the equation to locate an image — you can draw it using three standard rays from the tip of the object. Where any two rays intersect (or appear to diverge from) is the image tip.
- Ray 1 (Parallel ray): Travels parallel to the optical axis → reflects through focal point F
- Ray 2 (Focal ray): Travels through F first → reflects parallel to the optical axis
- Ray 3 (Center ray): Travels through center of curvature C → reflects straight back through C
"Pick any two rays — they must agree on where the image is. Use the third as a check."
The mirror law (angle of incidence = angle of reflection) applies to every point on the mirror surface. But to calculate the reflection angle, you need the normal to the surface at that point — which is always along the radius direction (pointing toward C). The three special rays exploit known geometry: a ray hitting the vertex reflects at the same angle about the optical axis; a ray through C hits perpendicular and reverses; a ray parallel to the axis reflects through F by definition of the focal point.
You can verify all three rays agree — if your diagram has them converging at different points, you've made a drawing error.
Case 1 — Object Beyond C (xo > 2f): Real, Inverted, Reduced
When the object is more than twice the focal length away, the image forms between F and C. It is real, inverted, and smaller than the object.
Worked Example
Given: f = 5 cm, xo = 15 cm (note: 15 > 2×5 = 10, so object is beyond C). Find xi and m.
1/xᵢ = 1/f − 1/x₀ = 1/5 − 1/15 = 3/15 − 1/15 = 2/15
→ xᵢ = 15/2 = 7.5 cm
m = −xᵢ / x₀ = −7.5 / 15 = −0.5
m < 0 → inverted | |m| = 0.5 < 1 → reduced
So: xi = +7.5 cm (real — in front of mirror), image is inverted and half the size. Note 5 < 7.5 < 10, so image is between F and C. ✓
As x₀ → ∞, the mirror equation gives 1/xᵢ → 1/f, so xᵢ → f. The image compresses toward F. Meanwhile m = −xᵢ/x₀ → −f/∞ = 0. The image becomes a point at F. This is exactly how a parabolic telescope mirror works: starlight from "infinity" focuses to a point at F where the detector sits.
Case 2 — Object at C (xo = 2f): Real, Inverted, Same Size
When the object sits exactly at the center of curvature, the image also forms at C. It is real, inverted, and exactly the same size. This is the symmetric case.
Worked Example
Given: f = 5 cm, xo = 2f = 10 cm. Find xi and m.
1/xᵢ = 1/f − 1/x₀ = 1/5 − 1/10 = 2/10 − 1/10 = 1/10
→ xᵢ = 10 cm = 2f = x₀
m = −xᵢ / x₀ = −10 / 10 = −1
m = −1: inverted, same size
The image lands right back at C, inverted and same size. Practical use: put an object at C, and the image appears at C — engineers use this to calibrate optics precisely (the "cat's eye" retroreflector principle).
To find the focal length of an unknown concave mirror in the lab: place an illuminated object in front of the mirror and slide it back and forth. When the image appears the same size as the object and is inverted, the object is at C. Measure that distance, divide by 2, and you have f. This works because at x₀ = 2f → xᵢ = 2f and m = −1 — the "same size" condition uniquely identifies the C position.
Case 3 — Object Between F and C (f < xo < 2f): Real, Inverted, Magnified
When the object is between the focal point and the center of curvature, the image jumps to the far side — beyond C. The image is real, inverted, and larger than the object.
Worked Example
Given: f = 5 cm, xo = 7 cm (between f=5 and 2f=10). Find xi and m.
1/xᵢ = 1/f − 1/x₀ = 1/5 − 1/7 = 7/35 − 5/35 = 2/35
→ xᵢ = 35/2 = 17.5 cm
m = −xᵢ / x₀ = −17.5 / 7 = −2.5
m < 0 → inverted | |m| = 2.5 > 1 → magnified (2.5× bigger)
xi = 17.5 cm > 2f = 10 cm. Image is beyond C. ✓
This is exactly how a movie projector (or slide projector) works: the film is placed just beyond F, and the lens projects a large, inverted, real image onto a screen far away. The image is inverted — which is why film slides are mounted upside-down in projectors! The screen sees the image right-side up because the film was flipped. Next time you see a projector, remember: the object is between F and 2F of the projection lens.
Full Exam Problem: R = 10 cm, Object 16 mm Tall at 10 cm
This is the style of problem that appears directly on Exam 3. Work through every step — do not skip.
