A lens bends light at two curved surfaces so that rays from one point of an object meet again at one point of an image. Today you’ll measure a lens’s focal length, test the lens equation with real images, and see why a lens in front of your eye can make a blurry world sharp.
🎯 Goals
Find a converging lens’s focal length by focusing a distant object.
Form real images at different object distances, and test the lens equation with your own data.
Predict whether an image is real or virtual, upright or inverted, larger or smaller.
Convert focal length to lens power in diopters, and explain how glasses fix blurry vision.
🧰 Materials
🔍A converging lens with a focal length of about 10–20 cm, such as a magnifying glass
🔭A diverging lens, if you have one
💡A light source with an object, such as a flashlight or phone lamp behind a card with an arrow or “F” cut in it
🗒️A white card to use as a screen, and a lens holder or modeling clay
📏A meter stick or optical bench, and a ruler
📰A page of printed text
🧮Calculator
📓Physics notebook
Section 1 of 4
Find the Focal Length
1
Focus a distant scene onto a card.
Stand across the room from a window, or look out into the hall, so the bright scene is far away. Hold the converging lens and a white card in line with the window, and move the card back and forth until you see a small, sharp, upside-down image of the window. Measure the distance from the lens to the card. Because the scene is so far away, its rays are nearly parallel, and a lens brings parallel rays to the focal point, so this distance is f.
⚠️ Never use the Sun for this. A lens can focus sunlight hot enough to burn skin, paper, and eyes. Use a window, ceiling lights, or the room beyond.
2
Do three trials and average.
Record the three distances in Table A. Their average is your lens’s focal length.
Sample data (use only if you can’t run the lab)
Lens-to-card distances: 15.2 cm, 14.8 cm, and 15.0 cm, so f = 15.0 cm.
Trial
Lens-to-card distance (cm)
📓 Physics Notebook
What is your lens’s average focal length? Why is it fair to say that the image forms at the focal point when the object is very far away?
Section 2 of 4
Test the Lens Equation
The thin-lens equation: 1/f = 1/d_o + 1/d_i Magnification: m = h_i ÷ h_o = −d_i ÷ d_o f is positive for a converging lens and negative for a diverging one. d_i is positive for a real image (on the far side of the lens, where a screen can catch it) and negative for a virtual one. A negative m means the image is inverted.
3
Set up the lens and the screen.
Line up the object, the lens, and the screen along a meter stick. Put the object at about 3f from the lens (about 45 cm for a 15 cm lens). Slide the screen back and forth until the image is as sharp as you can get it. Record the object distance d_o, the image distance d_i, and whether the image is upright or inverted. Then repeat at about 2f and at about 1.5f.
4
Calculate f and m for each setting.
Work out 1/d_o + 1/d_i, then flip it to get f, and find m = −d_i ÷ d_o. Your f should match your Section 1 value.
Sample data (use only if you can’t run the lab)
d_o → d_i, for the 15 cm lens: 45 → 22, 30 → 31, 22.5 → 44 (cm). All three images were inverted.
d_o (cm)
d_i (cm)
Upright or inverted?
1/d_o + 1/d_i (1/cm)
f = 1 ÷ (1/d_o + 1/d_i) (cm)
m = −d_i ÷ d_o
Lens Checker
Work a row by hand first. Then enter its object and image distances to check 1/d_o + 1/d_i, the focal length, and the magnification. Add your Section 1 focal length to see how well the two methods agree.
What good data looks like
Your three values of f should land within a few percent of each other, and of your Section 1 value. As the object moves closer to the lens, the image should move farther away and get larger. At d_o = 2f, d_i should also be 2f, and the image should be the same size as the object.
📓 Physics Notebook
How did the image change as you moved the object closer to the lens (its distance, size, and orientation)? What happened at d_o = 2f? What do you think would happen if you moved the object to exactly f?
Section 3 of 4
Magnifiers and Diverging Lenses
5
Use the converging lens as a magnifier.
Hold the lens about 5 cm above a line of printed text and look through it. Describe the image. Now try to catch that image on the white card. Move the card anywhere you like, and see what happens.
6
Look through a diverging lens.
If you have a diverging lens, look at the text through it from several distances, and try to catch an image on the card. If you don’t, use the description in the Predict the Image tool.
7
Predict the image for four cases.
Use everything you’ve seen to choose whether each image is real or virtual, upright or inverted, and larger or smaller.
Predict the Image
Use what you saw with the screen and with the lenses. For each case, choose whether the image is real or virtual, how it is oriented, and how big it is compared with the object.
A converging lens, with the object beyond 2f (far away)
A converging lens, with the object between f and 2f
A converging lens, with the object inside f (a magnifying glass)
A diverging lens, at any distance
📓 Physics Notebook
Why can’t you catch the magnified image on a screen? Why can a converging lens make a real image on a screen but a diverging lens never can?
Section 4 of 4
Lenses, Power, and Glasses
8
Solve seven lens problems.
The first five use the lens equation. The last two use lens power, which is how eyeglass prescriptions are written. Power in diopters (D) is P = 1 ÷ f, with f in meters. A correct answer is within 2%.
Lens Problems
Distances are in centimeters unless a problem says meters.
A converging lens has f = +10 cm, and an object is 30 cm away. What is the image distance?1/d_i = 1/f − 1/d_o
What is the magnification for that image?m = −d_i ÷ d_o
The same lens, but the object is now only 5.0 cm away. What is the image distance?A negative d_i means a virtual image
What is the magnification for that image?
A diverging lens has f = −20 cm, and an object is 40 cm away. What is the image distance?
What is the power, in diopters, of a lens with a focal length of +0.50 m?P = 1 ÷ f, with f in meters
A nearsighted student’s prescription is −2.5 D. What is the focal length of the lens, in meters?f = 1 ÷ P
How glasses work
Your eye is a converging lens that focuses the world onto the retina at the back. If the eye focuses light in front of the retina, distant things are blurry. That is nearsightedness, and a diverging lens in front of the eye spreads the light just enough to push the focus back onto the retina. If the eye focuses behind the retina, close things are blurry. That is farsightedness, and a converging lens pulls the focus forward.
📓 Physics Notebook
A student’s glasses are labeled −2.5 D. Is that a converging or a diverging lens, and is the student nearsighted or farsighted? Explain how the lens helps, using where the image forms relative to the retina. (This is the unit’s second question.)
Question 1
Compare the focal length you found with a distant object to the one you found with the lens equation. What is the percent difference, and what explains it?
Question 2
Describe how the image changed as the object moved closer to the lens, and what happens to the image when the object is inside f.
Question 3
Explain why a converging lens can make a real image on a screen, but a diverging lens can’t.
Question 4
Explain how a pair of glasses takes a blurry image and makes it sharp. Use both nearsightedness and farsightedness.
📤 Turn In
Completed Table A and your average focal length
Completed Table B (the lens equation) and your notebook note on how the image changed
Your Predict the Image results and your notebook note on magnifiers
Your Lens Problems results and your answer about the −2.5 D glasses