🔍 Types of Lenses

Part 1 of 7 — Converging vs Diverging

Lenses are transparent optical elements that refract light to form images. Every camera, microscope, telescope, and pair of eyeglasses relies on lenses. In this part, you'll learn to distinguish the two fundamental lens types and trace the principal rays through each.

Converging (Convex) Lenses

A converging lens (also called a convex lens) is thicker at the center than at the edges. It bends parallel light rays inward so they meet at the focal point on the far side of the lens.

Key Terminology

Three Principal Rays (Converging Lens)

1. Parallel ray: Enters parallel to the optical axis → refracts through the focal point $F$ on the far side
2. Focal ray: Passes through $F$ on the near side → refracts parallel to the optical axis
3. Central ray: Passes straight through the center of the lens → continues undeviated

Where any two of these rays intersect on the far side, a real image forms.

Diverging (Concave) Lenses

A diverging lens (also called a concave lens) is thinner at the center than at the edges. It spreads parallel light rays outward so they appear to originate from a focal point on the same side as the incoming light.

Key Facts

• Focal length is negative: $f < 0$
• Parallel rays diverge after passing through the lens
• The focal point is virtual (on the same side as the incoming light)
• A diverging lens always produces a virtual, upright, reduced image (for real objects)

Three Principal Rays (Diverging Lens)

1. Parallel ray: Enters parallel to the axis → refracts as if coming from $F$ on the near side
2. Focal ray: Aimed toward $F$ on the far side → refracts parallel to the axis
3. Central ray: Passes straight through the center → undeviated

The refracted rays diverge, but their backward extensions meet on the same side as the object — that intersection is the virtual image.

Real vs Virtual Images

Quick Rule

• Converging lens: can produce either real or virtual images depending on object distance
• Diverging lens: always produces virtual images (for real objects)

Lens Basics Quiz 🔍

Identify the Lens Type 🎯

Exit Quiz

🧮 The Thin Lens Equation

Part 2 of 7 — Quantitative Image Formation

Now that you know the two lens types, it's time to calculate exactly where images form and how big they are. The thin lens equation and magnification formula are the workhorses of geometric optics.

The Thin Lens Equation

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

Solving for $d_i$:

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

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

This equation works for both converging and diverging lenses — the sign of $f$ takes care of everything.

Sign Conventions (Critical!)

Magnification

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

⚠️ The negative sign in $m = -d_i/d_o$ is essential. A real image ($d_i > 0$) gives $m < 0$ → inverted. A virtual image ($d_i < 0$) gives $m > 0$ → upright.

Worked Example

An object is placed 30 cm from a converging lens with $f = 10$ cm.

Step 1: Find $d_i$

$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{10} - \frac{1}{30} = \frac{3 - 1}{30} = \frac{2}{30} = \frac{1}{15}$

$d_i = 15 \text{ cm}$

Since $d_i > 0$: the image is real (on the opposite side from the object).

Step 2: Find magnification

$m = -\frac{d_i}{d_o} = -\frac{15}{30} = -0.5$

Since $m < 0$: the image is inverted.

Since $|m| = 0.5 < 1$: the image is reduced (half the object height).

Thin Lens Equation Practice

Calculation Drill 🧮

A converging lens has $f = 12$ cm. An object of height 6.0 cm is placed 18 cm from the lens.

1. Image distance $d_i$ (in cm)
2. Magnification $m$
3. Image height $h_i$ (in cm, negative if inverted)

Exit Quiz

📐 Ray Diagrams — Converging Lenses

Part 3 of 7 — Five Cases for Convex Lenses

A converging lens produces dramatically different images depending on where you place the object relative to the focal point $F$ and the point $2F$ (twice the focal length). Mastering these five cases is essential for the AP exam.

Case 1: Object Beyond $2F$ ($d_o > 2f$)

Ray Diagram

• Parallel ray → refracts through $F$
• Focal ray → refracts parallel
• Central ray → straight through center

The three rays converge between $F$ and $2F$ on the opposite side.