Given
- R = 10 cm (radius of curvature)
- xo = 10 cm (object distance)
- ho = 16 mm (object height)
- Concave mirror (f > 0)
Step 1: Find f from R
f = R/2 = 10/2 = 5 cm
⚠ TRAP: R ≠ f. ALWAYS divide R by 2.
Step 2: Find xi
1/xᵢ = 1/f − 1/x₀ = 1/5 − 1/10 = 2/10 − 1/10 = 1/10
→ xᵢ = 10 cm
Step 3: Find magnification m
m = −xᵢ / x₀ = −10/10 = −1
Step 4: Find image height hi
hᵢ = m × h₀ = (−1) × 16 mm = −16 mm
The negative sign means inverted. Magnitude = 16 mm.
Summary of answers
- xi = +10 cm → Real image (positive), in front of mirror
- m = −1 → Inverted, same size
- hi = −16 mm → Inverted image, 16 mm tall
- Image location: At C (xi = xo = 2f = 10 cm)
Exam problems often give R (not f) specifically to trap students who set f = R. The very first step is f = R/2. For R = 10 cm, f = 5 cm. Then x₀ = 10 cm = 2f, which puts the object at C. The algebra then gives xᵢ = 10 cm, m = −1, and hᵢ = −16 mm automatically. Students who set f = R = 10 cm get 1/xᵢ = 1/10 − 1/10 = 0 → xᵢ = ∞ (wrong!). This is the #1 error on Chapter 24 exams.
What Happens When xo ≤ f? No Image, Then Virtual
Case A: Object at F (xo = f = 5 cm)
1/xᵢ = 1/5 − 1/5 = 0/5 = 0
→ xᵢ = ∞ (image at infinity — no image forms)
Reflected rays emerge perfectly parallel. No image forms at any finite distance. This is how a flashlight or car headlight works: put the bulb at the focal point of a concave mirror → parallel beam of light emerges.
Case B: Object Inside F (xo = 3 cm < f = 5 cm)
1/xᵢ = 1/5 − 1/3 = 3/15 − 5/15 = −2/15
→ xᵢ = −7.5 cm (VIRTUAL image — behind mirror)
m = −(−7.5)/3 = +2.5
m > 0 → upright | |m| = 2.5 → magnified
The image is virtual, upright, and magnified. This is the makeup mirror / shaving mirror effect: hold your face close to a concave mirror (inside F) and you see an enlarged, upright reflection.
The exam variant asks: "same mirror (R=10 cm, f=5 cm), but object placed at x₀ = 5 cm." Plugging in: 1/xᵢ = 1/5 − 1/5 = 0 → xᵢ = ∞. No image forms at a finite distance — the answer is literally "no image" or "image at infinity." This comes up because students confuse "what happens at x₀=f" with "what happens just inside F." They are qualitatively different: at x₀=f you get no image; just inside F (x₀ slightly less than f) you get a virtual, upright, magnified image far behind the mirror.
Convex Mirror: Always Virtual, Upright, Reduced
A convex mirror curves away from the incoming light. Its focal length is negative (f = −|R|/2). For any real object (xo > 0), the image is always:
- Virtual (xi < 0 — behind the mirror)
- Upright (m > 0)
- Reduced (|m| < 1)
This is why security mirrors in stores and the passenger-side car mirror give a wide-angle view — they compress a large field of view into a small image. The warning "objects are closer than they appear" is because the image is reduced.
Worked Example
Given: R = −20 cm (convex, so R is negative), f = R/2 = −10 cm, xo = 30 cm.
1/xᵢ = 1/f − 1/x₀ = 1/(−10) − 1/30 = −3/30 − 1/30 = −4/30
→ xᵢ = −30/4 = −7.5 cm (virtual)
m = −xᵢ / x₀ = −(−7.5)/30 = +0.25
m > 0 → upright | |m| = 0.25 < 1 → reduced (1/4 size)
The passenger-side car mirror is a convex mirror with a warning because the reduced image makes objects appear smaller — and smaller objects look farther away by depth-perception instinct. The car behind you is actually closer than your brain interprets from the small image size. The flat driver's-side mirror gives a true-size image and no distortion. The convex passenger mirror trades accuracy for field of view. Some cars now use cameras instead.
