Unit 13 · Topic 13.4

13.4 Images Formed by Lenses

A thin lens bends light by refraction at its two curved surfaces. A converging (convex) lens brings parallel rays together at a focal point on the far side, while a diverging (concave) lens spreads them out as if they came from a focal point on the near side. The same equation as for mirrors, $\frac{1}{s_i} + \frac{1}{s_o} = \frac{1}{f}$ , and the same kind of principal rays find the image.

Key terms

converging lens
diverging lens
thin-lens equation
principal rays
focal length
image distance

Converging and diverging lenses

A converging lens, also called a convex lens, is thicker in the middle than at the edges. Rays that come in parallel to the principal axis refract and meet at the focal point on the far side of the lens. A magnifying glass and a camera lens are converging.

A diverging lens, also called a concave lens, is thinner in the middle. Parallel rays spread out after passing through it, and if you trace them backward they seem to come from a focal point on the near side, the side the light came from. Glasses for nearsighted people use diverging lenses.

Every lens has a focal point on each side. For a thin lens in air, both are the same distance f from the lens. The \"thin lens\" model treats all the bending as happening at the lens's center line, and you measure all distances from that line.

Principal rays for lenses

Draw two or three of these rays from the top of the object. Where they cross after the lens, or where their backward extensions cross, is the image.

A ray parallel to the axis refracts through the far focal point (converging), or bends outward as if it came from the near focal point (diverging).
A ray through the center of the lens goes straight through without bending.
A ray through the near focal point (converging), or aimed at the far focal point (diverging), comes out parallel to the axis.

The thin-lens equation and signs

$\frac{1}{s_i} + \frac{1}{s_o} = \frac{1}{f}$

The sign rules match mirrors, with one twist: a real image from a lens forms on the far side, where the light actually goes. f is positive for a converging lens and negative for a diverging one. $s_o$ is positive for an object on the incoming side. A positive $s_i$ means a real image on the far side; a negative $s_i$ means a virtual image on the same side as the object.

Magnification works the same way: $\lvert M \rvert = \left\lvert \frac{h_i}{h_o} \right\rvert = \left\lvert \frac{s_i}{s_o} \right\rvert$ . For a single lens, real images are inverted and virtual images are upright.

Image patterns

A converging lens behaves like a concave mirror. With the object beyond 2f, the image is real, inverted and reduced (a camera). Between f and 2f, the image is real, inverted and enlarged (a projector). At f, no image forms because the rays leave parallel. Inside f, the image is virtual, upright and enlarged (a magnifying glass).

A diverging lens behaves like a convex mirror: for any real object, the image is virtual, upright, reduced and on the same side as the object, closer to the lens than the object is.

Finding f in the lab

Put a bright object, a lens and a screen on a meter stick. For several object distances, slide the screen until the image is sharp and record $s_o$ and $s_i$ . This only works for a converging lens with the object beyond f, since you need a real image to catch on the screen.

Rearranging the thin-lens equation gives $\frac{1}{s_i} = -\frac{1}{s_o} + \frac{1}{f}$ . So a graph of $\frac{1}{s_i}$ against $\frac{1}{s_o}$ is a straight line with slope −1, and its vertical intercept is $\frac{1}{f}$ .