Image Characteristics

📷 Application: This is how a camera works — the object is far away, and a small, inverted, real image forms on the film/sensor.

Case 2: Object at $2F$ ($d_o = 2f$)

Image Characteristics

Verification: $\frac{1}{d_i} = \frac{1}{f} - \frac{1}{2f} = \frac{2-1}{2f} = \frac{1}{2f}$, so $d_i = 2f$. $m = -\frac{2f}{2f} = -1$.

Case 3: Object Between $F$ and $2F$ ($f < d_o < 2f$)

Image Characteristics

🎥 Application: This is how a projector works — the slide is placed between $F$ and $2F$, producing a large, inverted, real image on the screen.

Case 4: Object at $F$ ($d_o = f$)

Image Characteristics

Verification: $\frac{1}{d_i} = \frac{1}{f} - \frac{1}{f} = 0$, so $d_i \to \infty$. The refracted rays are parallel — they neither converge nor diverge.

🔦 Application: Flashlights and searchlights place the bulb at the focal point to produce a parallel beam.

Case 5: Object Inside $F$ ($d_o < f$)

Image Characteristics

Verification ($d_o = f/2$): $\frac{1}{d_i} = \frac{1}{f} - \frac{2}{f} = -\frac{1}{f}$, so $d_i = -f$. $m = -(-f)/(f/2) = +2$.

🔎 Application: This is how a magnifying glass works — you hold the object inside the focal length to see an enlarged, upright, virtual image.

Summary Table: All Five Cases

Key Patterns

• As the object moves toward $F$ from the left, the real image moves farther away and gets larger
• At $F$, the image is at infinity
• Inside $F$, the image flips to virtual and upright

Match the Case to the Image 🎯

For a converging lens, select the correct image type for each object position:

Exit Quiz

📐 Ray Diagrams — Diverging Lenses

Part 4 of 7 — Concave Lenses and Calculation Practice

Diverging lenses are simpler than converging lenses because there is only one case: the image is always virtual, upright, and reduced. Let's master the ray diagram and then combine it with calculation practice.

Ray Diagram for a Diverging Lens

Three Principal Rays

1. Parallel ray: Enters parallel to the axis → refracts so it appears to come from $F$ on the incoming side
2. Central ray: Passes through the center of the lens → continues straight (undeviated)
3. Focal ray: Aimed toward $F$ on the far side → refracts parallel to the axis

The refracted rays diverge on the far side. Tracing them backward, their extensions meet on the same side as the object — this intersection is the virtual image.

Image Characteristics (Always!)

No matter where you place the object, a diverging lens produces the same type of image. The image is always closer to the lens than the object is.

Worked Example

An object is placed 24 cm from a diverging lens with $f = -8$ cm.

Step 1: Find $d_i$

$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{-8} - \frac{1}{24} = \frac{-3 - 1}{24} = \frac{-4}{24} = -\frac{1}{6}$

$d_i = -6 \text{ cm}$

✓ Negative → virtual image (on the same side as the object).

✓ $|d_i| = 6 < 8 = |f|$ → image is between the lens and $F$. ✓

Step 2: Find magnification

$m = -\frac{d_i}{d_o} = -\frac{-6}{24} = +\frac{1}{4} = +0.25$

✓ Positive → upright. ✓

✓ $|m| = 0.25 < 1$ → reduced. ✓

Diverging Lens Concepts

Diverging Lens Drill 🧮

A diverging lens has $f = -12$ cm. An object of height 9.0 cm is placed 36 cm from the lens.

1. Image distance $d_i$ (in cm)
2. Magnification $m$ (as a fraction or decimal)
3. Image height $h_i$ (in cm)

Round all answers to 3 significant figures.

Converging vs Diverging Comparison 🔍

An object is placed 20 cm from a lens. Calculate $d_i$ for each lens:

1. Converging lens with $f = +10$ cm: $d_i$ = ? (in cm)
2. Diverging lens with $f = -10$ cm: $d_i$ = ? (in cm)
3. What is the magnification for the diverging lens?