Drawing Ray Diagrams for Thin Lenses
Thin lens ray diagrams use the same 3-ray logic, but refraction replaces reflection. The three standard rays for a converging (convex) lens are:
- Ray 1 (Parallel ray): Enters parallel to axis → refracts through the far focal point F2
- Ray 2 (Focal ray): Passes through the near focal point F1 → exits parallel to axis
- Ray 3 (Center ray): Passes through the center of the lens → no deviation (straight through)
For a diverging (concave) lens: Ray 1 refracts as if coming from F1 (on same side as object). Ray 2 aims toward F2 but refracts parallel.
A ray through the optical center of a thin lens hits both lens surfaces (front and back) at points where they are locally parallel to each other. The ray is refracted at the front surface and then refracted back by an equal and opposite amount at the back surface. For a thin lens, these two surfaces are so close together that the net lateral displacement is negligible — the ray emerges on the same straight line it entered. This is only true for the exact center. Rays through off-center portions of the lens converge toward F.
Exam: Given Object 8 cm Left, Image 3 cm Left (Same Side) → Find f
The exam gives a diagram where the object is 8 cm to the left of the lens, and the image appears 3 cm to the left of the lens (same side as the object). Image on same side as object for a lens = virtual image → xi is negative.
Step 1: Set up sign conventions
- xo = +8 cm (always positive)
- xi = −3 cm (virtual — same side as object for a lens)
Step 2: Find f using thin lens equation
1/f = 1/x₀ + 1/xᵢ = 1/8 + 1/(−3) = 1/8 − 1/3
= 3/24 − 8/24 = −5/24
f = −24/5 = −4.8 cm
f < 0 → DIVERGING lens (confirmed by virtual image)
Step 3: Find magnification
m = −xᵢ / x₀ = −(−3)/8 = +3/8 = +0.375
m > 0 → upright | |m| < 1 → reduced
Step 4: Find image height (if ho = 6.5 mm)
hᵢ = m × h₀ = 0.375 × 6.5 mm = 2.44 mm
Positive → upright image, 2.44 mm tall
On an exam diagram: if the image is shown on the same side as the object relative to the lens, xᵢ is negative (virtual). If the image is on the opposite side, xᵢ is positive (real). For mirrors: if image is on the same side as the object (in front of mirror), xᵢ is positive (real). If behind the mirror, xᵢ is negative (virtual). The key is knowing which side is "positive" for each element type.
Converging Lens (f > 0): Five Distinct Cases
As the object distance decreases from infinity down through 2f, f, and below f, the image makes a dramatic journey:
Summary Table (f = 10 cm as example)
x₀ > 2f (e.g., 30 cm): 1/xᵢ = 1/10−1/30 = 2/30 → xᵢ = 15 cm. m = −0.5. Real, inverted, reduced. Image between F and 2F.
x₀ = 2f (20 cm): xᵢ = 20 cm. m = −1. Real, inverted, same size. At 2F.
f < x₀ < 2f (e.g., 15 cm): 1/xᵢ = 1/10−1/15 = 1/30 → xᵢ = 30 cm. m = −2. Real, inverted, magnified. Beyond 2F.
x₀ = f (10 cm): xᵢ = ∞. No image (parallel rays).
x₀ < f (e.g., 7 cm): 1/xᵢ = 1/10−1/7 = −3/70 → xᵢ = −23.3 cm. m = +3.3. Virtual, upright, magnified (magnifying glass).
When you hold a magnifying glass close to an object (x₀ < f), the lens produces a virtual, upright, enlarged image on the same side as the object. Your eye, looking through the lens from the other side, sees this virtual image — which appears much larger than the object. The key: you must hold the object inside the focal length. Move it outside f and the image flips to real and inverted (and you'd need a screen to see it). This is why there's a "sweet spot" for a magnifier — too close and it's blurry, too far and the image flips.
Diverging Lens (f < 0): Always Virtual, Upright, Reduced
A diverging (concave) lens always produces a virtual, upright, reduced image — for any object position. This mirrors the behavior of the convex mirror. There are no special cases with distinct regions like the converging lens.
Why always virtual?
A diverging lens spreads parallel rays outward. No matter where the object is, the refracted rays always diverge after passing through the lens — they never converge to form a real image. The image is always virtual (the diverging rays appear to come from a point behind the lens, on the object side).
Worked Example
Given: f = −10 cm, xo = 20 cm.
1/xᵢ = 1/f − 1/x₀ = 1/(−10) − 1/20 = −2/20 − 1/20 = −3/20
→ xᵢ = −20/3 = −6.67 cm (VIRTUAL)
m = −xᵢ / x₀ = −(−6.67)/20 = +0.33
m > 0 → upright | |m| = 0.33 < 1 → reduced
Check: xi is negative → virtual ✓, m is positive → upright ✓, |m| < 1 → reduced ✓.