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Projector setup
A 4.0 cm tall slide is placed 20 cm from a converging lens with a focal length of 15 cm. Where does the image form, how tall is it, and what kind of image is it?
Show the solutionHide the solution
1. Step 1: f = +15 cm and $s_o$ = +20 cm: $\dfrac{1}{s_i} = \dfrac{1}{15} - \dfrac{1}{20} = \dfrac{4 - 3}{60} = \dfrac{1}{60}$ , so $s_i$ = +60 cm.
2. Step 2: Positive $s_i$ : a real image on the far side of the lens, so it's inverted. You could focus it on a screen 60 cm away.
3. Step 3: $\lvert M \rvert = \dfrac{60}{20} = 3.0$ , so $h_i$ = 3.0 × 4.0 cm = 12 cm.
4. Step 4: That fits the pattern: the object is between f and 2f, so the image is real, inverted and enlarged. That's why slides go into a projector upside down.
Answer: Real, inverted image 60 cm beyond the lens, 12 cm tall.
Example 2Calculator allowed
Diverging lens
An object sits 24 cm from a diverging lens whose focal length has magnitude 12 cm. Find the image distance and magnification.
Show the solutionHide the solution
1. Step 1: Diverging means f = −12 cm. $\dfrac{1}{s_i} = -\dfrac{1}{12} - \dfrac{1}{24} = -\dfrac{2 + 1}{24} = -\dfrac{3}{24}$ , so $s_i$ = −8.0 cm.
2. Step 2: The image is virtual, on the same side as the object, 8.0 cm from the lens, and upright.
3. Step 3: $\lvert M \rvert = \dfrac{8.0}{24} = 0.33$ : one-third the object's size.
Answer: Virtual, upright image 8.0 cm from the lens on the object's side, one-third the size.
Example 3Calculator allowed
Focal length from lab data
A student finds sharp images with a converging lens for object distances of 20, 30 and 60 cm, at image distances of 30, 20 and 15 cm. Use a linear graph to find the focal length.
Show the solutionHide the solution
1. Step 1: Compute reciprocals (in cm⁻¹). $\frac{1}{s_o}$ : 0.0500, 0.0333, 0.0167. $\frac{1}{s_i}$ : 0.0333, 0.0500, 0.0667.
2. Step 2: Plot $\frac{1}{s_i}$ against $\frac{1}{s_o}$ . The points fall on a line with slope $\dfrac{0.0333 - 0.0667}{0.0500 - 0.0167} = -1.0$ , as the thin-lens equation predicts.
3. Step 3: The vertical intercept is where $\frac{1}{s_o} = 0$ : 0.0667 + 0.0167 = 0.0833 cm⁻¹. That equals $\frac{1}{f}$ .
4. Step 4: So $f = \dfrac{1}{0.0833} \approx 12$ cm.
Answer: f ≈ 12 cm

Common mistakes

Using a positive f for a diverging lens. Concave (diverging) lenses always take a negative focal length.
Putting a lens's real image on the same side as the object, as with a mirror. Light passes through a lens, so real images form on the far side.
Thinking that covering half of a lens removes half of the image. Every part of the lens helps form the whole image, so you get the full image, just dimmer.
Bending the center ray. A ray through the center of a thin lens passes straight through.

On the exam

You may be asked to sketch a ray diagram and then confirm it with the thin-lens equation. Make sure the two agree on the side, orientation and rough size of the image.
Experimental design questions often ask for the focal length from object and image distances. Describe the screen method and a linearized graph of $\frac{1}{s_i}$ against $\frac{1}{s_o}$ , with $\frac{1}{f}$ as the intercept.

Connected topics

Videos

Converging (Convex) Lenses Explained | AP Physics 2 - Unit 13 - Lesson 5
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Ray Diagrams - Lenses
Bozeman ScienceWatch on YouTube (opens in a new tab)
How to Draw Ray Diagrams for Convex and Concave Lenses
The Physics UniverseWatch on YouTube (opens in a new tab)
Diverging (Concave) Lenses Made Easy | AP Physics 2 - Unit 13 - Lesson 6
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Thin lens equation and problem solving | Geometric optics | Physics | Khan Academy
khanacademymedicineWatch on YouTube (opens in a new tab)
Thin Lens Equation, Optics, Converging Lens & Diverging Lens - Physics
The Organic Chemistry TutorWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 13.4 Images Formed by Lenses. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

An object is 25 cm from a converging lens with a focal length of 15 cm. How far from the lens does the image form?

Question 2 of 4Calculator allowed

An object is 25 cm from a converging lens with a focal length of 15 cm, and its image forms 37.5 cm from the lens on the far side. Which describes the image?

Question 3 of 4Calculator allowed

A stamp is held 10 cm from a converging lens with a focal length of 15 cm. What is the magnification?

Question 4 of 4Calculator allowed

An object is 24 cm from a diverging lens whose focal length has magnitude 12 cm. Where is the image?

0 of 4 answered