Round all answers to 3 significant figures.

Exit Quiz

🔬 Multi-Lens Systems

Part 5 of 7 — Compound Optics

Real optical instruments — microscopes, telescopes, cameras — use multiple lenses working together. The key principle: the image from the first lens becomes the object for the second lens.

Two-Lens Systems: Step-by-Step Method

Algorithm

1. Lens 1: Use the thin lens equation to find $d_{i1}$ from $d_{o1}$ and $f_1$
2. Transfer: The object distance for lens 2 is $d_{o2} = L - d_{i1}$, where $L$ is the distance between the two lenses
3. Lens 2: Use the thin lens equation again to find $d_{i2}$ from $d_{o2}$ and $f_2$
4. Total magnification: $m_{\text{total}} = m_1 \times m_2$

Important Notes

• If $d_{i1} > L$, then $d_{o2} < 0$ — the image from lens 1 is a virtual object for lens 2 (it's on the far side of lens 2)
• Each magnification: $m_1 = -d_{i1}/d_{o1}$ and $m_2 = -d_{i2}/d_{o2}$
• Total magnification is the product: $m_{\text{total}} = m_1 \times m_2$

Worked Example

Two converging lenses are separated by 50 cm. Lens 1 has $f_1 = 20$ cm. Lens 2 has $f_2 = 10$ cm. An object is placed 30 cm in front of lens 1.

Lens 1:

$\frac{1}{d_{i1}} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$

$d_{i1} = 60 \text{ cm}$

$m_1 = -\frac{60}{30} = -2$

Transfer to Lens 2:

$d_{o2} = L - d_{i1} = 50 - 60 = -10 \text{ cm}$

Negative $d_{o2}$ means the image from lens 1 is 10 cm past lens 2 — a virtual object for lens 2.

Lens 2:

$\frac{1}{d_{i2}} = \frac{1}{10} - \frac{1}{-10} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}$

$d_{i2} = 5 \text{ cm}$

$m_2 = -\frac{5}{-10} = +0.5$

Total magnification:

$m_{\text{total}} = m_1 \times m_2 = (-2)(+0.5) = -1$

The final image is 5 cm past lens 2, real, inverted, and the same size as the object.

Optical Instruments

Compound Microscope

A microscope uses two converging lenses:

• Objective lens (short $f$): placed so the specimen is just outside $f_1$ → produces a real, enlarged, inverted image inside the tube
• Eyepiece (longer $f$): acts as a magnifying glass on the real image → produces a virtual, enlarged image for the eye

$M_{\text{microscope}} \approx -\frac{L}{f_{\text{obj}}} \times \frac{25\text{ cm}}{f_{\text{eye}}}$

where $L$ is the tube length and 25 cm is the near point of the eye.

Refracting Telescope

A telescope uses two converging lenses:

• Objective lens (long $f$): captures light from a distant object → forms a real, inverted image at its focal point
• Eyepiece (short $f$): magnifies that real image

$M_{\text{telescope}} = -\frac{f_{\text{obj}}}{f_{\text{eye}}}$

The negative sign indicates the image is inverted.

Multi-Lens Concepts

Two-Lens Calculation 🧮

Two converging lenses are 55 cm apart. Lens 1: $f_1 = 15$ cm. Lens 2: $f_2 = 10$ cm. An object is 30 cm in front of lens 1.

1. Image distance from lens 1: $d_{i1}$ (in cm)
2. Object distance for lens 2: $d_{o2}$ (in cm)
3. Image distance from lens 2: $d_{i2}$ (in cm, round to 1 decimal)
4. Total magnification $m_{\text{total}}$ (round to 2 decimals)

Round all answers to 3 significant figures.

Exit Quiz

👓 Lens Aberrations & Corrective Lenses

Part 6 of 7 — Real-World Optics

Ideal thin lenses produce perfect images, but real lenses have imperfections called aberrations. Understanding these flaws — and how to correct them — is essential for AP Physics 2 and connects optics to everyday life.