Nearsighted (myopic) people have eyes that focus too strongly — the focal point falls in front of the retina for distant objects. The correction: a diverging lens in the eyeglass. It pre-diverges the incoming rays so that the eye's lens can then converge them onto the retina. Prescription "−3.5 diopters" means f = 1/(−3.5) = −0.286 m = −28.6 cm. The prescription is literally the focal length of the diverging corrective lens. This is Ch. 24 optics applied to human physiology.
Magnification — Complete Guide
Magnification relates the image size to the object size and tells you the orientation:
m = hᵢ / h₀ = −xᵢ / x₀
Both ratios must give the same number. Use either to find the other.
Reading the sign and magnitude
- m negative → image is inverted (upside-down relative to object)
- m positive → image is upright (same orientation as object)
- |m| > 1 → image is larger (magnified)
- |m| < 1 → image is smaller (reduced)
- |m| = 1 → image is same size as object
Finding image height hi
hᵢ = m × h₀
Keep the sign! Negative hᵢ means inverted.
Worked Example (from Slide 11 exam problem)
xo = 8 cm, xi = −3 cm, ho = 6.5 mm
m = −xᵢ / x₀ = −(−3)/8 = +3/8 = +0.375
hᵢ = m × h₀ = +0.375 × 6.5 mm = +2.44 mm
Positive → upright. 2.44 mm tall.
The negative sign comes from similar triangles in the ray diagram. When rays cross through the focal region, the image forms on the opposite side of the optical axis from the object tip. This geometric crossing forces hᵢ = −(xᵢ/x₀)×h₀. For a real image (xᵢ > 0), the image tip is on the opposite side of the axis (inverted), hence the negative. For a virtual image (xᵢ < 0), the negatives cancel and m > 0 (upright). The sign convention encodes the geometry.
What-If: How xi and m Change as xo Varies
Using f = 5 cm (from R = 10 cm). Systematically vary xo from half-F to 10f:
Concave Mirror (f = +5 cm): xᵢ and m as x₀ varies
x₀ = 2.5 cm (< f): 1/xᵢ = 1/5−1/2.5 = −1/5 → xᵢ = −5 cm (virtual, behind mirror). m = +2 (upright, 2× magnified).
x₀ = 5 cm (= f): xᵢ = ∞ (image at infinity — no image). Flashlight case.
x₀ = 7.5 cm (between F and C): 1/xᵢ = 1/5−1/7.5 = 1/15 → xᵢ = 15 cm. m = −2 (inverted, 2× magnified).
x₀ = 10 cm (= 2f = C): 1/xᵢ = 1/10 → xᵢ = 10 cm. m = −1 (same size, inverted).
x₀ = 15 cm (> 2f): 1/xᵢ = 2/15 → xᵢ = 7.5 cm. m = −0.5 (reduced, inverted).
x₀ = 50 cm (10f): 1/xᵢ = 1/5−1/50 = 9/50 → xᵢ = 5.56 cm. m = −0.11 (very small).
x₀ → ∞: xᵢ → 5 cm = f. m → 0 (point image at F). Telescope mirror.
At x₀ = 2f (the C position), m = −1. As x₀ increases beyond C, xᵢ decreases toward f, and m = −xᵢ/x₀ → 0. The image gets smaller and smaller, approaching a point at F. As x₀ decreases from C toward F, xᵢ increases toward ∞, and m = −xᵢ/x₀ → −∞. The image grows without bound — this divergence is what happens near the focal plane in high-magnification instruments.
What-If: Same Numbers, Different Lens Types
Scenario A: Same xo=8, xi=+3 (image on far side) — what is f?
1/f = 1/8 + 1/3 = 3/24 + 8/24 = 11/24
→ f = 24/11 ≈ +2.18 cm → CONVERGING lens
With xi = +3 (real, far side), f comes out positive. Completely different result from Slide 11 where xi = −3 (virtual, same side).
Scenario B: Move object inside f for a converging lens
f = +10 cm, xo = 6 cm (inside f):
1/xᵢ = 1/10 − 1/6 = 3/30 − 5/30 = −2/30
→ xᵢ = −15 cm (virtual, same side as object)
m = −(−15)/6 = +2.5
Virtual, upright, 2.5× magnified → magnifying glass!
Scenario C: Two lenses in series
Treat the image of lens 1 as the object of lens 2. If lens 1 forms an image 12 cm to its right, and lens 2 is 8 cm to the right of lens 1, then lens 2's object is 4 cm to its right → but that's on the far side → xo for lens 2 is negative (virtual object). Apply the lens equation again with that value.
A compound microscope uses exactly the two-lens setup above. The objective lens (short focal length) creates a highly magnified real image of the specimen. This real image then acts as the object for the eyepiece lens, which acts as a magnifying glass (object inside f) and produces a further magnified virtual image. The total magnification is mtotal = m₁ × m₂. A 10× eyepiece on a 40× objective gives 400× total magnification. The physics is just Chapter 24 applied twice.
The 6 Most Dangerous Exam Traps in Chapter 24
Trap 1 — R is NOT f
Wrong: "R = 10 cm, so f = 10 cm." Right: f = R/2 = 5 cm. A mirror with R = 10 cm focuses at 5 cm — not 10 cm. This is the #1 source of wrong answers. If you use f = 10 and the problem gave R = 10, every subsequent calculation is wrong. Step 1 of every mirror problem: convert R → f = R/2.
Trap 2 — Wrong sign for xᵢ
When xᵢ comes out negative for a lens: the image is virtual (same side as object). This is expected and correct for diverging lenses and for converging lenses with object inside f. Do not drop the negative sign or take the absolute value. The sign is information.
Trap 3 — Forgetting the negative in m = −xᵢ/x₀
There IS a negative sign in the magnification formula. m = −xᵢ/x₀, not +xᵢ/x₀. For a real image (xᵢ > 0), m is negative → inverted. For a virtual image (xᵢ < 0), the two negatives cancel → m is positive → upright. Forgetting the minus sign makes real images appear upright (wrong) and virtual images appear inverted (wrong).
Trap 4 — Object at f gives no image, not an image at f
"Object at f → image at f?" No. Object at f → image at ∞ (no image). Use the equation: 1/xᵢ = 1/f − 1/f = 0 → xᵢ = ∞. This is the flashlight/headlight configuration. The answer is "no image forms."
Trap 5 — Confusing mirror and lens sign conventions for xᵢ
Mirror: xᵢ > 0 = real, in front of mirror (same side as object). Lens: xᵢ > 0 = real, on far side of lens (opposite side from object). The sign convention for xᵢ is defined differently. Real images form on opposite sides for mirrors vs. lenses.
Trap 6 — Not checking reasonableness
After every calculation: (a) Is xᵢ sign consistent with the setup? (b) Does the image type (real/virtual) match? (c) Does |m| direction (magnified/reduced) make physical sense? Object beyond C → reduced. Object between F and C → magnified. Object inside F → virtual and magnified. If your answer contradicts these qualitative rules, recheck your arithmetic.
Chapter 24 Master Reference — Everything on One Slide
Master Equations
1/x₀ + 1/xᵢ = 1/f = 2/R (mirrors: f = R/2)
m = hᵢ / h₀ = −xᵢ / x₀
Works for both mirrors and lenses — only sign conventions differ
3-Ray Rules (quick list)
- Mirrors: Ray ∥ axis → reflects through F | Ray through F → reflects ∥ axis | Ray through C → reflects back through C
- Lenses: Ray ∥ axis → refracts through F₂ | Ray through F₁ → exits ∥ axis | Ray through center → no deviation
Common exam numbers to memorize
- R = 10 cm → f = 5 cm (not 10!)
- Object at C (xo = 2f): xi = 2f, m = −1 (same size, inverted)
- Object at F (xo = f): xi = ∞ (no image)
- Object inside F: virtual, upright, magnified (|m| > 1)
- Convex mirror or diverging lens: always virtual, upright, |m| < 1
"In physics: explain = Words + Diagrams + Equations. All three must agree."
The most elegant fact about Chapter 24 is that one equation governs everything: 1/x₀ + 1/xᵢ = 1/f. It applies to concave mirrors (f > 0), convex mirrors (f < 0), converging lenses (f > 0), and diverging lenses (f < 0). The difference between mirrors and lenses is just the sign convention for what "real" means for xᵢ. The mathematical structure is identical.
This is a recurring theme in physics: a single equation captures all behavior once you establish sign conventions. The mirror/lens equation is the optical analog of Ohm's law or F = ma — one formula that subsumes all cases through the careful use of signs. Master the signs, and you master the chapter.
Good luck on Exam 3! Remember: f = R/2, m = −xᵢ/x₀, and check the sign of xᵢ before calling the image real or virtual.