Spherical Aberration

Rays hitting the edges of a spherical lens are refracted more strongly than rays near the center. The result: edge rays focus at a slightly different point than central rays, creating a blurry image.

Cause

• Spherical lens surfaces are not the ideal shape for perfect focusing
• The paraxial (near-axis) approximation breaks down for wide lenses

Corrections

• Use a lens stop (aperture) to block edge rays — reduces light but sharpens image
• Use aspherical lenses with precisely shaped non-spherical surfaces
• Use a combination of lenses that compensate for each other's aberration

Chromatic Aberration

Different wavelengths (colors) of light refract by different amounts — a phenomenon called dispersion. This means a lens has a slightly different focal length for each color:

• Violet light: refracted most → shorter focal length
• Red light: refracted least → longer focal length

The result: colored fringes around the image, especially noticeable at high magnification.

Correction: Achromatic Doublet

An achromatic lens (or doublet) combines:

• A converging lens made of crown glass (low dispersion)
• A diverging lens made of flint glass (high dispersion)

The diverging lens cancels the chromatic spread of the converging lens without fully canceling its focusing power. The result is a lens system that brings two wavelengths to the same focus.

Corrective Lenses for Vision

The human eye uses a converging lens (the crystalline lens) to focus images on the retina. Vision defects occur when the image doesn't fall exactly on the retina.

Myopia (Nearsightedness)

• Problem: Eye focuses distant objects in front of the retina (eyeball too long or lens too strong)
• Correction: Diverging (concave) lens — spreads rays slightly before they enter the eye
• Focal length: negative

Hyperopia (Farsightedness)

• Problem: Eye focuses nearby objects behind the retina (eyeball too short or lens too weak)
• Correction: Converging (convex) lens — converges rays slightly before they enter the eye
• Focal length: positive

Lens Power

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

• Converging lens: $P > 0$ (positive diopters)
• Diverging lens: $P < 0$ (negative diopters)

Example: A lens with $f = +0.50$ m has $P = +2.0$ D. A lens with $f = -0.25$ m has $P = -4.0$ D.

Aberrations & Vision Quiz 👓

Match the Condition 🎯

Exit Quiz

🎯 Lenses — Synthesis & AP Review

Part 7 of 7 — Complete Lens Mastery

This final part ties together everything about lenses: the thin lens equation, sign conventions, ray diagrams, multi-lens systems, and corrective optics. Master these concepts and you're ready for the AP exam.

Complete Lens Concept Map

The Core Equations

Sign Convention Summary

Top 5 AP Mistakes with Lenses

Comprehensive Quiz

AP-Style Calculation Drill 🧮

A 4.0 cm tall object is placed 24 cm from a converging lens with $f = 8.0$ cm.

1. Image distance $d_i$ (in cm)
2. Magnification $m$
3. Image height $h_i$ (in cm, negative if inverted)
4. Is the image real or virtual? (enter "real" or "virtual")

Round all answers to 3 significant figures.

Sign Convention Mastery 🎯

Determine the sign of each quantity:

AP FRQ Preview

Typical Lens FRQ Structure

Part (a): Draw a ray diagram showing two principal rays for the given lens and object position. Label the image.

Part (b): Use the thin lens equation to calculate $d_i$ and $m$. State whether the image is real or virtual, upright or inverted, enlarged or reduced.

Part (c): A second lens is placed at a specified distance. Find the final image position and total magnification.

Part (d): Explain a real-world application (microscope, telescope, corrective lens) using your results.

Strategy Tips

1. Always state sign conventions at the start of your solution
2. Show all algebra — don't skip steps in the thin lens equation
3. Check your answer against the ray diagram — they must agree
4. State the image characteristics explicitly: real/virtual, upright/inverted, enlarged/reduced
5. Watch units — if the problem gives cm, work in cm; convert to meters only for diopters

Final Mastery Quiz 🏆